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kolbaska11 [484]
2 years ago
5

A function f(x) has x intercepts of -3 and -5. What is the constant term in the function?

Mathematics
1 answer:
Anon25 [30]2 years ago
3 0

Answer:

The constant term in the function is 5

Step-by-step explanation:

we have

f(x)=x^{2}+8x+b

where

b is the y-intercept or the constant term of the function

Remember that

The x-intercept is the value of x when the value of the function is equal to zero

so

For x=-3 ----> f(x)=0

For x=-5 ----> f(x)=0

substitute any of the intercepts in the function

For x=-3

0=(-3)^{2}+8(-3)+b

0=9-24+b

0=-15+b

b=15

f(x)=x^{2}+8x+15

Verify with the other intercept

For x=-5

0=(-5)^{2}+8(-5)+15

0=25-40+15

0=0 ---> is true

therefore

The constant term in the function is 5

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Answer: \int\limits^a_E {\int\limits^a_E {\int\limits^a_E {xy} } \, dV = 1087.5

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An equation of a plane is found with a point and a normal vector. <u>Normal</u> <u>vector</u> is a perpendicular vector on the plane.

Given the points, determine the vectors:

P = (5,0,0); Q = (0,9,0); R = (0,0,4)

vector PQ = (5,0,0) - (0,9,0) = (5,-9,0)

vector QR = (0,9,0) - (0,0,4) = (0,9,-4)

Knowing that cross product of two vectors will be perpendicular to these vectors, you can use the cross product as normal vector:

n = PQ × QR = \left[\begin{array}{ccc}i&j&k\\5&-9&0\\0&9&-4\end{array}\right]\left[\begin{array}{ccc}i&j\\5&-9\\0&9\end{array}\right]

n = 36i + 0j + 45k - (0k + 0i - 20j)

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a(x-x_{0}) + b(y-y_{0}) + c(z-z_{0}) = 0

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36(x-5) + 20(y-0) + 45(z-0) = 0

The equation is: 36x + 20y + 45z - 180 = 0

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In terms of z:

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\int\limits^5_0 {\int\limits {\int\ {xyz}  \, dy } \, dx

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\frac{1}{45} \int\limits^5_0  {90xy^{2}-18x^{2}y^{2}-\frac{20}{3} xy^{3} } \, dx

\frac{1}{45} \int\limits^5_0  {2430x-1458x^{2}+\frac{94770}{125} x^{3}-\frac{23490}{375}x^{4}  } \, dx

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