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4 years ago

I only need the answer for number 8 in this picture! please and thank you!!

Mathematics

2 answers:

Natali [406]4 years ago

8 0

Answer:

I cant see the full picture of the bottom triangle

Step-by-step explanation:

Leona [35]4 years ago

8 0

Answer:

Step-by-step explanation: you divide 11 by 5 because you know the length for the same side on both of those triangles therefore you divide the number for the larger figure by thesmaller one and you get you get your scale factor (the number you divide or multiply by to shrink ir stretch a figure to equal the size a similar figure). Now knowing what you scale factor is you can multiply 6 times 2.2 to get 13.2. Also you multiply 6 and 2.2 because on both triangles if you were to rotate them 6 and x are the same side of both triangles. I hope I helped.

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What is the following number in decimal form? 1.68 × 10^8

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3 years ago

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Solving Multi-Step Equations

KengaRu [80]

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$3.\\2z+9.75-7z=5.15\\-5z=5.15-9.75\\-5z=-4.6\\z=(-4.6):(-5)\\\boxed{z=0.92}\\------------------\\4.\\3h-5h+11=17\\-2h=17-11\\-2h=6\\h=6:(-2)\\\boxed{h=-3}\\------------------$

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$7.\\3c-8c+7=-18\\-5c=-18-7\\-5c=-25\\c=(-25):(-5)\\\boxed{c=5}\\------------------\\8.\\7g+14-5g=-8\\2g=-8-14\\2g=-22\\g=-22:2\\\boxed{g=-11}\\------------------$

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3 years ago

Find the poin on the graph 0f f(x)=1-x^2 that are closest to0(0,0)

zalisa [80]

Represent any point on the curve by (x, 1-x^2). The distance between (0, 0) and (x, 1-x^2) is

$\sqrt{(x-0)^2+(1-x^2-0)^2}=\sqrt{x^2+(1-x^2)^2}=\sqrt{x^2+1-2x^2+x^4}$

To make this easier, let's minimize the SQUARE of this quantity because when the square root is minimal, its square will be minimal.

So minimize $L=x^4-x^2+1$

Find the derivative of L and set it equal to zero.

$\frac{d}{dx}(L)=4x^3-2x \\ 4x^3-2x=0 \\ 2x(2x^2-1)=0$

This gives you $x=0$ or $x^2=\frac{1}{2} \\ x=\pm\sqrt{2}/2$

You can use the Second Derivative Test to figure out which value(s) produce the MINIMUM distance.

$\frac{d^2}{dx}=12x^2-2$

When x = 0, the second derivative is negative, indicating a relative maximum. When $x=\pm\frac{\sqrt{2}}{2}$ , the second derivative is positive, indicating a relative MINIMUM.

The two points on the curve closest to the origin are $\left( \pm\frac{\sqrt{2}}{2},\frac{1}{2} \right)$

7 0

4 years ago

2 |n + 10|= -2<br><br> solve the equation

MissTica

2n + 20 = -2

20 + 2 = -2n

22 = -2n

22/2 = n

11 = n

3 0

3 years 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

3 years ago