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Komok [63]
2 years ago
14

PLEASE HELP AND KNOW YOU ARE CORRECT BEFORE ANSWERING PLEASE AND THANK YOU.

Mathematics
1 answer:
swat322 years ago
8 0

Answer:

Both are false

Step-by-step explanation:

B is 3 units below P

After dilation B' will be 5 units below P so cannot be on the same horizontal line.

similarly segment CD is 3 units long,

After dilation segment C'D' will be 5 units long so is not equal to segment CD.

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Plsss help will mark brainliest
DiKsa [7]
C, because 7 divided, or over nine, is 7/9
Hope this helps!
3 0
3 years ago
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28/36 reduced fraction
ratelena [41]
28/4 = 7,
36/4 = 9,
Place the 7 over the 9 and you get 7/9!
6 0
3 years ago
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Can someone actually answer this question pls its my 3rd time posting it and ill ive gotten was wrong or answers that make no se
andre [41]

Answer:

(-2+-5) equals -7.

Because the signs are the same, we dont change it.

(3+5) equals 8. Same like the other equation, the signs are the same so we do not change it.

Now we do (-7+8) The answer will be 1. Lets use a little graph since its hard to explain.

-8 -7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8

Mark -7 anyway you want. (Circle it, draw a line under it, what ever you want.)

Now Move the line 8 times.

When you do that your line will stop at 1.

And thats how you do it.

Step-by-step explanation:

7 0
2 years ago
Read 2 more answers
For which equation would x = 8 not be a solution?
Olegator [25]

Answer:

b. 4x = 48

Step-by-step explanation:

If you look closely an multiply 4 x 8, it equals 32, not 48

The other equations:

8 x 8 does equal 64

10 x 8 does equal 80

9 x 8 does equal 72

It's pretty simple once you try each problem first

Hope this helps!

5 0
3 years ago
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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
3 years ago
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