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4 years ago

Solve the equation. Show your work. 3x – 10 = 5x + 32

Mathematics

2 answers:

liq [111]4 years ago

4 0

Answer:

-21 =x

Step-by-step explanation:

3x – 10 = 5x + 32

Subtract 3x from each side

3x-3x – 10 = 5x-3x + 32

-10 =2x +32

Subtract 32 from each side

-10 -32 = 2x+32-32

-42 = 2x

Divide each side by 2

-42/2 = 2x/2

-21 =x

Marta_Voda [28]4 years ago

3 0

Answer: $x=-21$

Step-by-step explanation:

To solve the equation you need to find the value of "x".

First, you need to subtract 32 from both sides of the equation:

$3x - 10 -32= 5x + 32-32\\\\3x - 42= 5x$

Now you must subtract $3x$ from both sides of the equation:

$3x - 42-3x= 5x-3x\\\\- 42= 2x$

And finally divide both sides of the equation by 2. Then:

$\frac{-42}{2}=\frac{2x}{2}\\\\x=-21$

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Which equations have the same value of x as Three-fifths (30 x minus 15) = 72? Select three options.

lozanna [386]

The equivalent expressions are 3(6x - 3) = 72 and 18x - 9 = 72

<h3>Equations and expressions</h3>

Given the equation below as shown;

3/5(30x - 15) = 72

Expand

3/5(30x) - 3/5(15) = 72

90x/5 - 15/5 = 72

3(6x - 3) = 72

Expand

18x - 9 = 72

Hence the equivalent expressions are 3(6x - 3) = 72 and 18x - 9 = 72

Learn more on equations here: brainly.com/question/13763238

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HURRYYY i neeeeeddd helppppppppp look at the photo ^^^

Oliga [24]

Can you show a better photo some of the question is cut off

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3 years ago

Read 2 more answers

A model rocket is launched from the top of a building. The height(in meters) of the rocket above the ground is given by h(t)=-6t

Shalnov [3]

Answer:

Step-by-step explanation:

I don't know if you are a calculus student, but since calculus is the easiest way to solve this, that's what I used. The position equation is

$s(t)=-6t^2+24t+14$

and the velocity function, the first derivative, is

$v(t)=-12t+24$

From Physics, you should know that the velocity of an object is 0 at its highest point. That means that if we sub in a 0 for the velocity and solve for t, we find the time at which the object is at its highest point. That is:

$0=-12t+24$ and

$0=-12(t-2)$

By the Zero Product Property,

-12 definitely does not equal 0, so that means that

t - 2 = 0 and

t = 2 seconds.

So at 2 seconds, the object is at its highest point. Go back to the position function now and sub in a 2 for t to find the height of the object at 2 seconds.

$s(2)=-6(2)^2+24(2)+14$ so