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2 years ago

Hi can you please help me with my work and please no Link please no Link please

Mathematics

1 answer:

STatiana [176]2 years ago

8 0

Answer:

The second one, -11/12

Step-by-step explanation:

I used a website so sorry I don't have an explanation

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Using the answers from z^4=16 determine the solutions of the equation (x+1)^4=16(x-1)^4

Artemon [7]

i hope i have been useful buddy.

good luck ♥️♥️♥️♥️♥️.

4 0

3 years ago

what is the coordinate of the vertex of the parabola y-2=1/12(x+10) a. ( 10,-2) b. (-10,2) c. (-2,10) d. (2.-10)

solniwko [45]

Answer:

$y-a= c(x-b)^2$

With the vertex $V= (b,a)$

We see that b = -10 and a = 2 and then the vertes wuld be:

$V= (-10,2)$

And the best option is:

b. (-10,2)

Step-by-step explanation:

For this problem we have the following function:

$y-2 = \frac{1}{2} (x+10)$

And if we compare this expression with the general expression for a parabola given by:

$y-a= c(x-b)^2$

With the vertex $V= (b,a)$

We see that b = -10 and a = 2 and then the vertes wuld be:

$V= (-10,2)$

And the best option is:

b. (-10,2)

4 0

3 years ago

A right triangle has side lengths 4 units, 5 units, and x units what is the difference between the difference values of x?

irina [24]

Use the Pythagorean theorem.

If x is a hypotenuse, then we have:

$x^2=4^2+5^2\\\\x^2=16+25\\\\x^2=41\to x=\sqrt{41}$

If x is a leg, then we have:

$x^2+4^2=5^2\\\\x^2+16=25\ \ \ \ |-16\\\\x^2=9\to x=\sqrt9\to x=3$

Answer: $\sqrt{41}-3$

6 0

3 years ago

Sum is the answer to what type of problem

Julli [10]

The answer will be Addition Problem

5 0

2 years ago

A box has a perimeter of 27 1/2 ft. if the lenght is 9 ft. what is the area?

gavmur [86]

Answer:

Perimeter = (Length x 2) + (Width x 2)

27.5 = (9 x 2) + (Width x 2)

27.5 = 18 + (Width x 2)

9.5 = (Width x 2)

4.75 = Width

check:

27.5 = (9 x 2) + (4.75 x 2)

27.5 = 18 + 9.5

27.5 = 27.5

Since the width is 4.75 and the length is 9, we can multiply the two to get the area. 4.75 x 9 = 42.75 sq. ft.

8 0

2 years ago

Hi can you please help me with my work and please no Link please no Link please​

Hi can you please help me with my work and please no Link please no Link please