Step-by-step explanation:
no it does not form a right triangle.
a+b+c = 250
1002 + 502 does not equal 1002
nor 1002 + 1002 does not equal 502
so therefore the sides are 100, 100, and 50 but these do not make a right triangle
solve pls brainliest
Answer:
Yes, This is a repeating decimal
No, This is a decimal that neither terminates or repeats
No, This is a decimal that neither terminates or repeats
Yes, This is a terminating decimal
6p - 5 =13
3
-3
12
15
Answer:
6p=13+5
6p=18
p=18/6
p=3
Which of the following is true?
|−5| < 4
|−4| < |−5|
|−5| < |4|
|−4| < −5
Answer:
|-4| < |-5|
Step-by-step explanation:
because if modules is given sub sign will be deducate
make a conjecture (prediction) about the table. what will happen if we continue the table, going down?
Can someone help me with this please
Answer:
y=1/2x
Step-by-step explanation:
Answer:
10
9
8
7
6
5
4
3
2
1
0 1 2 3 4 5 6 7 8 9 10
Step-by-step explanation:
So the arrow is pointing at 10 and 5
Answer 10 5
Not 5
its 10 5
So the answer is 10 5
Consider this function.
h(x) = (x - 2)^2+3
Which of the following domain restrictions would enable h(x) to have an inverse function?
a. x < 1
b. x ＞5
c. x < 3
d. x > 4
(Ps: all four answer and larger equal then or smaller equal then
Answer:
No inverse function: (a), (b), (c)
Inverse function exists: (d)
Step-by-step explanation:
The graph of h(x) = (x - 2)^2 + 3 is a parabola that opens upward and has vertex at (2, 3). If the entire graph is drawn, and the horizontal line test then applied, h(x) would not have an inverse, because the horizontal line would intersect the parabolic graph twice.
Note that if we restricted the domain to x ≥ 2, the resulting graph would pass the horizontal line test. This would also be true for x ≥ 3, x ≥ 4, and so on. Not so for (a) x < 1. False for x > -5. True for x < 3. True for x > 4.
No inverse function: (a), (b), (c)
Inverse function exists: (d)
You plan on joining a gym. The joining fee is $30 and you must pay $12 a month fee. If you have $100, how many months can you use the gym? Write and solve an inequality to represent the solution.
I need the process for this otherwise I won't get credit for it!!
Answer:
Inequality: 30+12x < 100
x ≈ 5 months
Step-by-step explanation:
To solve, first create the inequality. In this problem, 30 dollars is the flat fee for joining the gym and thus a constant. Additionally, 12 is the monthly payment and thus the coefficient for a variable because it changes with the number of months. Finally, 100 is the most that can be spent (the max), so it should be on the other side of the inequality and the sign should be less than or equal to 100.
This makes the inequality: 30+12x < 100
To find the number of months solve the inequality for x.
30+12x<10012x<70x<5.8So, the inequality equals x<5.8. However, the question asks for the number of full months. And, no more than 100 dollars can be spent. So, while you would normally round up. In this case, you must round down to 5 months.
which rotation about its center will carry a regular hexagon onto itself
Can someone help plz
For which function is the ordered pair (5, 10) not a solution?
y = 15 - x
y = x - 5
y = x + 5
y = 2 x
Answer:
y = x - 5
Step-by-step explanation:
y = x - 5
10 = 5 - 5
10 is not equal 0
Answer:
y = x - 5
Step-by-step explanation:
y = x - 5
10 = 5 - 5
10 is not equal 0
A delivery company's charge for an overnight package weighing in excess of one pound is given by the formula c= 2.53w + 15, 16
where w is the weight of package in pounds. Find the following.
solve pls brainliest
Answer:
[tex]18 {m}^{2} [/tex]
Step-by-step explanation:
[tex]area \: = 6m \times 4m \\ = 24 {m}^{2} \\ \\ grass \: area = 3m \times 2m \\ = 6 {m}^{2} \\ \\ cement \: area \: = 24 {m}^{2} - 6 {m}^{2} \\ = 18 {m}^{2} [/tex]
Answer:
18 m^2
Step by step explanation:
In these types of math problems, we have two ways to solve.
1) Directly find the area of the shaded area.
2)Find the unshaded area and then minus that from the total area.
In this case, I will use the second way.
The grass area (unshaded) is 3 x 2 = 6 m^2 ( 6 square meters )
The total area (grass + cement) is 4 x 6 = 24 m^2 ( 24 square meters )
Now, we want the area of the cement part but the grass's area is also in the total.
So, we minus 6m^2 from 24m^2.
Then we get 18m^2.
And that is the answer.
I hope it helps.
