All categories

3 years ago

An equation is shown : 5x + 3x = 5x + 15/2 what value of 3x makes the equation true?

Mathematics

1 answer:

MrMuchimi3 years ago

7 0

Answer:

15/2

Step-by-step explanation:

Subtract 5x from both sides of the equation to find out.

5x +3x -5x = 5x +15/2 -5x . . . . . . subtract 5x

3x = 15/2 . . . . . . . . . . . . . . . . . . . . collect terms

The required value of 3x is 15/2.

You might be interested in

Determine the value of x.

harina [27]

Answer:

Step-by-step explanation:

7 0

2 years ago

What is the gradient of the blue line Attachment below!!!!!!!!!!!

mr_godi [17]

Answer:

-3 Or 3 it could be any one of them

Step-by-step explanation:

-3 or 3

Rise/run so an

6 0

3 years ago

*please help and explain how you got the answer!*

stepan [7]

Answer:

Step-by-step explanation:

A= (4-1)^3

simplify A to be 3^3

Which gives us 27

B=(2*3)^2-9

simplify B

first multiply the 2 numbers in paranthesis which gives us 6. raise it to the power of 2 which is 39 and then subtract 9. Gives us 27.

C=15^3*4-12

Simplify the exponent first. 3*4 gives us 12 and 12-12 equals 0. Anything raised to the power of 0=1

If A-B^C is the equation we can write 27-27 raised to the power of 1 which is 0

7 0

3 years ago

This question is realy hard so helpp

LUCKY_DIMON [66]

Answer:

Algebrically the result can be written as -

$9 {x}^{2} + 24x + 16$

Step-by-step explanation:

Let the number be 'x'.

When multiplied by 3 , algebrically it is written as "3x".
When 4 is added to 3x , algebrically it is written as "3x + 4".
When (3x + 4) is squared completely , algebrically it is written as - ${(3x + 4)}^{2} = 9 {x}^{2} + 24x + 16$

8 0

2 years ago

Find the EXACT value of sin(A−B) if cos A = 3/5 where A is in Quadrant IV and cos B = 12/13 where B is in Quadrant IV. Assume al

MissTica

$\bf \textit{Sum and Difference Identities} \\\\ sin(\alpha - \beta)=sin(\alpha)cos(\beta)- cos(\alpha)sin(\beta)$

well, for both angles A and B we're on the IV Quadrant, meaning, the sine is negative, the cosine is positive, likewise, the opposite side is negative and the adjacent side for the angle is positive.

$\bf cos(A)=\cfrac{\stackrel{adjacent}{3}}{\underset{hypotenuse}{5}}\qquad \qquad \stackrel{\textit{getting the opposite side}}{b=\pm\sqrt{5^2-3^2}}\implies b = \pm 4 \\\\\\ \stackrel{IV~Quadrant}{b = -4}\qquad \qquad sin(A)=\cfrac{\stackrel{opposite}{-4}}{\underset{hypotenuse}{5}} \\\\[-0.35em] ~\dotfill\\\\ cos(B)=\cfrac{\stackrel{adjacent}{12}}{\underset{hypotenuse}{13}}\qquad \qquad \stackrel{\textit{getting the opposite side}}{b=\pm\sqrt{13^2-12^2}}\implies b = \pm 5$

$\bf \stackrel{IV~Quadrant}{b = -5}\qquad \qquad sin(B)=\cfrac{\stackrel{opposite}{-5}}{\underset{hypotenuse}{13}} \\\\[-0.35em] ~\dotfill\\\\ sin(A-B)=\cfrac{-4}{5}\cdot \cfrac{12}{13}-\left( \cfrac{3}{5}\cdot \cfrac{-5}{13} \right)\implies sin(A-B)=\cfrac{-48}{65} - \left( \cfrac{-15}{65} \right) \\\\\\ sin(A-B)=\cfrac{-48}{65} + \cfrac{15}{65}\implies sin(A-B)=\cfrac{-33}{65}$

4 0

2 years ago