The difference of course is the symbol between the f and g letters.
The circle [tex]\circ[/tex] notation means we're doing a function composition.
Writing [tex](f \circ g)(x)[/tex] is the same as saying [tex]f(g(x))[/tex] where g is the inner function.
Here's an example
f(x) = x^2
g(x) = 3x
f( g(x) ) = ( g(x) )^2 ... note how x is replaced with g(x)
f( g(x) ) = ( 3x )^2
f( g(x) ) = 9x^2
-------------------
On the other hand, the dot notation means we multiply the f(x) and g(x) functions.
Going back to the previous example, we could say
[tex]f(x) = x^2\\\\g(x) = 3x\\\\(f \cdot g)(x) = f(x)*g(x)\\\\(f \cdot g)(x) = x^2*3x\\\\(f \cdot g)(x) = 3x^3\\\\[/tex]
Assume the random variable X is normally distributed with mean μ = 50 and standard deviation σ = 7. Compute the probability
P(35 < X < 58)= ________
Answer:
Probability-Between .8574 = 85.74%
Step-by-step explanation:
Z1=-2.14 Z2=1.14
*x-1 35
*x-2 58
*µ 50
*σ 7
find the measure of d
Which of the following inequalities matches the graph?
10
6
-10
Oxs-1
Ox>-1
Oys-1
Oy 2-1
Answer:
y > -1
Step-by-step explanation:
the line is going across the y axis and is everything above -1
f=((-1,1),(1,-2),(3,-4)) g=((5,0),(-3,4),(1,1),(-4,1)) find (fg)(1)
Answer:
f(g(1)) = - 2
Step-by-step explanation:
Find g(1) then use the value obtained to find f(x)
g(1) = 1 ← value of y when x = 1 (1, 1 ) , then
f(1) = - 2 ← value of y when x = 1 (1, - 2 )
Which of the following theorems verifies that AABC - ASTU?
A. AA
B. HL
C. HA
D. LL
Answer:
AA
Step-by-step explanation:
See In Triangle ABC and Triangle STU
[tex]\because\begin{cases}\sf \angle A=\angle S=90° \\ \sf \angle B=\angle T=31°\end{cases}[/tex]
Hence
[tex]\sf \Delta ABC~\Delta STU(Angle-Angle)[/tex]
By AA similarity triangle ABC is similar to triangle SUT. Therefore, option A is the correct answer.
What are similar triangles?Two triangles are similar if the angles are the same size or the corresponding sides are in the same ratio. Either of these conditions will prove two triangles are similar.
In the given triangle ABC, ∠C=180°-90°-31°
∠C=59°
In the given triangle SUT, ∠U=180°-90°-31°
∠U=59°
Here, ∠B=∠T (Given)
∠C=∠U (Obtained using angle sum property of a triangle)
So, by AA similarity ΔABC is similar to ΔSUT.
Therefore, option A is the correct answer.
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what is the volume of a cube with a length of a 10cm,
a width of 8cm and a height of 8cm
Answer:
640
Step-by-step explanation:
muntiply all the number length width height
Find the value of a.
A. 58
B. 130
C. 86
D. 65
Answer:
[tex]C. \ \ \ 86[/tex]°
Step-by-step explanation:
1. Approach
In order to solve this problem, one must first find a relationship between arc (a) and arc (c). This can be done using the congruent arcs cut congruent segments theorem. After doing so, one can then use the secants interior angle to find the precise measurement of arc (a).
