find the least common multiple of two two-digit numbers. SHOW WORK

Answer:

8

Step-by-step explanation:

5 = 40

1 = x

Then we multiply by the rule of crisscrossing

5 x X = 40 x 1

5x = 40 then divide both by 5

X = 8

Answer:

3x+4

Step-by-step explanation:

When you factor 9x^2+24x+16, it factors to (3x+4)^2

Factoring 9x^2 - 16 factors to (3x+4)(3x-4)

Therefore the common factor is 3x+4

I hope this helps!

Answer:

There are no shortcuts to scaling successfully. It takes just as much — or more — work as it did to start the company in the first place. Take control of the transition. Get organized with tools that support your employees in their day-to-day roles, making it easier for them to get more done as the company grows.

9514 1404 393

Answer:

  b. Victoria's method is linear because the number of minutes increased by an equal number every month

Step-by-step explanation:

Zach's reading doubled each month, a characteristic of an exponential function.

Victoria's reading increased by 15 minutes each month, characteristic of a linear function.

The only reasonable description among those offered is the one shown above: choice B.

Answer:

sorry can't understand the language

the answer is c i just had this question your welcome

Answer:

12

Step-by-step explanation:

PEMDAS is the order

P = parenthesis

E = exponent

M = *

D = division

A = +

S = -

so first 8*2 = 16

and then -8/2 = -4

and then -4 + 16

= 12

Answer:

[tex]14x cos(\frac{1}{15}x^{2})=14 \sum _{k=0} ^{\infty} \frac{(-1)^{k}x^{4k+1}}{(2k)!15^{2k}}[/tex]

Step-by-step explanation:

In order to find this Maclaurin series, we can start by using a known Maclaurin series and modify it according to our function. A pretty regular Maclaurin series is the cos series, where:

[tex]cos(x)=\sum _{k=0} ^{\infty} \frac{(-1)^{k}x^{2k}}{(2k)!}[/tex]

So all we need to do is include the additional modifications to the series, for example, the angle of our current function is: [tex]\frac{1}{15}x^{2}[/tex] so for

[tex]cos(\frac{1}{15}x^{2})[/tex]

the modified series will look like this:

[tex]cos(\frac{1}{15}x^{2})=\sum _{k=0} ^{\infty} \frac{(-1)^{k}(\frac{1}{15}x^{2})^{2k}}{(2k)!}[/tex]

So we can use some algebra to simplify the series:

[tex]cos(\frac{1}{15}x^{2})=\sum _{k=0} ^{\infty} \frac{(-1)^{k}(\frac{1}{15^{2k}}x^{4k})}{(2k)!}[/tex]

which can be rewritten like this:

[tex]cos(\frac{1}{15}x^{2})=\sum _{k=0} ^{\infty} \frac{(-1)^{k}x^{4k}}{(2k)!15^{2k}}[/tex]

So finally, we can multiply a 14x to the series so we get:

[tex]14xcos(\frac{1}{15}x^{2})=14x\sum _{k=0} ^{\infty} \frac{(-1)^{k}x^{4k}}{(2k)!15^{2k}}[/tex]

We can input the x into the series by using power rules so we get:

[tex]14xcos(\frac{1}{15}x^{2})=14\sum _{k=0} ^{\infty} \frac{(-1)^{k}x^{4k+1}}{(2k)!15^{2k}}[/tex]

And that will be our answer.

Answer:

53.08 ft

Step-by-step explanation:

The pressure at the bottom of the tank, P = ρgh where ρ = density of water = 1000 kg/m³, g = acceleration due to gravity = 9.8 m/s² and h = depth of water in tank.

Since the pressure at the bottom of the tank, P = ρgh, making h subject of the formula, we have

h = P/ρg

Since P = 23 psig = 23 × 1 psig = 23 × 6894.76 Pa = 158579.48 Pa

Substituting the values of the other variables into h, we have

h = P/ρg

h = 158579.48 Pa/(1000 kg/m³ × 9.8 m/s²)

h = 158579.48 Pa/(9800 kg/m²s²)

h = 16.18 m

h = 16.18 × 1 m

h = 16.18 × 3.28 ft

h = 53.08 ft

Answer:

-2/3

Step-by-step explanation:

A rational number is a number that can be expressed by a fraction so when y add to fractions it’s a rational number.

Answer:

7 brown M&Ms.

