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sesenic [268]
3 years ago
8

Find the x-intercepts of the parabola with vertex (7,-12) and y-intercept (0,135).

Mathematics
1 answer:
Ira Lisetskai [31]3 years ago
8 0

Answer:

x=9 and x=5

OR

(9,0) and (5,0)

Step-by-step explanation:

<u>Use vertex form to solve for the unknown variable</u>

y=a(x-h)^2+k\\\\135=a(0-7)^2-12\\\\147=a(-7)^2\\\\147=49a\\\\3=a

<u>Use the new equation to find the x-intercepts</u>

<u />y=3(x-7)^2-12\\\\0=3(x-7)^2-12\\\\12=3(x-7)^2\\\\4=(x-7)^2

2=x-7\\9=x

-2=x-7\\5=x

Therefore, the x-intercepts of the parabola are (9,0) and (5,0). Feel free to review the attached graph for more information.

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3 years ago
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Solve for u. 20 − (u · 0.9) = 4.7
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Answer:

20 - 0.9u = 4.7

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Hope this helps.

3 0
3 years ago
Are the expressions 3{y+1} and 3y+3 equivalent for any value? explain
andre [41]

Answer:

Yes

Step-by-step explanation:

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3 years ago
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A triangle has side lengths of 9 in, 13 in, and 20 in. What is the measurement of this triangle’s largest angle?
zvonat [6]
1. Given any triangle ABC with sides BC=a, AC=b and AB=c, the following are true :

      i) the larger the angle, the larger the side in front of it, and the other way around as well. (Sine Law) Let a=20 in, then the largest angle is angle A.

       ii) Given the measures of the sides of a triangle. Then the cosines of any of the angles can be found by the following formula:

       a^{2}=b ^{2}+c ^{2}-2bc(cosA)

2. 

20^{2}=9 ^{2}+13 ^{2}-2*9*13(cosA)&#10;&#10;400=81+169-234(cosA)   150=-234(cosA)&#10;&#10;cosA=150/-234= -0.641

3. m(A) = Arccos(-0.641)≈130°, 

4. Remark: We calculate Arccos with a scientific calculator or computer software unless it is one of the well known values, ex Arccos(0.5)=60°, Arccos(-0.5)=120° etc
4 0
3 years ago
Read 2 more answers
1. Identify the lateral area and surface area of a regular square pyramid with base edge length 5 in. and slant height 9 in.
zlopas [31]
1. C
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L=1/2(20x9)
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L=90

SA=1/2(Pxl)+B
SA=90+5^2
SA=115

B, A, C, B.
6 0
3 years ago
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