2. Arc (a) and arc (c)
A secant is a line or line segment that intersects a circle in two places. The congruent segments cut congruent arcs theorem states that when two secants are congruent, meaning the part of the secant that is within the circle is congruent to another part of a secant that is within that same circle, the arcs surrounding the congruent secants are congruent. Applying this theorem to the given situation, one can state the following:
[tex]a = c[/tex]
3. Finding the degree measure of arc (a),
The secants interior angle theorem states that when two secants intersect inside of a circle, the measure of any of the angles formed is equal to half of the sum of the arcs surrounding the angles. One can apply this here by stating the following:
[tex]86=\frac{a+c}{2}[/tex]
Substitute,
[tex]86=\frac{a+c}{2}[/tex]
[tex]86=\frac{a+a}{2}[/tex]
Simplify,
[tex]86=\frac{a+a}{2}[/tex]
[tex]86=\frac{2a}{2}[/tex]
[tex]86=a[/tex]
Question
(X-5y/y3)-1=
Answer:
[tex]x = y^3+5y[/tex]
Step-by-step explanation:
Complete question
[tex]\frac{x - 5y}{y^3} - 1=0\\[/tex]
Required
Solve for x
We have:
[tex]\frac{x - 5y}{y^3} - 1=0[/tex]
Collect like terms
[tex]\frac{x - 5y}{y^3} = 1[/tex]
Multiply through by [tex]y^3[/tex]
[tex]x - 5y = y^3[/tex]
Make x the subject
[tex]x = y^3+5y[/tex]
Team A scored 30 points less than four times the number of points that Team B scored. Team C scored 61 points more than half of the number of points that Team B scored. If Team A and Team C shared in the victory, having earned the same number of points, how many more points did each team have than Team B?
Answer:
team a and team c scored 74 points which is 48 points more than team b, scoring 26 points.
Step-by-step explanation:
The sum of the 3rd and 7th terms of an A.P. is 38, and the 9th term is 37. Find the A.P.
Answer:
The AP is 1, 11/2, 10, 29/2, 19, ....
Step-by-step explanation:
Let the first term be a and d be the common difference of the arithmetic progression.
ATQ, a+2d+a+6d=38, 2a+8d=38 and a+8d=37. Solving this, we will get a=1 and d=9/2. The AP is 1, 11/2, 10, 29/2, 19, ....
Plz help a beggar I don’t get it
Answer: 3
happy learning
Answer:
B.
Step-by-step explanation:
From the point (-1,0) the next point on the graph is up 3, right 1, making the slope a positive 3.
If Bobby drinks 5 waters in 10 hours how many does he drink in 1 hour ?
Water drunk by Bobby in 10 hours = 5 units
So, water drink by Bobby in 1 hour
= 5/10 units
= 1/2 units
= 0.5 units
Answer:
1/2 water
Step-by-step explanation:
We can use a ratio to solve
5 waters x waters
----------- = ------------
10 hours 1 hours
Using cross products
5*1 = 10 *x
5 = 10x
Divide by 10
5/10 = x
1/2 waters =x
HELP ASAP!!
The equation (blank) has no solution
Answer:
Just to recap, an equation has no solution when it results in an incorrect "equation".
For example:
Equation: x+3 = x+4
Subtract x: 3 = 4???
But clearly, 3 is not equal to 4, so this equation has NO SOLUTION.
Now onto our problem:
13y+2-2y = 10y+3-y
11y+2 = 9y+3
2y=1
y=1/2
9(3y+7)-2 = 3(-9y+9)
27y+61 = -27y+27
54y = -34
y = -34/54
32.1y+3.1+2.4y-8.2=34.5y-5.1
34.5-5.1=34.5y-5.1
5.1=5.1
infinite solutions
5(2.2y+3.4) = 5(y-2)+6y
11y+17 = 11y-10
17 = -10??
That's not true, so the option "5(2.2y+3.4) = 5(y-2)+6y" has no solution.
Let me know if this helps
It is estimated that 75% of all young adults between the ages of 18-35 do not have a landline in their homes and only use a cell phone at home.
(a) On average, how many young adults do not own a landline in a random sample of 100?
(b) What is the standard deviation of probability of young adults who do not own a landline in a simple random sample of 100?
(c) What is the proportion of young adults who do not own a landline?
(d) What is the probability that no one in a simple random sample of 100 young adults owns a landline?
(e) What is the probability that everyone in a simple random sample of 100 young adults owns a landline?
(f) What is the distribution of the number of young adults in a sample of 100 who do not own a landline?
(g) What is the probability that exactly half the young adults in a simple random sample of 100 do not own a landline?
