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aniked [119]
3 years ago
10

What's the equation of the parabola that has its vertex at (4,–1) and a y-intercept at (0,7)?

Mathematics
1 answer:
xz_007 [3.2K]3 years ago
8 0

Answer:

C

because it is given that the vertex is (4, - 1) an an equation C it is clearly seen that the vertex is same.

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DVD cases are sold in packages of 20. Padded mailing envelopes are sold in packets of 12. What is the least number of cases and
Alex777 [14]
I think it might be 2 because 20 divided by 2 there is no left over
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3 years ago
aisha and brendan are watching a hot air balloon. the angle of elevation for aisha is 65 degrees. what is the angle of elevation
astra-53 [7]

Using the law of sins:

B =sin^−1((sin(A)⋅a)/b) = sin^−1((sin(65)⋅110)/155)=40.03 degrees

B= 40.03 degrees (round as needed)

3 0
4 years ago
Sampson has a list of items that are on sale to pick up at the store. The sale items are the following:
olchik [2.2K]

Answer:

the answer is A) Yes, Sampson's receipt is correct.

Step-by-step explanation:

8 0
3 years ago
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An 8-sided die is formed when two square pyramids are attached by their bases. The perimeter of the base shape is 6 centimeters.
lawyer [7]

Answer:

7.5cm²

Complete question:What is the surface area of the 8-sided figure?

Step-by-step explanation:

Consider the half of the 8-sided die i.e the pyramid ABCDE:

As the base is 6 of the perimeter, so each side of the base is 6/4=3/2

Now, the distance between the tallest points is 2 cm, so the height of Pyramid ABCDE is 1 cm.

Considering the right triangle EOM:

EO=1, OM= 3/4 (it is half of AB=3/2) and EM is the hypothenuse, so by the pythagorean theorem length of EM is 

\sqrt{1+ ( \frac{3}{4} )^{2} }=  \sqrt{1+ \frac{9}{16} }= \sqrt{ \frac{25}{16} }= \frac{5}{4}   (cm)

5. So side EBC has area  

\frac{1}{2}BC*EM=  \frac{1}{2}* \frac{3}{2}* \frac{5}{4}= \frac{15}{16}

6. The total area is 8*Area(EBC)=8* \frac{15}{16}= \frac{15}{2}=7.5 ( cm^{2} )

3 0
3 years ago
S = 1 / g t ² π, solve for t​
Finger [1]

s =   \frac{1}{g {t}^{2}\pi }

multiply through by

g {t}^{2} \pi

s \times g {t}^{2} \pi =  \frac{1}{g {t}^{2} \pi}  \times g {t}^{2} \pi

sg {t}^{2} \pi = 1

dividing through by

sgπ

{t}^{2}  =  \frac{1}{sg\pi}

t =  \sqrt{ \frac{1}{sg\pi} }

4 0
3 years ago
Read 2 more answers
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