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3 years ago

Suppose a parabola has vertex (5,-3) and also passes through the point (6,1). Write the equation of the parabola in vertex form.

Mathematics

1 answer:

Alla [95]3 years ago

4 0

Look it up on desmos, it’s an online graphing calculator, you’re welcome!

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The Acme Company manufactures widgets. The distribution of widget weights is bell-shaped. The widget weights have a mean of 60 o

Gala2k [10]

Answer:

Step-by-step explanation:

Given that the Acme Company manufactures widgets, which have a mean of 60 ounces and a standard deviation of 7 ounces

We know that 95% of the area lie between -2 and 2 std deviations from the mean.

i.e. Probability for lying in the middle of 95%

Z score $(60+7(-2), 60+7(2))\\=(46, 74)$

Between 46 and 74 oz.

b) Between 12 and 57

convert into Z score

$(\frac{12-60}{7}$

P(-6.86<z<-0.43)

=0.5-0.1664=0.3336

c) X<30 gives Z<-4.83

i.e. P(X<30) =0.00

3 0

2 years ago

What does he mean on question one

Katarina [22]

You plug -2 in for the function k(p) and add it to the function g(w), getting
(-2+3)*(-2-7)+(-2-5)^2=1*-9+49=40 for a - I challenge you to do B on your own!

6 0

2 years ago

Choose the equation of the horizontal line that passes through the point (−2, −1). (4 points)

Anni [7]

Answer:

y=-1 : )

Step-by-step explanation:

(Brainliest Please)

8 0

1 year ago

Please helpppppppppp<br> include step by step

balandron [24]

Answer:

507

Step-by-step explanation:

As they said, I will be using 3 as pi.

$3$ (pi) $* (\frac{26}{2})^2$ (radius squared)

Without annotations: $3*\frac{26}{2}^2$

$3*\frac{26}{2}^2$

$3*13^2$ ; $13*13=169$

$3*169$

$507$

5 0

1 year ago

Use the Taylor series you just found for sinc(x) to find the Taylor series for f(x) = (integral from 0 to x) of sinc(t)dt based

Marina CMI [18]

In this question (brainly.com/question/12792658) I derived the Taylor series for $\mathrm{sinc}\,x$ about $x=0$ :

$\mathrm{sinc}\,x=\displaystyle\sum_{n=0}^\infty\frac{(-1)^nx^{2n}}{(2n+1)!}$

Then the Taylor series for

$f(x)=\displaystyle\int_0^x\mathrm{sinc}\,t\,\mathrm dt$

is obtained by integrating the series above:

$f(x)=\displaystyle\int\sum_{n=0}^\infty\frac{(-1)^nx^{2n}}{(2n+1)!}\,\mathrm dx=C+\sum_{n=0}^\infty\frac{(-1)^nx^{2n+1}}{(2n+1)^2(2n)!}$

We have $f(0)=0$ , so $C=0$ and so

$f(x)=\displaystyle\sum_{n=0}^\infty\frac{(-1)^nx^{2n+1}}{(2n+1)^2(2n)!}$

which converges by the ratio test if the following limit is less than 1:

$\displaystyle\lim_{n\to\infty}\left|\frac{\frac{(-1)^{n+1}x^{2n+3}}{(2n+3)^2(2n+2)!}}{\frac{(-1)^nx^{2n+1}}{(2n+1)^2(2n)!}}\right|=|x^2|\lim_{n\to\infty}\frac{(2n+1)^2(2n)!}{(2n+3)^2(2n+2)!}$

Like in the linked problem, the limit is 0 so the series for $f(x)$ converges everywhere.

7 0

2 years ago