Answer:
a) 75
b) 4.33
c) 0.75
d) [tex]3.2 \times 10^{-13}[/tex] probability that no one in a simple random sample of 100 young adults owns a landline
e) [tex]6.2 \times 10^{-61}[/tex] probability that everyone in a simple random sample of 100 young adults owns a landline.
f) Binomial, with [tex]n = 100, p = 0.75[/tex]
g) [tex]4.5 \times 10^{-8}[/tex] probability that exactly half the young adults in a simple random sample of 100 do not own a landline.
Step-by-step explanation:
For each young adult, there are only two possible outcomes. Either they do not own a landline, or they do. The probability of an young adult not having a landline is independent of any other adult, which means that the binomial probability distribution is used to solve this question.
Binomial probability distribution
The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.
[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]
In which [tex]C_{n,x}[/tex] is the number of different combinations of x objects from a set of n elements, given by the following formula.
[tex]C_{n,x} = \frac{n!}{x!(n-x)!}[/tex]
And p is the probability of X happening.
The expected value of the binomial distribution is:
[tex]E(X) = np[/tex]
The standard deviation of the binomial distribution is:
[tex]\sqrt{V(X)} = \sqrt{np(1-p)}[/tex]
75% of all young adults between the ages of 18-35 do not have a landline in their homes and only use a cell phone at home.
This means that [tex]p = 0.75[/tex]
(a) On average, how many young adults do not own a landline in a random sample of 100?
Sample of 100, so [tex]n = 100[/tex]
[tex]E(X) = np = 100(0.75) = 75[/tex]
(b) What is the standard deviation of probability of young adults who do not own a landline in a simple random sample of 100?
[tex]\sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{100(0.75)(0.25)} = 4.33[/tex]
(c) What is the proportion of young adults who do not own a landline?
The estimation, of 75% = 0.75.
(d) What is the probability that no one in a simple random sample of 100 young adults owns a landline?
This is P(X = 100), that is, all do not own. So
[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]
[tex]P(X = 100) = C_{100,100}.(0.75)^{100}.(0.25)^{0} = 3.2 \times 10^{-13}[/tex]
[tex]3.2 \times 10^{-13}[/tex] probability that no one in a simple random sample of 100 young adults owns a landline.
(e) What is the probability that everyone in a simple random sample of 100 young adults owns a landline?
This is P(X = 0), that is, all own. So
[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]
[tex]P(X = 0) = C_{100,0}.(0.75)^{0}.(0.25)^{100} = 6.2 \times 10^{-61}[/tex]
[tex]6.2 \times 10^{-61}[/tex] probability that everyone in a simple random sample of 100 young adults owns a landline.
(f) What is the distribution of the number of young adults in a sample of 100 who do not own a landline?
Binomial, with [tex]n = 100, p = 0.75[/tex]
(g) What is the probability that exactly half the young adults in a simple random sample of 100 do not own a landline?
This is P(X = 50). So
[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]
[tex]P(X = 50) = C_{100,50}.(0.75)^{50}.(0.25)^{50} = 4.5 \times 10^{-8}[/tex]
[tex]4.5 \times 10^{-8}[/tex] probability that exactly half the young adults in a simple random sample of 100 do not own a landline.
help pls! I need the answer quickly and pls explain. thank you!
Answer:
h = 6[tex]\sqrt{3}[/tex]
Step-by-step explanation:
The given is the special right triangle with angle measures : 90-60-30
and the side lengths for the given angles are represented by :
2a-a[tex]\sqrt{3}[/tex]-a
the side length that sees 60 degrees is represented by a[tex]\sqrt{3}[/tex] (h in this case)
the area of a triangle is calculated by multiplying height and base and that is divided by 2
a[tex]\sqrt{3}[/tex]*a/2 = 18[tex]\sqrt{3}[/tex] multiply both sides by 2
a^2[tex]\sqrt{3}[/tex] = 36[tex]\sqrt{3}[/tex] divide both sides by [tex]\sqrt{3}[/tex]
a^2 = 36 find the roots for both sides
a = 6
since h sees angle measure 60 and is represented by a[tex]\sqrt{3}[/tex]
h = 6[tex]\sqrt{3}[/tex]
If it takes 5 years for an animal population to double, how many years will it take until the population
triples?