Step-by-step explanation:

This question is not complete, but I will assume that the final question is how many brown M&Ms will be in this package.

0.14 × 52 is our equation.

The answer is 7.28. We cannot have .28 of a brown M&M in a package (unless you count the broken ones) so there will be, on average, 7 brown M&Ms in a package.

(again, the question is incomplete, so this may not be the answer)

Answer:

B

Step-by-step explanation:

The coeffecients (I totally didn't spell that right) and variables match up.

Answer:

[tex]y=-\frac{3}{2}x+1[/tex]

Step-by-step explanation:

Hi there!

What we need to know:

1) Determine the slope (m)

[tex]3x+2y = 10[/tex]

First, we must organize this given equation in slope-intercept form. This will help us identify its slope.

[tex]3x+2y = 10[/tex]

Subtract 3x from both sides

[tex]2y = -3x+10[/tex]

Divide both sides by 2

[tex]y = -\frac{3}{2} x+5[/tex]

Now, we can identify clearly that [tex]-\frac{3}{2}[/tex] is in the place of m in [tex]y=mx+b[/tex], making it the slope. Because parallel lines have the same slope, this makes the slope of the line we're currently solving for [tex]-\frac{3}{2}[/tex] as well. Plug this number into [tex]y=mx+b[/tex]:

[tex]y=-\frac{3}{2}x+b[/tex]

2) Determine the y-intercept (b)

[tex]y=-\frac{3}{2}x+b[/tex]

Plug in the given point (8,-11) and solve for b

[tex]-11=-\frac{3}{2}(8)+b\\-11=-\frac{24}{2}+b\\-11=-12+b[/tex]

Add 12 to both sides

[tex]1=b[/tex]

Therefore, the y-intercept of the line is 1. Plug this back into [tex]y=-\frac{3}{2}x+b[/tex]:

[tex]y=-\frac{3}{2}x+1[/tex]

I hope this helps!

Answer:

a = 0.0040 m/s², v = 14.4 m/s.

Step-by-step explanation:

Given that,

The distance between Kathmandu and Khanikhola, d = 26 km = 26000 m

Time, t = 1 hour = 3600 seconds

Let a is the acceleration of the bus. Using second equation of motion,

[tex]d=ut+\dfrac{1}{2}at^2[/tex]

Where

u is the initial speed of the bus, u = 0

So,

[tex]d=\dfrac{1}{2}at^2\\\\a=\dfrac{2d}{t^2}\\\\a=\dfrac{2\times 26000}{(3600)^2}\\\\a=0.0040\ m/s^2[/tex]

Now using first equation of motion.

Final velocity, v = u +at

So,

v = 0+0.0040(3600)

v = 14.4 m/s

Hence, this is the required solution.

Answer:

TAMA PO

Step-by-step explanation:

Paki brainliest nadin po

Answer:

a) Figure B and Figure C

b) Figure C

c) Figure A, Figure B, and Figure C

Step-by-step explanation:

a) Rectangles are shapes that have four sides, and four right angels. Right angles are angles that are 90 degrees.

The only shapes with four sides AND four right angles are Figure C and Figure B.

b) Squares are shapes with four EQUAL sides and four right angles. The only shape with four equal sides and four right angles is Figure C.

c) Parallelograms are any shapes with four sides. All of these shapes have four sides.

Hope this helps!

Answer:

[tex]\lim_{(x,y) \to (0,0)} \frac{x^2 \sin^2y}{x^2+2y^2} = 0[/tex]

Step-by-step explanation:

Given

[tex]f(x,y) = \frac{x^2 \sin^2y}{x^2+2y^2}[/tex]

Required

[tex]\lim_{(x,y) \to (0,0)} f(x,y)[/tex]

[tex]\lim_{(x,y) \to (0,0)} f(x,y)[/tex] becomes

[tex]\lim_{(x,y) \to (0,0)} \frac{x^2 \sin^2y}{x^2+2y^2}[/tex]

Multiply by 1

[tex]\lim_{(x,y) \to (0,0)} \frac{x^2 \sin^2y}{x^2+2y^2}\cdot 1[/tex]

Express 1 as

[tex]\frac{y^2}{y^2} = 1[/tex]

So, the expression becomes:

[tex]\lim_{(x,y) \to (0,0)} \frac{x^2 \sin^2y}{x^2+2y^2} \cdot \frac{y^2}{y^2}[/tex]

