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katovenus [111]
3 years ago
9

Which polynomial could have the following graph?

Mathematics
2 answers:
zubka84 [21]3 years ago
8 0
Hi,
Roots are 2, -1,-3.So y=(x-3)(x+1)(x+3)
Answer B
babunello [35]3 years ago
8 0

Answer: y=(x-2)(x+1)(x+3)

Step-by-step explanation:

From the given graph, it can be seen that the x intercepts of the graph are

x=-3,\ x=-1,\ x=2

It implies the value of y at the above points must be zero.

[x-intercept is of the form (a,0), where the value of y is zero.]

It implies x=-3,\ x=-1,\ x=2 are the zeroes of the polynomial graphed.

It implies (x-2),\ (x+1),\ (x+3) are the factors of the polynomial graphed.

[By Factor theorem]

Thus the polynomials should be :  y=(x-2)(x+1)(x+3) which is the correct polynomial for the given graph.

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ben drives from a 36- foot platform. The equation h= -16t^2+14t+36 models the dive. How long will it take Ben to reach the water
Valentin [98]
So you want to plug in the height for h and solve for t

36 = -16t^2+14t+36
So you get 14/16 which simplifies to
7/8 seconds

Hope this helped!
6 0
2 years ago
In 12 hours, how many more kilometers could the yellow car go than the red car? show your work
trasher [3.6K]

Answer:

144

Step-by-step explanation:

12 times 12

3 0
2 years ago
6+6y=-30 solve for y​
kakasveta [241]

Answer:

y = -6

Step-by-step explanation:

6 + 6y = -30

6y = -30 - 6

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7 0
2 years ago
Help please explain ​
MatroZZZ [7]

Given:

\sin^{-1}\left(\dfrac{5}{8}\right)=A

To find:

The correct equivalent equation.

Solution:

We have,

\sin^{-1}\left(\dfrac{5}{8}\right)=A

Taking sin on both sides, we get

\sin \sin^{-1}\left(\dfrac{5}{8}\right)=\sin A

\dfrac{5}{8}=\sin A                      [\because \sin (\sin^{-1}\theta )=\theta]

Interchanging the sides, we get

\sin A=\dfrac{5}{8}

Therefore, the correct option is 4.

8 0
3 years ago
5. Firewood is often sold by the cord. A cord of wood is a bundle measuring 4 feet by 4 feet by 8 feet. What's the volume of a c
Eduardwww [97]
V = L *  W * H
V = 4 * 4 * 8
V = 16 * 8
V = 128 cubic ft
7 0
3 years ago
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