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Answer:
7.92 years
Step-by-step explanation:
We want to find t such that ...
3 = 2^(t/5)
where 2^(t/5) is the annual multiplier when doubling time is 5 years.
Taking logs, we have ...
log(3) = (t/5)log(2)
t = 5·log(3)/log(2) ≈ 7.92 . . . years
It will take about 7.92 years for the population to triple.
[tex] \frac{3x - 2}{7} - \frac{5x - 8}{4} = \frac{1}{14} [/tex]
Answer:
[tex]x=2[/tex]
Step-by-step explanation:
[tex]\frac{3x-2}{7}-\frac{5x-8}{4}=\frac{1}{14}[/tex]
In order to factor an integer, we need to divide it by the ascending sequence of primes 2, 3, 5.
The number of times that each prime divides the original integer becomes its exponent in the final result.
In here, Prime number 2 to the power of 2 equals 4.
[tex]\frac{3x-2}{7}-\frac{5x-8}{2^{2} }=\frac{1}{14}[/tex]
First, We need to add fractions-
Rule:-
[tex]\frac{A}{B} +\frac{C}{D} =\frac{\frac{LCD}{B}+\frac{LCD}{D}C }{LCD}[/tex]
LCD = [tex]7 \cdot 2^{2}[/tex]
[tex]\frac{4(3x-2)+7(-(5x-8))}{7*2^{2} } =\frac{1}{14}[/tex]
[tex]x=2[/tex]
OAmalOHopeO
Differentiate y=2x+200/x with respect to x
Answer:
Hello,
[tex]\boxed{y'=2-\dfrac{200}{x^2} }\\[/tex]
Step-by-step explanation:
[tex](f(x)+g(x))'=f'(x)+g'(x)\\\\(2x)'=2*(x)'=2*1=2\\\\(\dfrac{200}{x} )'=200*(x^{-1})'=200*(-1)*x^{-1-1})=-\dfrac{200}{x^2} \\\\\\\boxed{y'=2-\dfrac{200}{x^2} }\\[/tex]
Answer:
[tex] \frac{dy}{dx } = 2 - \frac{200}{ {x}^{2} } [/tex]
Step-by-step explanation:
[tex]the \: equation \: can \: be \: rewriten \: as \\ y = 2x + 200 {x}^{ - 1} \\ \\ now \: differentiate \: the \: equation\ \\ \frac{dy}{dx} = 2 - 200 {x}^{ - 2} \\ \frac{dy}{dx} = 2 - \frac{200}{ {x}^{2} } [/tex]
A student found the solution below for the given inequality.Which of the following explains whether the student is correct?The student is completely correct because the student correctly wrote and solved the compound inequality.The student is partially correct because only one part of the compound inequality is written correctly.The student is partially correct because the student should have written the statements using “or” instead of “and.”The student is completely incorrect because there is “ no solution “ to this inequality.
Answer:
The student is completely incorrect because there is no solution to this inequality.
Answer:
D on edge
Step-by-step explanation:
A telescope contains both a parabolic mirror and a hyperbolic mirror. They share focus , which is 46feet above the vertex of the parabola. The hyperbola's second focus is 6 ft above the parabola's vertex. The vertex of the hyperbolic mirror is 3 ft below . Find the equation of the hyperbola if the center is at the origin of a coordinate system and the foci are on the y-axis. Complete the equation.
the center is at the origin of a coordinate system and the foci are on the y-axis, then the foci are symmetric about the origin.
The hyperbola focus F1 is 46 feet above the vertex of the parabola and the hyperbola focus F2 is 6 ft above the parabola's vertex. Then the distance F1F2 is 46-6=40 ft.
In terms of hyperbola, F1F2=2c, c=20.
The vertex of the hyperba is 2 ft below focus F1, then in terms of hyperbola c-a=2 and a=c-2=18 ft.
Use formula c^2=a^2+b^2c
2
=a
2
+b
2
to find b:
\begin{gathered} (20)^2=(18)^2+b^2,\\ b^2=400-324=76 \end{gathered}
(20)
2
=(18)
2
+b
2
,
b
2
=400−324=76
.
