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Nitella [24]
2 years ago
11

Please help!!! picture below

Mathematics
1 answer:
iVinArrow [24]2 years ago
6 0

Answer:

4^3x=4^(4x+4)

Step-by-step explanation:

You can eliminate the first 2 options since those don’t make sense already, and the last option changed 64 to 2^6 which is correct, but when they changed the other side. They should have multiplied 8 to both x and 1 which would have made the whole term 2^(8x+8) and not 2^(8x+1). Left with the third option, 4 to the power of 3 is 64 so the left side makes sense, 4^3x=64^x, on the right side, 4^4 equals 256, so the whole exponent should multiply by 4 so from 256^(x+1) you get 4^4(x+1) giving 4^(4x+4) making the third choice the right answer.

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10. simplify the rational expression by rationalizing the denominator.( 1 point ) 4√150/√189x
ddd [48]
Rationalizing the denominator, simply means "getting rid of that pesky root at the bottom", and we do so by simply multiplying it by something to take it out, of course, we multiply the bottom, we have to also multiply the top,

\bf \cfrac{4\sqrt{150}}{\sqrt{189x}}\cdot \cfrac{\sqrt{189x}}{\sqrt{189x}}\implies \cfrac{4\sqrt{150}\sqrt{189x}}{(\sqrt{189x})^2}\implies \cfrac{4\sqrt{(150)({189x})}}{189x}

\bf \cfrac{4\sqrt{28350x}}{189x}\qquad 
\begin{cases}
28350=2\cdot 3\cdot 3\cdot 3\cdot 3\cdot 5\cdot 5\cdot 7\\
\qquad 2\cdot 3^2\cdot 3^2\cdot 5^2\cdot 7\\
\qquad 2\cdot (3^2)^2\cdot 5^2\cdot 7\\
\qquad 2\cdot (3^2\cdot 5)^2\cdot 7\\
\qquad 2\cdot (45)^2\cdot 7\\
\qquad 14\cdot 45^2
\end{cases}\implies \cfrac{4\sqrt{14\cdot 45^2}}{189x}
\\\\\\
\cfrac{4\cdot 45\sqrt{14}}{189}\implies \cfrac{180\sqrt{14}}{189x}\implies \cfrac{20\sqrt{14}}{21x}
4 0
3 years ago
Help please thanks!!!!!!!!!!
sergeinik [125]
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6 0
3 years ago
Manuel created a factor tree and wrote the prime factorization of 60 shown below.
bagirrra123 [75]

Answer:

The second one or B- "He did not find the prime factors of 4."

Step-by-step explanation:

I got it right on Edge2020

7 0
2 years ago
Read 2 more answers
Which input value produces the sane output value for the two functions on the gragh f(x)=-2/3x+1 g(x)=1/3x-2
zimovet [89]

Answer:

X=3

Step-by-step explanation:

We have two linear functions which intersect at a point. This point is shown in the attached graph. Linear functions are lines which are made of points that satisfy the function or relationship.  This means at the intersection, this point (3,-1), both functions have the same values. An input of x=3 produces y=-1 in both functions.


5 0
3 years ago
Pls explain How do do all of these and say answers
lyudmila [28]

Answer:

QUESTION ONE-

Cross multiply:

5 * 15 = 12 * y

Simplifying

5 * 15 = 12 * y

Multiply 5 * 15

75 = 12 * y

Solving

75 = 12y

Solving for variable 'y'.

Move all terms containing y to the left, all other terms to the right.

Add '-12y' to each side of the equation.

75 + -12y = 12y + -12y

Combine like terms: 12y + -12y = 0

75 + -12y = 0

Add '-75' to each side of the equation.

75 + -75 + -12y = 0 + -75

Combine like terms: 75 + -75 = 0

0 + -12y = 0 + -75

-12y = 0 + -75

Combine like terms: 0 + -75 = -75

-12y = -75

Divide each side by '-12'.

y = 6.25

Simplifying

y = 6.25

QUESTION TWO-

Cross multiply:

8 * w = 20 * 6

Simplifying

8 * w = 20 * 6

Multiply 20 * 6

8w = 120

Solving

8w = 120

Solving for variable 'w'.

Move all terms containing w to the left, all other terms to the right.

Divide each side by '8'.

w = 15

Simplifying

w = 15

QUESTION 3-

Cross multiply:

s + 1 * 8 = 4 * 4

Simplifying

s + 1 * 8 = 4 * 4

Reorder the terms:

1 + s * 8 = 4 * 4

Reorder the terms for easier multiplication:

8 * 1 + s = 4 * 4

1 * 8 + s * 8 = 4 * 4

8 + 8s = 4 * 4

Multiply 4 * 4

8 + 8s = 16

Solving

8 + 8s = 16

Solving for variable 's'.

Move all terms containing s to the left, all other terms to the right.

Add '-8' to each side of the equation.

8 + -8 + 8s = 16 + -8

Combine like terms: 8 + -8 = 0

0 + 8s = 16 + -8

8s = 16 + -8

Combine like terms: 16 + -8 = 8

8s = 8

Divide each side by '8'.

s = 1

Simplifying

s = 1

Step-by-step explanation:

6 0
3 years ago
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