Rewrite as:

[tex]\lim_{(x,y) \to (0,0)} \frac{x^2 y^2}{x^2+2y^2} \cdot \frac{\sin^2y}{y^2}[/tex]

In limits:

[tex]\lim_{(x,y) \to (0,0)} \frac{\sin^2y}{y^2} \to 1[/tex]

So, we have:

[tex]\lim_{(x,y) \to (0,0)} \frac{x^2 y^2}{x^2+2y^2} *1[/tex]

[tex]\lim_{(x,y) \to (0,0)} \frac{x^2 y^2}{x^2+2y^2}[/tex]

Convert to polar coordinates; such that:

[tex]x = r\cos\theta;\ \ y = r\sin\theta;[/tex]

So, we have:

[tex]\lim_{(x,y) \to (0,0)} \frac{(r\cos\theta)^2 (r\sin\theta;)^2}{(r\cos\theta)^2+2(r\sin\theta;)^2}[/tex]

Expand

[tex]\lim_{(x,y) \to (0,0)} \frac{r^4\cos^2\theta\sin^2\theta}{r^2\cos^2\theta+2r^2\sin^2\theta}[/tex]

Factor out [tex]r^2[/tex]

[tex]\lim_{(x,y) \to (0,0)} \frac{r^4\cos^2\theta\sin^2\theta}{r^2(\cos^2\theta+2\sin^2\theta)}[/tex]

Cancel out [tex]r^2[/tex]

[tex]\lim_{(x,y) \to (0,0)} \frac{r^2\cos^2\theta\sin^2\theta}{\cos^2\theta+2\sin^2\theta}[/tex]

[tex]\lim_{(x,y) \to (0,0)} \frac{r^2\cos^2\theta\sin^2\theta}{\cos^2\theta+2\sin^2\theta}[/tex]

Express [tex]2\sin^2 \theta[/tex] as [tex]\sin^2\theta+\sin^2\theta[/tex]

So:

[tex]\lim_{(x,y) \to (0,0)} \frac{r^2\cos^2\theta\sin^2\theta}{\cos^2\theta+\sin^2\theta+\sin^2\theta}[/tex]

In trigonometry:

[tex]\cos^2\theta + \sin^2\theta = 1[/tex]

So, we have:

[tex]\lim_{(x,y) \to (0,0)} \frac{r^2\cos^2\theta\sin^2\theta}{1+\sin^2\theta}[/tex]

Evaluate the limits by substituting 0 for r

[tex]\frac{0^2 \cdot \cos^2\theta\sin^2\theta}{1+\sin^2\theta}[/tex]

[tex]\frac{0 \cdot \cos^2\theta\sin^2\theta}{1+\sin^2\theta}[/tex]

[tex]\frac{0}{1+\sin^2\theta}[/tex]

Since the denominator is non-zero; Then, the expression becomes 0 i.e.

[tex]\frac{0}{1+\sin^2\theta} = 0[/tex]

So,

[tex]\lim_{(x,y) \to (0,0)} \frac{x^2 \sin^2y}{x^2+2y^2} = 0[/tex]

Option D. The set of transformation that is performed on this triangle (ABCD)' is A translation 5 units to the right, followed by a 180-degree counterclockwise rotation about the origin

The answer choices have been shown to have all solutions to be rotated around their origin.

To shift the triangle, It has to be done through the connection the points ABC ' to the origin in such a way that the line is extended as we can seen in the diagram.

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Answer:

1. x + 4 = 9

Hint: the word 'sum' generally refers to addition.

2. 10a = 70

3. [tex]\frac{3}{4} t[/tex] = 15

4. [tex]\frac{1}{4} x[/tex] - 4 = 4

Answer:

34m = c

Step-by-step explanation:

For every month (m) you pay 34 dollars. However many months youu use that service time 34 equals your total cost (c).

Answer:

[tex]let \: cost \: be \: { \bf{c}} \: and \: months \: be \: { \bf{n}} \\ { \bf{c \: \alpha \: n}} \\ { \bf{c = kn}} \\ 34 = (k \times 1) \\ k = 34 \: dollars \\ \\ { \boxed{ \bf{c = 34n}}}[/tex]

9514 1404 393

Answer:

Step-by-step explanation:

Of the items sold, sodas were 2/(2+1) = 2/3 of the total.