The branches of hyperbola go in y-direction, so the equation of hyperbola is
\dfrac{y^2}{b^2}- \dfrac{x^2}{a^2}=1
b
2
y
2
−
a
2
x
2
=1 .
Substitute a and b:
\dfrac{y^2}{76}- \dfrac{x^2}{324}=1
76
y
2
−
324
x
2
=1 .
The weekly amount of money spent on maintenance and repairs by a company was observed, over a long period of time, to be approximately normally distributed with mean $440 and standard deviation $20. How much should be budgeted for weekly repairs and maintenance so that the probability the budgeted amount will be exceeded in a given week is only 0.1
Answer:
$465.6 should be budgeted.
Step-by-step explanation:
Normal Probability Distribution
Problems of normal distributions can be solved using the z-score formula.
In a set with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the z-score of a measure X is given by:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the p-value, we get the probability that the value of the measure is greater than X.
Normally distributed with mean $440 and standard deviation $20.
This means that [tex]\mu = 440, \sigma = 20[/tex]
How much should be budgeted for weekly repairs and maintenance so that the probability the budgeted amount will be exceeded in a given week is only 0.1?
The 100 - 10 = 90th percentile should be budgeted, which is X when Z has a p-value of 0.9, so X when Z = 1.28. Then
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]1.28 = \frac{X - 440}{20}[/tex]
[tex]X - 440 = 1.28*20[/tex]
[tex]X = 465.6[/tex]
$465.6 should be budgeted.
on the same graph draw line 2y-x=10 and y=3x
Answer:
Step-by-step explanation:
Which figure always has exactly one line of symmetry?
A. rectangle
B. trapezoid
C. isosceles right triangle
D. circle
A shape of a trapezoid has exactly one line of symmetry. The correct option is B.
What is a trapezoid?An open, flat object with four straight sides and one pair of parallel sides is referred to as a trapezoid or trapezium.
A balanced and proportionate likeness between an object's two halves is referred to as symmetry in geometry. It implies that one half is the other's mirror image.
A trapezium's non-parallel sides are referred to as the legs, while its parallel sides are referred to as the bases. The legs of a trapezium can also be parallel. The parallel sides may be vertical, horizontal, or angled.
Therefore, the shape of a trapezoid has exactly one line of symmetry. The correct option is B.
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In an assembly-line production of industrial robots, gearbox assemblies can be installed in one minute each if holes have been properly drilled in the boxes and in ten minutes if the holes must be redrilled. Twenty-four gearboxes are in stock, 6 with improperly drilled holes. Five gearboxes must be selected from the 24 that are available for installation in the next five robots. (Round your answers to four decimal places.) (a) Find the probability that all 5 gearboxes will fit properly. (b) Find the mean, variance, and standard deviation of the time it takes to install these 5 gearboxes.
Answer:
The right answer is:
(a) 0.1456
(b) 18.125, 69.1202, 8.3139
Step-by-step explanation:
Given:
N = 24
n = 5
r = 7
The improperly drilled gearboxes "X".
then,
⇒ [tex]P(X) = \frac{\binom{7}{x} \binom {17}{5-x}}{\binom{24}{5}}[/tex]
(a)
P (all gearboxes fit properly) = [tex]P(x=0)[/tex]
= [tex]\frac{\binom{7}{0} \binom{17}{5}}{\binom{24}{5}}[/tex]
= [tex]0.1456[/tex]
(b)
According to the question,
[tex]X = 91+5[/tex]
Mean will be:
⇒ [tex]\mu = E(x)[/tex]
[tex]=E(91+5)[/tex]
[tex]=9E(1)+5[/tex]
[tex]=9.\frac{nr}{N}+5[/tex]
[tex]=9.\frac{5.7}{24} +5[/tex]
[tex]=18.125[/tex]
Variance will be:
⇒ [tex]\sigma^2=Var(X)[/tex]
[tex]=V(9Y+5)[/tex]
[tex]=81.V(Y)[/tex]
[tex]=81.n.\frac{r}{N}.\frac{N-r}{N}.\frac{N-n}{N-1}[/tex]
[tex]=81.5.\frac{7}{24}.\frac{24-7}{24}.\frac{24-5}{24-1}[/tex]
[tex]=69.1202[/tex]
Standard deviation will be:
⇒ [tex]\sigma = \sqrt{69.1202}[/tex]
[tex]=8.3139[/tex]
The legend on a map states that 1 inch is 20 miles. If you measure 5 inches on the map, how many miles would the actual distance be? Actual distance = [ ? ] miles
Answer:
1 inch= 20 miles. 5*20=100 miles. The answer is 100 miles.