(Note : because this problem is easy, you can use both ways but most use the second way. There may also be problems where we can use only the first or second way.)
3. A gym charges a fee of $15 per month plus an additional charge for every group class
attended. The total monthly gym cost T can be represented by this equation: T = 15+c*n,
where c is the additional charge for a group class, and n is the number of group classes
attended
Which equation can be used to find the number of group classes a customer attended if we
know c and T?
a. n = I - 15
N
b. n=1 – 150
c. n = (T - 15) - C.
(T-15)
d. n=
1
Answer:
Option D) [tex]\huge\sf{n\:=\:\frac{(T\:-\:15)}{c}}[/tex]
Step-by-step explanation:
Given the equation, T = 15 + c × n, where:
T = represents the total monthly gym cost
c = represents the additional charge for a group class, and
n = represents the number of group classes attended
Solution:In order to determine which equation can be used to find the number of group classes a customer attended, if there are given values for c and T, we must isolate the variable, n algebraically.
The first step is to subtract 15 from both sides:
T = 15 + c × n
T - 15 = 15 - 15 + c × n
T - 15 = c × n
Next, divide both sides by c to isolate n :
[tex]\huge\mathsf{\frac{({T\:-\:15})}{c}\:=\:\frac{{c\:\times\:n}}{c}}[/tex]
[tex]\huge\sf{n\:=\:\frac{(T\:-\:15)}{c}}[/tex]
Therefore, the correct answer is Option D) [tex]\huge\sf{n\:=\:\frac{(T\:-\:15)}{c}}[/tex].
a number of students are standing in a circle. they are evenly spaced and the fith student is directly opposite the eleventh student. how many students are there all together
Answer:
There would be 12
Step-by-step explanation:
21. A square park has an area of 120 m2
a) What are the dimensions of the park? Give your answer to the nearest metre.
b) If fencing costs $18.50/m, how much would it cost to install a fence around the park?
Show your work
plsss help me quick :((
Answer:
120m2
$6.25
$4.50
85
57
12
12
32
54
69
87
89
12
34
58
the common multiple of 4 and 20 is? a.3 b.4 c.8 d.20
Answer:
i belive its A
Step-by-step explanation: hope this helps :)
Answer:
B. 4
Step-by-step explanation:
4 times 5 is 20 and 20 divided by 5 is 4
Today everything at a store is on sale the store offers a 20
% discount the regualr price of a t shirt is 18 what is the discount price
Answer:
$14.40 is the discount price.
Step-by-step explanation:
0.2 x 18 = 3.6
18 - 3.6 = 14.4
(27/8)^1/3×[243/32)^1/5÷(2/3)^2]
Simplify this question sir pleasehelpme
Step-by-step explanation:
[tex] = {( \frac{27}{8} )}^{ \frac{1}{3} } \times ( \frac{243}{32} )^{ \frac{1}{5} } \div {( \frac{2}{3} )}^{2} [/tex]
[tex] = { ({ (\frac{3}{2} )}^{3}) }^{ \frac{1}{3} } \times {( {( \frac{3}{2}) }^{5} )}^{ \frac{1}{5} } \div {( \frac{2}{3} )}^{2} [/tex]
[tex] = {( \frac{3}{2} )}^{3 \times \frac{1}{3} } \times {( \frac{3}{2} )}^{5 \times \frac{1}{5} } \times {( \frac{3}{2} )}^{2} [/tex]
[tex] = \frac{3}{2} \times \frac{3}{2} \times {( \frac{3}{2} )}^{2} [/tex]
[tex] = {( \frac{3}{2} )}^{1 + 1 + 2} [/tex]
[tex] = {( \frac{3}{2} )}^{4} \: or \: \frac{81}{16} [/tex]
[tex]\large\underline{\sf{Solution-}}[/tex]
[tex]\sf{\longmapsto{\bigg( \dfrac{27}{8} \bigg)^{\frac{1}{3}} \times \Bigg[\bigg( \dfrac{243}{32} \bigg)^{\frac{1}{5}} \div \bigg(\dfrac{2}{3} \bigg)^{2}\Bigg]}} \\[/tex]
