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3 years ago

15

The least common multiple of two whole numbers is 12. What is the ratio of the greater number to the lesser number is 3:2 what a

re the two numbers

1 answer:

Evgen [1.6K]3 years ago

7 0

Answer:

36 and 24

.............

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I NEED HELP ASAP PLEASEEE !!!!!

Volgvan

Answer:

3

Step-by-step explanation:

Using Pythagorean Theorem, 3² + 4² = 5²

8 0

3 years ago

<img src="https://tex.z-dn.net/?f=%5Csf%2023%289x%2B6%29%3D6x%2B4" id="TexFormula1" title="\sf 23(9x+6)=6x+4" alt="\sf 23(9x+6)=

ozzi

The value of x from the given expression is; x= -2/3

<h3>Solving one-variable equations</h3>

The given expression is;

23(9x+6) = 6x + 4.

Hence, when we expand the equation; we have;

207x + 138 = 6x +4

207x - 6x = 4 - 138

201x = -134

x = -134/201

x = -2/3

Read more on one-variable equations;

brainly.com/question/21528555

4 0

2 years ago

Read 2 more answers

What is the radius of a circle in which a 30° arc is 2 inches long?

marysya [2.9K]

It would be 1 inch and 15 degree arc due to the act that the radius is half.

5 0

3 years ago

Read 2 more answers

Point Q' is the image of Q(-7, -6) under the translation (x, y) + (x + 12, y + 8).

Lina20 [59]

Answer:

The co-ordinates of Q' is (5,2).

Step-by-step explanation:

Given:

Pre-image point

Q(-7,-6)

To find Image point Q' after following translation.

$(x,y)\rightarrow (x+12,y+8)$

Solution:

Translation rules:

Horizontal shift:

$(x,y)\rightarrow (x+k,y)$

when $K>0$ the point is translated $k$ units to the right.

when $K$ the point is translated $k$ units to the left.

Vertical shift:

$(x,y)\rightarrow (x,y+k)$

when $K>0$ the point is translated $k$ units up.

when $K$ the point is translated $k$ units down.

Given translation $(x,y)\rightarrow (x+12,y+8)$ shows the point is shifted 12 units to the right and 8 units up.

The point Q' can be given as:

Q'= $(-7+12,-6+8)=(5,2)$

So, the co-ordinates of Q' is (5,2). (Answer)

7 0

3 years ago

zmey [24]

Answer:

Part 1) The inequality that represent this situation is $5(x+7)^{2} \geq1,050$ or $5x^{2}+70x+245 \geq1,050$

Part 2) Yes, 8 inches is a reasonable width for his tablet

Step-by-step explanation:

Part 1)

Let

L -----> the length of the screen television

W ----> the width of the screen television

x ----> the width of Andrew's tablet

we know that

$L=5W$ ------> equation A

$W=x+7$ ----> equation B

The area of the television is

$A=LW$ -----> equation C

Substitute equation A and equation B in equation C

$A=5(x+7)(x+7)$

$A=5(x+7)^{2}$

$5(x+7)^{2} \geq1,050$

$5(x^{2}+14x+49) \geq1,050$

$5x^{2}+70x+245 \geq1,050$ ------> inequality that represent this situation

Part 2) Determine if 8 inches is a reasonable width for his tablet

For x=8 in

Substitute in the inequality

$5(8+7)^{2} \geq1,050$

$5(15)^{2} \geq1,050$

$1,125 \geq1,050$ -----> is true

therefore

Yes, 8 inches is a reasonable width for his tablet

3 0

3 years ago