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3 years ago

Identify the y-intercept of the graph below

Mathematics

1 answer:

emmasim [6.3K]3 years ago

3 0

Answer:

I want to say it's A

Step-by-step explanation:

the y-intercept is where the line crosses through the y-axis.

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Find the measure of the indicated angle to the nearest degree

uranmaximum [27]

51 the other person was right

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3 years ago

D. 10 + 11k can someone plzz help meeee plzzzz

torisob [31]

Answer:

(A). $\frac{15}{2}$ + $\frac{33}{4}$ <em>k</em>

Step-by-step explanation:

$\frac{3}{4}$ × 2(2 + 4k + 3 + $\frac{3}{2}$ k ) = $\frac{3}{2}$ ( 5 + $\frac{11}{2}$ k ) = $\frac{15}{2}$ + $\frac{33}{4}$ k

8 0

3 years ago

In Mr. Diaz’s class, 20% of the students received a grade of “A” on their report cards. There are 25 students in

Simora [160]

25/x=100/20 criss cross and u will get 500/100=5 then 5 students resivs A.

7 0

4 years ago

Read 2 more answers

The length of a rectangular garden is 2 feet more than twice it's width. If the area is 40 square feet, how much fencing would b

Westkost [7]

Answer:

w =4 ft and l =10 ft

Step-by-step explanation:

Let w = width

l = 2w+2

A = l*w

A = (2w+2)(w)

40 = (2w+2) w

Distribute

40 = 2w^2 +2w

Subtract 40

0 = 2w^2 +2w - 40

Dividing each side by 2

0 = w^2 +w -20

Factor

0 = (w+5) (w-4)

Using the zero product property

w+5 =0 w-4=0

w= -5 w=4

Since we cant have a negative length

w=4

l = 2w+2 = 2(4)+2 = 8+2 =10

7 0

3 years ago

A rectangular tank that is 5324 ft cubed with a square base and open top is to be constructed of sheet steel of a given thicknes

marishachu [46]

Answer:

Side of 22 and height of 11

Step-by-step explanation:

Let s be the side of the square base and h be the height of the tank. Since the tank volume is restricted to 5324 ft cubed we have the following equation:

$V = s^2h = 5324$

$h = 5324 / s^2$

As the thickness is already defined, we can minimize the weight by minimizing the surface area of the tank

Base area with open top $s^2$

Side area 4sh

Total surface area $A = s^2 + 4sh$

We can substitute $h = 5324 / s^2$

$A = s^2 + 4s\frac{5324}{s^2}$

$A = s^2 + 21296/s$

To find the minimum of this function, we can take the first derivative, and set it to 0

$A' = 2s - 21296/s^2 = 0$

$2s = 21296/s^2$

$s^3 = 10648$

$s = \sqrt[3]{10648} = 22$

$h = 5324 / s^2 = 5324 / 22^2 = 11$

4 0

3 years ago