We can write as :
27 = 3 × 3 × 3 = 3³
8 = 2 × 2 × 2 = 2³
243 = 3 × 3 × 3 × 3 × 3 = 3⁵
32 = 2 × 2 × 2 ×2 × 2 = 2⁵
[tex]\sf{\longmapsto{\bigg( \dfrac{3 \times 3 \times 3}{2 \times 2 \times 2} \bigg)^{\frac{1}{3}} \times \Bigg[\bigg( \dfrac{3 \times 3 \times 3 \times 3 \times 3}{2 \times 2 \times 2 \times 2 \times 2} \bigg)^{\frac{1}{5}} \div \bigg(\dfrac{2}{3} \bigg)^{2}\Bigg]}} \\[/tex]
[tex]\sf{\longmapsto{\bigg( \dfrac{{(3)}^{3}}{{(2)}^{3}} \bigg)^{\frac{1}{3}} \times \Bigg[\bigg( \dfrac{({3}^{5})}{{(2)}^{5}} \bigg)^{\frac{1}{5}} \div \bigg(\dfrac{2}{3} \bigg)^{2}\Bigg]}} \\[/tex]
Now, we can write as :
(3³/2³) = (3/2)³
(3⁵/2⁵) = (3/2)⁵
[tex]\sf{\longmapsto{\left\{\bigg(\frac{3}{2} \bigg)^{3} \right\}^{\frac{1}{3}} \times \Bigg[\left\{\bigg(\frac{3}{2} \bigg)^{5} \right\}^{\frac{1}{5}} \div \bigg(\dfrac{2}{3} \bigg)^{2}\Bigg]}} \\[/tex]
Now using law of exponent :
[tex]{\sf{({a}^{m})^{n} = {a}^{mn}}}[/tex]
[tex]\sf{\longmapsto{\bigg( \frac{3}{2} \bigg)^{3 \times \frac{1}{3}} \times \Bigg[\bigg(\frac{3}{2} \bigg)^{5 \times \frac{1}{5}} \div \bigg(\dfrac{2}{3} \bigg)^{2}\Bigg]}} \\[/tex]
[tex] \sf{\longmapsto{\bigg( \frac{3}{2} \bigg)^{\frac{3}{3}} \times \Bigg[\bigg(\frac{3}{2} \bigg)^{\frac{5}{5}} \div \bigg(\dfrac{2}{3} \bigg)^{2}\Bigg]}} \\[/tex]
[tex]\sf{\longmapsto{\bigg( \frac{3}{2} \bigg)^{1} \times\Bigg[\bigg(\frac{3}{2} \bigg)^{1} \div \bigg(\dfrac{2}{3} \bigg)^{2}\Bigg]}} \\[/tex]
[tex]\sf{\longmapsto{\bigg( \frac{3}{2} \bigg)^{1} \times \Bigg[\bigg(\frac{3}{2} \bigg)^{1} \times \bigg(\dfrac{3}{2} \bigg)^{2}\Bigg]}} \\[/tex]
[tex]\sf{\longmapsto{\bigg( \frac{3}{2} \bigg)^{1} \times \Bigg[\bigg(\frac{3}{2} \bigg)^{1} \times \bigg(\dfrac{3}{2} \times \dfrac{3}{2} \bigg)\Bigg]}} \\[/tex]
[tex]\sf{\longmapsto{\bigg( \dfrac{3}{2} \bigg)^{1} \times \Bigg[\bigg(\dfrac{3}{2} \bigg)^{1} \times \bigg(\dfrac{3 \times 3}{2 \times 2}\bigg)\Bigg]}} \\[/tex]
[tex]\sf{\longmapsto{\bigg( \dfrac{3}{2} \bigg)^{1} \times \Bigg[\bigg(\dfrac{3}{2} \bigg)^{1} \times \bigg(\dfrac{9}{4}\bigg)\Bigg]}} \\[/tex]
[tex] \sf{\longmapsto{\bigg( \frac{3}{2} \bigg)\times \Bigg[\bigg(\frac{3}{2} \bigg)\times \bigg(\dfrac{9}{4}\bigg)\Bigg]}} \\[/tex]
[tex]\sf{\longmapsto{\bigg( \dfrac{3}{2} \bigg)\times \Bigg[ \: \: \dfrac{3}{2} \times \dfrac{9}{4} \: \: \Bigg]}}\\[/tex]
[tex]\sf{\longmapsto{\bigg( \dfrac{3}{2} \bigg)\times \Bigg[ \: \: \dfrac{3 \times 9}{2 \times 4} \: \: \Bigg]}} \\[/tex]
[tex]\sf{\longmapsto{\bigg(\dfrac{3}{2} \bigg)\times \Bigg[ \: \: \dfrac{27}{8} \: \: \Bigg]}} \\[/tex]
[tex]\sf{\longmapsto{\dfrac{3}{2} \times \dfrac{27}{8}}} \\[/tex]
[tex]\sf{\longmapsto{\dfrac{3 \times 27}{2 \times 8}}} \\[/tex]
[tex] \sf{\longmapsto{\dfrac{81}{16}}\: ≈ \:5.0625\:\red{Ans.}} \\[/tex]
e
Week 6: Culmination Knowledge Check
CLEAR MY CHOICE
Question 8
You buy milk in 1-gallon containers. One portion of waffles requires 2 ounces of milk. How many portions can be made with one container? Select one
a 128
b. 16
c. 32
d. 64
CLEAR MY CHOICE
Question 9
You have 24 quarts of brown stock. You need 75 cups to make one serving of kidney beans. How many servings can you make? Select one
the sum of three whole numbers in a row is 57. what are the three numbers?
