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3 years ago

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Which expressions are equivalent to 16x4 − 64? Check all that apply. 16x4 + 16x – 16x – 64 16x4 – 8x – 8x – 64 (4x2 + 8)(4x2 – 8

) 16(x2 + 2)(x2 – 2) 4(4x4 – 8)

1 answer:

KengaRu [80]3 years ago

7 0

Answer:

16x4 + 16x – 16x – 64

16x4 – 8x – 8x – 64

(4x2 + 8)(4x2 – 8)

16(x2 + 2)(x2 – 2)

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Help me someone!!!!!!!!!!!!!!!!!!!!!!!

Julli [10]

Substitute answers in, a = 3 and b = 2.

(3^2 - 2^2)/(3+2)
= (9-4)/5
= 5/5
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3 years ago

Arrange the systems of equations that have a single solution in Increasing order of the x-values in their solutions.

NemiM [27]

Complete question is;

Arrange the systems of equations that have a single solution in Increasing order of the x-values in their solutions.

1) 2x + y = 10

x - 3y = -2

2) x + 2y = 5

2x + y = 4

3) x + 3y = 5

6x - y = 11

4) 2x + y = 10

-6x - 3y = -2

Answer:

Arranging the systems in increasing value of x gives;

System 2, system 3, system 1

Step-by-step explanation:

1) 2x + y = 10 - - - (eq 1)

x - 3y = -2 - - - (eq 2)

Multiply eq 1 by 3 to get;

6x + 3y = 30

Add it to eq 2 to get;

7x = 28

x = 28/7

x = 4

2) x + 2y = 5 - - - (eq 1)

2x + y = 4 - - - (eq 2)

Multiply eq 2 by 2 to get;

4x + 2y = 8

Subtract eq 1 from it to get;

3x = 3

x = 3/3

x = 1

3) x + 3y = 5 - - - (eq 1)

6x - y = 11 - - - (eq 2)

Multiply eq 2 by 3 to get;

18x - 3y = 33

Add to eq(1) to get;

19x = 38

x = 38/19

x = 2

4) 2x + y = 10 - - - (eq 1)

-6x - 3y = -2 - - - (eq 2)

Make y the subject in eq 1 to get;

y = 10 - 2x

Put in eq (2);

-6x - 3(10 - 2x) = -2

-6x - 30 + 6x = -2

0 = 0

Thus,no solution exists.

Arranging the systems in increasing value of x gives;

System 2, system 3, system 1

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3 years ago

Consider the following system of equations:

Delicious77 [7]

Answer:

The system has infinitely solutions

Step-by-step explanation:

we have

$-\frac{1}{3}x^{2}=-\frac{5}{6}+\frac{1}{3}y^{2}$

$\frac{1}{3}x^{2}+\frac{1}{3}y^{2}=\frac{5}{6}$

Multiply by 3 both sides

$x^{2}+y^{2}=\frac{5}{2}$ ----> equation A

The equation A is a circle centered at origin with radius $r=\sqrt{5/2}\ units$

and

$5y^{2} =\frac{25}{2}-5x^{2}$

$5x^{2}+5y^{2} =\frac{25}{2}$

Divide by 5 both sides

$x^{2}+y^{2} =\frac{5}{2}$ ----> equation B

The equation B is a circle centered at origin with radius $r=\sqrt{5/2}\ units$

Equation A and Equation B are the same

Therefore

The system has infinitely solutions

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3 years ago

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Find each quotient. Write in simplest form. keep answer as fraction - not decimal

kirill [66]

So we are given the expression:

$\frac{mn^{2}}{4}$ ÷ $\frac{m^{2}n}{8}$

When we divide fractions, we must flip the second term and change the sign to multiplication:

$\frac{mn^{2}}{4} *\frac{8}{m^{2}n}$

And then we multiply across:

$\frac{mn^{2}}{4} *\frac{8}{m^{2}n}= \frac{8mn^{2}}{4m^{2}n}$

Then we can break apart all of the like variables for simplification:

$\frac{8mn^{2}}{4m^{2}n}=\frac{8}{4} * \frac{m}{m^{2}} *\frac{n^{2}}{n}$

When we simplify variables through division, we subtract the exponent of the numerator from the exponent of the denominator. So we then have:

$\frac{8}{4} =2$

$\frac{m}{m^{2}}=m^{-1} =\frac{1}{m}$

$\frac{n^{2}}{n}=n^{1}=\frac{n}{1}$

So then we multiply all of these simplified parts together:

$\frac{2}{1}*\frac{1}{m}*\frac{n}{1} =\frac{2n}{m}$

So now we know that the simplified form of the initial expression is: $\frac{2n}{m}$ .

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3 years ago

A pentagon with one interior angle that measures 200 degrees is a ___________ polygon.

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A.concave. Any shape with an angle more than 180 is called concave.

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