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Nataliya [291]
2 years ago
10

How can you convert vertex form into standard form

Mathematics
1 answer:
loris [4]2 years ago
4 0
It would take a while to explain so here is an easier explanation to understand :http://www.virtualnerd.com/algebra-2/quadratics/transforming-functions/vertex-form/convert-vertex-to...

Hope this helps!
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Annette measured pepper plant sprouts and recorded the following lengths in
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Answer:

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shuan polleted a point on the number line by drawing 5 equally marks 0 and 1 and placing a point on the third mark he clams that
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2 years ago
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Find constants a and b such that the function y = a sin(x) + b cos(x) satisfies the differential equation y'' + y' − 7y = sin(x)
Elza [17]
The first thing we must do in this case is find the derivatives:
 y = a sin (x) + b cos (x)
 y '= a cos (x) - b sin (x)
 y '' = -a sin (x) - b cos (x)
 Substituting the values:
 (-a sin (x) - b cos (x)) + (a cos (x) - b sin (x)) - 7 (a sin (x) + b cos (x)) = sin (x)
 We rewrite:
 (-a sin (x) - b cos (x)) + (a cos (x) - b sin (x)) - 7 (a sin (x) + b cos (x)) = sin (x)
 sin (x) * (- a-b-7a) + cos (x) * (- b + a-7b) = sin (x)
 sin (x) * (- b-8a) + cos (x) * (a-8b) = sin (x)
 From here we get the system:
 -b-8a = 1
 a-8b = 0
 Whose solution is:
 a = -8 / 65
 b = -1 / 65
 Answer:
 
constants a and b are:
 
a = -8 / 65
 
b = -1 / 65
4 0
3 years ago
which expression is equivalent to 100 n2 − 1? (10n)2 − (1)2 (10n2)2 − (1)2 (50n)2 − (1)2 (50n2)2 − (1)2
givi [52]
Which expression is equivalent to 100 n2 − 1? (10n)2 − (1)2 (10n2)2 − (1)2 (50n)2 − (1)2 (50n2)2 − (1)2

Regroup:
100n^2-1

=10^2n^2-1
=(10n)^2-1^2

So the correct answer is : 
(10n)^2-1^2

To attempt to factor a polynomial of four or more terms with no common factor, first rewrite it in groups. Each group may possibly be separately factored, and the resulting expression may possibly lend itself to further factorization if a greatest common factor<span> or special form is created.</span>
5 0
3 years ago
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What is the surface area of this triangular prism? The base of each triangle is 42 m and the height of the triangular base is 20
shutvik [7]

<u>Given</u>:

The base of each triangular base is 42 m.

The height of each triangular base is 20 m.

The sides of the triangle are 29 m each.

The height of the triangular prism is 16 m.

We need to determine the surface area of the triangular prism.

<u>Surface area of the triangular prism:</u>

The surface area of the triangular prism can be determined using the formula,

SA=bh+(s_1+s_2+s_3)H

where b is the base of the triangle,

h is the height of the triangle,

s₁, s₂ and s₃ are sides of the triangle and

H is the height of the prism.

Substituting the values, we get;

SA=(42)(20)+(42+29+29)(16)

SA=840+(100)(16)

SA=840+1600

SA=2440

Thus, the surface area of the triangular prism is 2440 m²

8 0
3 years ago
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