Answer:
18, 19, 20
Step-by-step explanation:
(n) + (n + 1) + (n + 2) = 57
3n + 3 = 57
3n = 57 - 3 = 54
n = 54/3 = 18
three numbers are 18, 18+1, 18+2
18, 19, 20
Answer:
18,19,20
Step-by-step explanation:
x + (x+1) + (x+2) = 57
x + x + 1 + x + 2 = 57
Combine like terms
3x + 3 = 57
subtract 3 from both sides
3x = 54
divide both sides by 3
x = 18
The answer is 18, 19, 20
Since we know the value of x, we can look at it like this:
x + (x+1) + (x+2) --- > 18 + (18+ 1) + (18+2) --> 18 + 19 + 20
Help me this question is so hard i fried up my brain yesterday working on it for so long!!!!
Hello there! (:
The answer is 9.
3^4=3*3*3*3 (81)
3^2=9
81:9=9
So the answer is 9.
Hope it helps! If you have any question or query, feel free to ask! (:
~An excited gal
[tex]SparklingFlower[/tex]
The greatest possible number whose digits are all even numbers from 1 to 9
Answer:
8642Step-by-step explanation:
Our even numbers from 1-9 are:
2,4,6,8The largest possible number using the even numbers once is 8642.
Hoped this helped
HURRY PLEASE! Do questions 4,5,6 on paper
This should be right , if any doubt please comment
PLEASE HELP ME!!! jacky starts biking to meet up with her friend. she knows that if she bikes at a speed of 10 mph, she will be 3 minutes late. jacky also know that if she bikes at a speed of 12 mph, she will be there 2 minutes before they agree to meet. how much time does she have left before the appointed time?
Answer:
27s
Step-by-step explanation:
d = distance to rendezvous point in miles
t = time of arrival at 10mph, i.e. 3 mins later than appointed time
10 mph = 10 miles per hour = ¹/₆ miles per min
12 mph = ¹/₅ miles per min
t = d/(¹/₆)
t = 6d
5 min = ¹/₁₂ hr
t - ¹/₁₂ = d/(¹/₅)
t = 5d + ¹/₁₂
Elimination of t:
6d = 5d + ¹/₁₂
d = ¹/₁₂ mi
t = 6(¹/₁₂)
t = ¹/₂ min
Time left to appointed time = ¹/₂ - ³/₆₀
= ¹⁰/₂₀ - ¹/₂₀
= ⁹/₂₀.min → 27 seconds
Answer:
27 min
Step-by-step explanation:
rsm
Please help ASAP !!!
THANKS!!
Answer:
rotation
Step-by-step explanation:
Answer:
Reflection
Step-by-step explanation:
I think its reflection because if you reflect them across the y axis then the x axis I believe they will be in those positions.
Express the tan G as a fraction in simplest terms.
Answer:
[tex]\frac{\sqrt{70} }{5}[/tex]
how do you translate nine equals the quotient of a number and 54
Answer:
9 = [tex]\frac{n}{54}[/tex] or n ÷ 54
Step-by-step explanation:
nine equals the quotient of a number and 54
/\
9 = the quotient of a number and 54
/\
9 = will be division of a number and 54
/\
9 = [tex]\frac{n}{54}[/tex] or n ÷ 54
Have a nice day!
I hope this is what you are looking for, but if not - comment! I will edit and update my answer accordingly. (ノ^∇^)
- Heather
Using the appropriate Algebraic identity evaluate the following:(4a - 5b)²
[tex](4a - 5b)^{2} \\ by \: \: \: using \: \: \: (x - y)^{2} = {x}^{2} - 2xy + {y}^{2} \\ = {(4a)}^{2} - 2(4a)(5b) + {(5b)}^{2} \\ = {16a}^{2} - 40ab + 25 {b}^{2} [/tex]
Answer:[tex] {16a}^{2} - 40ab + {25b}^{2} [/tex]
Hope it helps.
Do comment if you have any query.
(2x+y)2-y2 if x=-3 y=4 and z=-5
pasagot po please
answer it please
Answer:
24 I hope help you yieeeeeee
Answer:
ummmmmmmmmmm itsssssssss
Step-by-step explanation:
