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3 years ago

What is the solution to this system of linear combination 3a+6b=45 2a-2b=-12 what is the solution to the system?

1 answer:

3 0

Ok i tried and ig youre looking for what a and b equal for both systems..... it'll be 1) a=-2b+15 b=7.5-(1/2)a <-- fraction........2) a=-6+b b=6-a

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7 − 58z = −23 ASPA A. 9.38 B. 48 C. −24 D. 24

Hitman42 [59]

Answer:

the given are not correct the real answer is 15/29 or 0.517...

Step-by-step explanation:

7-58z = -23

-58z = -23 -7

-58z =-30 the negatives will cancel each other out

-58. -58

z = 30/58 divide both by 2

z= 15/29 which can also be equal to 0.517...

5 0

3 years ago

Solve -3(-8b+7)=3(2b-1)

nekit [7.7K]

Answer: $b=1$

Step-by-step explanation:

Given the equation $-3(-8b+7)=3(2b-1)$ you need to solve for the variable "b".

You need to apply Distributive property on both sides of the equation:

$-3(-8b+7)=3(2b-1)\\\\24b-21=6b-3$

Now you need to add 21 to both sides of the equation:

$24b-21+21=6b-3+21\\\\24b=6b+18$

Subtract $6b$ from both sides:

$24b-6b=6b+18-6b\\\\18b=18$

And finally, you can divide both sides of the equation by 18:

$\frac{18b}{18}=\frac{18}{18}\\\\b=1$

4 0

3 years ago

Read 2 more answers

What are properties of right triangles? 3 minimum.

Readme [11.4K]

Answer:

One angle is always 90° or right angle. The side opposite angle 90° is the hypotenuse. The hypotenuse is always the longest side. The sum of the other two interior angles is equal to 90°.

Step-by-step explanation:

7 0

3 years ago

Pleaseeee help!!! I will mark you as brainlinest for correct answer!!!!!

DaniilM [7]

A would be your answer
f(x) = 20 $( \frac{3}{5} ) ^{x}$

8 0

3 years ago

Hunter and Brian are creating a poster for a project and they need to combine two different size boards to create their project.

shepuryov [24]

Answer:

The maximum area of the second board must be $(5x^2+3x-6)$ square inches.

Step-by-step explanation:

The total area of two different size boards cannot exceed $(7x^2-6x+2)$ square inches.

The area of one board is $(2x^2-9x+8)$ square inches.

Suppose, the area of the second board is $y$ square inches.

That means......

$y+(2x^2-9x+8)\leq 7x^2-6x+2\\ \\ y\leq (7x^2-6x+2)-(2x^2-9x+8)\\ \\y \leq 7x^2-6x+2-2x^2+9x-8\\ \\ y\leq 5x^2+3x-6$

So, the maximum area of the second board must be $(5x^2+3x-6)$ square inches.

7 0

3 years ago

Read 2 more answers