Ask question

Ask question

All categories

zhannawk [14.2K]

3 years ago

8

Solve for x.

1 answer:

Tamiku [17]3 years ago

5 0

The answer to the question is a

You might be interested in

1. What is the slope of the line in the graph shown below?

Agata [3.3K]

1. Answer( -2)
2. Answer is D. 4/3
3. Answer is either A or B

3 0

3 years ago

Read 2 more answers

If 3x-y=123x−y=123, x, minus, y, equals, 12, what is the value of \dfrac{8^x}{2^y} 2 y 8 x start fraction, 8, start superscrip

julia-pushkina [17]

Answer:

$2^{12}=4,096$

Step-by-step explanation:

You know that $3x-y=12$ and have to find

$\dfrac{8^x}{2^y}$

Use the main properties of exponents:

1. $(a^m)^n=a^{m\cdot n}$

2. $\dfrac{a^m}{a^n}=a^{m-n}$

Note that

$8=2^3,$

then

$7^x=(2^3)^x=2^{3\cdot x}=2^{3x}$

Now

$\dfrac{8^x}{2^y}=\dfrac{2^{3x}}{2^y}=2^{3x-y}$

Since $3x-y=12,$ then $2^{3x-y}=2^{12}=4,096$

3 0

3 years ago

Read 2 more answers

The four-member math team at Pecanridge Middle School is chosen from the math club, which has three girls and five boys. How man

cestrela7 [59]

Answer:

$Total\ Selection = 30\ ways$

Step-by-step explanation:

Given

Girls = 3

Boys = 5

Required

How many ways can 2 boys and girls be chosen?

The keyword in the question is chosen;

This implies combination and will be calculated as thus;

$Selection =\ ^nC_r = \frac{n!}{(n-r)!r!}$

For Boys;

n = 5 and r = 2

$Selection =\ ^5C_2$

$Selection = \frac{5!}{(5-2)!2!}$

$Selection = \frac{5!}{3!2!}$

$Selection = \frac{5 * 4 * 3!}{3!*2 * 1}$

$Selection = \frac{20}{2}$

$Selection = 10$

For Girls;

n = 3 and r = 2

$Selection =\ ^3C_2$

$Selection = \frac{3!}{(3-2)!2!}$

$Selection = \frac{3!}{1!2!}$

$Selection = \frac{3 * 2!}{1 *2!}$

$Selection = \frac{3}{1}$

$Selection = 3$

Total Selection is calculated as thus;

$Total\ Selection = Boys\ Selection * Girls\ Selection$

$Total\ Selection = 10 * 3$

$Total\ Selection = 30\ ways$

5 0

4 years ago

Find each f(c) using synthetic substitution.

love history [14]

f(3) = 69

Step-by-step explanation:

Given f(x) =5x5 + 10x4 + 3x3 + 8x2 – 6x – 3; c = 3

f(x) = 25+40+9+16 - 6x - 3

f(x) = - 6x +90-3

f(x) = - 6f(3) = 69x + 87

Taken c = 3

f(c) = - 6c +90-3

f(c) = - 6c + 87

substitute c=3

f(3) = - 6 (3) +90-3

= -18 +87

= 69

Hence .,f(3) = 69

3 0

3 years ago

Read 2 more answers

Kate begins solving the equation StartFraction 2 Over 3 EndFraction left-parenthesis 6 x minus 3 right-parenthesis equals StartF

zubka84 [21]

Answer:

(B)The equation has one solution: x = 0.

Step-by-step explanation:

The equation Kate is solving is: $\dfrac{2}{3} (6x-3)=\dfrac{1}{2}(6x-4)$

$\dfrac{2(6x-3)}{3}=\dfrac{(6x-4)}{2}\\\dfrac{2*3(2x-1)}{3}=\dfrac{2(3x-2)}{2}\\4x-2=3x-2\\$Adding two both sides$\\4x=3x\\4x-3x=0\\x=0$

Therefore, the equation has one solution: x = 0.

4 0

4 years ago