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2 years ago

Question 13 and I added a photo

Mathematics

1 answer:

Roman55 [17]2 years ago

6 0

The answer is C. 90 degrees.

This is because angle R is one of the four angles of a square.

I hope this helps.

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PLEASE HELP!!!!!!!!!!!!!!!

kotykmax [81]

What's the question?

7 0

3 years ago

Solve the equation x^2 - 4x + 6 = 9 by completing the square.

tatuchka [14]

Hello!

x² - 4x + 6 = 9 <=>

<=> x² - 4x + 6 - 9 = 0 <=>

<=> x² - 4x - 3 = 0 <=>

<=> x = -(-4)±√(-4)²-4×1×(-3)/2×1 <=>

<=> x = 4±√16+12/2 <=>

<=> x = 4±√28/2 <=>

<=> x = 4±2√7/2 =>

=> x = 4+2√7/2 and x = 4-2√7/2 =>

=> x = 2+√7 and x = 2-√7

The equation has two solutions.

Good luck! :)

7 0

2 years ago

Read 2 more answers

Can't get this one. Anyone know how to do it?

avanturin [10]

Function g takes 4 as input and return 2 as output
f takes input 2 and returns output 4
from the graph
g(4)=2
f(2)=4

7 0

2 years ago

What is the maximum volume in cubic inches of an open box to be made from a 12-inch by 16-inch piece of cardboard by cutting out

Romashka-Z-Leto [24]

If you start with a 12x16 rectangle and cut square with side length x, when you bend the sides you'll have an inner rectangle with sides $12-2x$ and $16-2x$ , and a height of x.

So, the volume will be given by the product of the dimensions, i.e.

$(12-2x)(16-2x)x = 4x^3-56x^2+192x$

The derivative of this function is

$12x^2-112x+192$

and it equals zero if and only if

$12x^2-112x+192=0 \iff x = \dfrac{14\pm 2\sqrt{13}}{3}$

If we evaluate the volume function at these points, we have

$f\left(\dfrac{14-2\sqrt{13}}{3}\right) = \dfrac{64}{27}(35+13\sqrt{13})> f\left(\dfrac{14-2\sqrt{13}}{3}\right) = -\dfrac{64}{27}(13\sqrt{13}-35)$

So, the maximum volume is given if you cut a square with side length

$x=\dfrac{14-2\sqrt{13}}{3}\approx 2.7$

5 0

2 years ago

What is (5x-2)(4x^2 -3x-2)

Ronch [10]

Answer: 20x^3 - 23x^2 - 4x + 4

Explanation:

Use distributive property:

(5x-2)(4x^2 - 3x-2)
= 20x^3 - 15x^2 - 10x - 8x^2 + 6x + 4
= 20x^3 - 23x^2 - 4x + 4

8 0

2 years ago

Read 2 more answers