All categories

1 year ago

Solve for x: 5x+7=3x−9.

Mathematics

1 answer:

ludmilkaskok [199]1 year ago

6 0

To solve for x in the given equation;

$5x+7=3x-9$

We start by collecting like terms, as follows;

$5x-3x=-9-7$

When a number crosses the sign of equality, the sign in front of it switches. Positive numbers becomes negative, and vice versa.

We would now have;

$\begin{gathered} 5x-3x=-9-7 \\ 2x=-16 \\ \text{Divide both sides by 2;} \\ \frac{2x}{2}=\frac{-16}{2} \\ x=-8 \end{gathered}$

ANSWER:

The answer is x = -8

You might be interested in

PLESSE HELPP I NEED HELPPP ASAPPPPPPPP

Akimi4 [234]

D. (2,3)

Explanation:

We already know that x=2, so of we plug in 2 for x, we should get the y value.

Once we plug in 2 for x we should get the equation:

2(2)+3y=13.

Simplify that and we get:

4+3y=13.

Subtract 4 from both sides and we get:

3y=9.

Divide both sides by 3 and we get:

y=3.

This means that your ordered pair should be (2,3).

Hope this helps :)

4 0

4 years ago

How do i help people by answering there question

Fudgin [204]

What do you mean by this?

3 0

3 years ago

a person kicks a ball from the ground into the air with an initial upward velocity of 8 feet per second. what is the maximum hei

KengaRu [80]

Chech the picture below.

if the person kicked it from the ground, that means its initial height is 0.

it reaches its maximum height at the y-coordinate of its vertex, and it will hit the ground when y = 0, as you see in the picture.

$\bf ~~~~~~\textit{initial velocity}
\\\\
\begin{array}{llll}
~~~~~~\textit{in feet}
\\\\
h(t) = -16t^2+v_ot+h_o
\end{array} 
\quad 
\begin{cases}
v_o=\stackrel{8}{\textit{initial velocity of the object}}\\\\
h_o=\stackrel{0}{\textit{initial height of the object}}\\\\
h=\stackrel{}{\textit{height of the object at "t" seconds}}
\end{cases}
\\\\\\
h(t)=-16t^2+8t+0$

now let's find the y-coordinate of its vertex

$\bf \textit{vertex of a vertical parabola, using coefficients}
\\\\
h(t)=\stackrel{\stackrel{a}{\downarrow }}{-16}t^2\stackrel{\stackrel{b}{\downarrow }}{+8}t\stackrel{\stackrel{c}{\downarrow }}{+0}
\qquad \qquad 
\left(-\cfrac{ b}{2 a}~~~~ ,~~~~ c-\cfrac{ b^2}{4 a}\right)
\\\\\\
\left(\qquad ,~~0-\cfrac{8^2}{4(-16)} \right)\implies (\quad ,~0+1)\implies (\quad ,~1)$

when will it hit the ground?

$\bf \stackrel{h(t)}{0}=-16t^2+8t\implies 16t^2-8t=0\implies 8t(2t-1)=0
\\\\\\
8t=0\implies \stackrel{\textit{0 seconds when it took off first}}{t=0}
\\\\\\
2t-1=0\implies 2t=1\implies \stackrel{\textit{half a second later it when it came back down}}{t=\cfrac{1}{2}}$

3 0

3 years ago

What is a buffer made from?

Leona [35]

Answer:

Step-by-step explanation:

A buffer must contain a weak acid and its conjugate base. There are several ways a solution containing these two components can be made: Buffers can be made from weak acids or base and their salts. ... Buffers can be made from two salts that provide a conjugate acid-base pair.

8 0

3 years ago

Read 2 more answers

Equations of lines in slope intercept form<br> Through: (1,5) m=2

Mrac [35]

Answer:

The equation would be

y = 2x + 3

5 0

4 years ago