Using the graph, determine the coordinate of the y-intercept of the parabola.

Mathematics

1 answer:

12345 [234]2 years ago

7 0

-8,0 i believe not entirely sure I would wait to see if someone else can help you but just felt like your being timed

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Quadrilateral PQRS is a square whos side length is 10. Let X and Y be points outside the square so that XQ = YS = 6 and XP = YR

erik [133]

Answer:

392

Step-by-step explanation:

Triangles XQP and YRS are right triangles because triples 6, 8, 10 are Pythagorean triples.

Extend lines XQ, YR, YS and XP and mark their intersection as A and B.

Quadrilateral XAYB is a square because all right triangles PXQ, QAR, RYS and SBP are congruent (by ASA postulate) and therefore

all angles of the quadrilateral XAYB are right angles

all sides of XAYB are congruent and equal to 6 + 8 = 14 units.

Segment XY is the diagonal of the square XAYB, by Pythagorean theorem,

$XY^2=XA^2+AY^2\\ \\XY^2=14^2+14^2\\ \\XY^2=196+196\\ \\XY^2=392$

3 0

2 years ago

A contractor is required by a county planning department to submit one, two, three, four, or five forms (depending on the nature

Anuta_ua [19.1K]

Answer:

a) c = 1/15

b) =2/5

c) = 3/5

d) No, it cannot be the PMF of y.

Step-by-step explanation:

We have that

$p_X(x)\geq0$

4 0

2 years ago

Find the Maclaurin series for f(x) using the definition of a Maclaurin series. [Assume that f has a power series expansion. Do n

aliya0001 [1]

Answer:

$f(x)=\sum_{n=1}^{\infty}(-1)^{(n-1)}2^{n}\dfrac{x^n}{n}$

Step-by-step explanation:

The Maclaurin series of a function f(x) is the Taylor series of the function of the series around zero which is given by

$f(x)=f(0)+f^{\prime}(0)x+f^{\prime \prime}(0)\dfrac{x^2}{2!}+ ...+f^{(n)}(0)\dfrac{x^n}{n!}+...$

We first compute the n-th derivative of $f(x)=\ln(1+2x)$ , note that

$f^{\prime}(x)= 2 \cdot (1+2x)^{-1}\\f^{\prime \prime}(x)= 2^2\cdot (-1) \cdot (1+2x)^{-2}\\f^{\prime \prime}(x)= 2^3\cdot (-1)^2\cdot 2 \cdot (1+2x)^{-3}\\...\\\\f^{n}(x)= 2^n\cdot (-1)^{(n-1)}\cdot (n-1)! \cdot (1+2x)^{-n}\\$

Now, if we compute the n-th derivative at 0 we get

$f(0)=\ln(1+2\cdot 0)=\ln(1)=0\\\\f^{\prime}(0)=2 \cdot 1 =2\\\\f^{(2)}(0)=2^{2}\cdot(-1)\\\\f^{(3)}(0)=2^{3}\cdot (-1)^2\cdot 2\\\\...\\\\f^{(n)}(0)=2^n\cdot(-1)^{(n-1)}\cdot (n-1)!$

and so the Maclaurin series for f(x)=ln(1+2x) is given by

$f(x)=0+2x-2^2\dfrac{x^2}{2!}+2^3\cdot 2! \dfrac{x^3}{3!}+...+(-1)^{(n-1)}(n-1)!\cdot 2^n\dfrac{x^n}{n!}+...\\\\= 0 + 2x -2^2 \dfrac{x^2}{2!}+2^3\dfrac{x^3}{3!}+...+(-1)^{(n-1)}2^{n}\dfrac{x^n}{n}+...\\\\=\sum_{n=1}^{\infty}(-1)^{(n-1)}2^n\dfrac{x^n}{n}$