Step-by-step explanation:
On these types of questions just do that every time, then you don't need to ask, for example:
1 foot = 50 miles
If it measures 3 feet.
3*50=150 miles.
If you have any questions regarding my answer, tell me in the comments, and I will answer them.
Each course at college X is worth either 2 or 3 credits. The members of the men's swim team are taking a total of 48 courses that are worth a total of 107 credits. How many 2-credit courses and how many 3-credit courses are being taken?
Answer:
Let the number of courses that are worth 3 credits each be x and those worth 4 credits be y. With the given information, you can write the following equations:
x + y = 48
3x + 4y = 155
You can solve the above equations by method of elimination/substitution
x + y = 48 ⇒ x = 48 - y (Now, substitution this equation into 3x + 4y = 155)
3(48 - y) + 4y = 155
144 -3y + 4y = 155
y + 144 = 155
y = 11
Now plug this solution back into x = 48 - y
x = 48 - 11 = 37
Check work (by plugging the solutions back into the 3x + 4y and see if it's equal to 155):
3(37) + 4(11) = 155
Answer: There are 37 of the 3-credit course and 11 of the 4-credit course
Consider an x distribution with standard deviation o = 34.
(a) If specifications for a research project require the standard error of the corresponding distribution to be 2, how
large does the sample size need to be?
B) If specifications for a research project require the standard error of the corresponding x distribution to be 1, how large does the sample size need to be?
Part (a)
The standard error (SE) formula is
[tex]\text{SE} = \frac{\sigma}{\sqrt{n}}\\\\[/tex]
where n is the sample size. We're given SE = 2 and sigma = 34, so,
[tex]\text{SE} = \frac{\sigma}{\sqrt{n}}\\\\2 = \frac{34}{\sqrt{n}}\\\\2\sqrt{n} = 34\\\\\sqrt{n} = \frac{34}{2}\\\\\sqrt{n} = 17\\\\n = 17^2\\\\n = 289\\\\[/tex]
So we need a sample size of n = 289 to have an SE value of 2.
Answer: 289========================================================
Part (b)
We'll use SE = 1 this time
[tex]\text{SE} = \frac{\sigma}{\sqrt{n}}\\\\1 = \frac{34}{\sqrt{n}}\\\\1*\sqrt{n} = 34\\\\\sqrt{n} = 34\\\\n = 34^2\\\\n = 1156\\\\[/tex]
Because we require greater precision (i.e. a smaller SE value), the sample size must be larger to account for this. In other words, as SE goes down, then n must go up, and vice versa.
Answer: 1156Place the labels in the chart
If you can draw this out for me or describe were they are that will be very helpful:)
Answer:
Check the image
Given the central angle, name the arc formed.
Major arc for ∠EQD
A. EQDˆ
B. GDFˆ
C. EGDˆ
D. EDˆ
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Answer:
C. EGD
Step-by-step explanation:
A major arc is typically named using the end points and a point on the arc. Here, the end points are E and D, and points on the major arc include C, G, and F. The major arc ED could be named any of
arc ECDarc EGD . . . . choice Carc EFDOf course, the reverse of any of these names could also be used: DCE, DGE, DFE.
The sum of three numbers is 124
The first number is 10 more than the third.
The second number is 4 times the third. What are the numbers?
Answer:
182/3,3 8/3, 152/3
Step-by-step explanation:
a+b+c=124
a trừ c= 10
4b=c
Answer:
a=29,b=79,c=19
Step-by-step explanation:
a=c+10
b=4c
=> a+b+c=c+10+4c+c=124
=> c=19
=> a= 29, b=79