  (2/3)(228) = 152 . . . sodas were sold

  152/2 = 76 . . . . hot dogs were sold

Answer:

a) The 60th percentile of the diameters is of 25.0177 millimeters.

b) The 67th percentile of the diameters is of 25.0308 millimeters.

c) The diameter of the hole should be of 24.8562 millimeters.

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 25.0 millimeters and standard deviation 0.07 millimeter.

This means that [tex]\mu = 25, \sigma = 0.07[/tex]

(a) Find the 60th percentile of the diameters.

This is X when Z has a p-value of 0.6, so X when Z = 0.253.

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

[tex]0.253 = \frac{X - 25}{0.07}[/tex]

[tex]X - 25 = 0.253*0.07[/tex]

[tex]X = 25.0177[/tex]

The 60th percentile of the diameters is of 25.0177 millimeters.

(b) Find the 67th percentile of the diameters.

This is X when Z has a p-value of 0.67, so X when Z = 0.44.

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

[tex]0.44 = \frac{X - 25}{0.07}[/tex]

[tex]X - 25 = 0.44*0.07[/tex]

[tex]X = 25.0308[/tex]

The 67th percentile of the diameters is of 25.0308 millimeters.

(c) A hole is to be designed so that 2% of the ball bearings will fit through it. The bearings that fit through the hole will be melted down and remade. What should the diameter of the hole be.

This is the 2nd percentile, which is X when Z has a p-value of 0.08, so X when Z = -2.054.

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

[tex]-2.054 = \frac{X - 25}{0.07}[/tex]

[tex]X - 25 = -2.054*0.07[/tex]

[tex]X = 24.8562[/tex]

The diameter of the hole should be of 24.8562 millimeters.

Answer:

1) has bigger answer

Step-by-step explanation:

1)

solving parenthesis first we get

5 × 10

so, the answer = 50

2)

solving 3 × 5 first as we have to see multiplication first then addition

2 + 15 + 5

22

comparing both

50 > 22

so problem 1 has a bigger answer

Answer:

(2,3)

Step-by-step explanation:

Answer:23

Step-by-step explanation:

Hi

we know that sin(A+B)=Sin(A)Cos(B)-Cos(A) Sin (B)

cos(2x)=1-2sin²(x)

sin(2x)=2sin(x)cos(x)

Exp:- Sin(3x)=Sin(x+2x)

Sin(x)cos(2x)+cos(x)sin(2x)

sin(x){1-2sin²(x)}+2cos²(x)sin(x)

sin(x)-2sin³(x)+2sin(x){1-sin²(x)}

sin(x)-2sin³(x)+2sin(x)-2sin³(x)

3sin(x)-4sin³(x)

Hope it helps....

Answer:

the first wheel ever built

Answer:

The first wheel ever built

Step-by-step explanation

(*) Sorry for my late answer but I hope this helps others that are looking for this.

100% in the test :)

Answer:

Step-by-step explanation:

Since, CD is an altitude, ∠CDB will be a right angle.

m∠CDB = m∠CDA = 90°

By applying triangle sum theorem in ΔABC,

m∠CAB + m∠CBA + m∠ACB = 180°

20° + m∠CBA + 90° = 180°

m∠CBA = 180° - 110°

             = 70°

Therefore, m∠CBD = 70°

By applying triangle sum theorem in ΔBCD,

m∠BCD + m∠CDB + m∠DBC = 180°

m∠BCD + 90° + 70° = 180°

m∠BCD + 160° = 180°

m∠BCD = 20°

m∠CAD = m∠A = 20°

m∠ACD = 90° - m∠BCD

             = 90° - 20°

m∠ACD = 70°

Answer:

SECOND ONE

Step-by-step explanation:

Answer: [tex]7\sqrt{42}[/tex]

Step-by-step explanation:

42 students divided equally into 7 groups means 42 divided by 7, and [tex]7\sqrt{42}[/tex] is the only choice that shows that.

7√42. is the expressions would represent a class of 42 students divided equally into 7 groups

A division is a process of splitting a specific amount into equal parts.

Given,

A class of forty two Forty two students divided equally into 7 groups.

Forty two students divided equally into seven groups means forty two divided by seven, and

this can be done by using 7√42.

42 students divided equally into 7 groups means forty two divided by seven

Hence  7√42. is the expressions would represent a class of 42 students divided equally into 7 groups

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