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steposvetlana [31]
3 years ago
13

PLEASE I NEED HELP SO I CAN MOVE ON WITH MY HW

Mathematics
1 answer:
AnnZ [28]3 years ago
7 0

At the 2/8 part, just use the quadratic formula.  If you use it, you should get the other two zeroes:

-4+\frac{\sqrt{14}}{2} \\ -4-\frac{\sqrt{14}}{2}

You might be interested in
Just 16 not the words below​
alekssr [168]

Answer:

Table F

Step-by-step explanation:

The equation is C = 0.75t. The variable c is for cost and t is for tickets. When we use substitution to solve the equation, we see that table F correctly represents the situation. In simpler terms, the equation is asking us to mutliply the cost PER ticket by the number of tickets.

0.75 * 1 = 0.75

0.75 * 2 = 1.50

0.75 * 3 = 2.25

0.75 * 4 = 3

And so on. As we can see, the only table that shows these values is in fact table F.

Hope this helps!

8 0
3 years ago
39,892 to the nearest 1000 on the number line
Iteru [2.4K]

I'm pretty sure it would be 40,000 because if i'm not mistaken you meant the nearest 10,000 since the number is 39,892.

7 0
3 years ago
Which statement describes the relationship between x and y?
stich3 [128]

Answer:

as x increases, y increases

Step-by-step explanation:

3 0
3 years ago
Compare the light gathering power of an 8" primary mirror with a 6" primary mirror. The 8" mirror has how much light gathering p
tankabanditka [31]

Answer:

1.778 times more or 16/9 times more

Step-by-step explanation:

Given:

- Mirror 1: D_1 = 8''

- Mirror 2: D_2 = 6"

Find:

Compare the light gathering power of an 8" primary mirror with a 6" primary mirror. The 8" mirror has how much light gathering power?

Solution:

- The light gathering power of a mirror (LGP) is proportional to the Area of the objects:

                                           LGP ∝ A

- Whereas, Area is proportional to the squared of the diameter i.e an area of a circle:

                                           A ∝ D^2

- Hence,                              LGP ∝ D^2

- Now compare the two diameters given:

                                           LGP_1 ∝ (D_1)^2

                                           LGP ∝ (D_2)^2

- Take a ratio of both:

                           LGP_1/LGP_2 ∝ (D_1)^2 / (D_2)^2

- Plug in the values:

                               LGP_1/LGP_2 ∝ (8)^2 / (6)^2

- Compute:             LGP_1/LGP_2 ∝ 16/9 ≅ 1.778 times more

6 0
3 years ago
The following graph describes function 1 and the equation below it describes function 2:
Tanya [424]
Function 1: 
f(x) = -x² + 8(x-15)f(x) = -x² <span>+ 8x - 120
Function 2:
</span>f(x) = -x² + 4x+1
Taking derivative will find the highest point of the parabola, since the slope of the parabola at its maximum is 0, and the derivative will allow us to find that.
Function 1 derivative: -2x + 8 ⇒ -2x + 8 = 0 ⇒ - 2x = -8 ⇒ x = -8/-2 = 4

Function 2 derivative:  -2x+4 ⇒ -2x + 4 = 0 ⇒ -2x = -4 ⇒ x = -4/-2 ⇒  x= 2
Function 1: f(x) = -x² <span>+ 8x - 120 ; x = 4
f(4) = -4</span>² + 8(4) - 120 = 16 + 32 - 120 = -72
<span>
Function 2: </span>f(x) = -x²<span> + 4x+1 ; x = 2
</span>f(2) = -2² + 4(2) + 1 = 4 + 8 + 1 = 13

Function 2 has the larger maximum.
8 0
3 years ago
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