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dedylja [7]
3 years ago
7

I need help with this one question please hope me! It is an inequality on both sides (solve and graph): -4(2-5)>(-3)2 [as in

the exponent] -h
Mathematics
1 answer:
laila [671]3 years ago
3 0
-4(2-5)>(-3)^2-h
-4(-3)>9-h
12>9-h
add h to both sides
h+12>9
minus 12 from both sides
h>-3
not sure if h is x or y or vertical axis or horizontal axis

anyway, what you would do is draw a dotted line at h=-3, then shade to the positive side
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What is the distance between the points (-6.5) and (8,-2) on a coordinate plane?
Kaylis [27]

Answer:

5√7 or 15.652

Step-by-step explanation:

is the numbers (-6,5) (8-2) or is it (-6.5) because of it's -6.5 i do not understand how to get that answer (-6,5) the answer is 5√7

6 0
2 years ago
Find T5(x) : Taylor polynomial of degree 5 of the function f(x)=cos(x) at a=0 . (You need to enter function.) T5(x)= Find all va
Burka [1]

Answer:

\bf cos(x)\approx1-\displaystyle\frac{x^2}{2}+\displaystyle\frac{x^4}{4!}=\\\\=1-\displaystyle\frac{x^2}{2}+\displaystyle\frac{x^4}{24}

The polynomial is an approximation with an error less than or equals to <em>0.002652</em> for x in the interval

[-1.113826815, 1.113826815]

Step-by-step explanation:

According to Taylor's theorem

\bf f(x)=f(0)+f'(0)x+f''(0)\displaystyle\frac{x^2}{2}+f^{(3)}(0)\displaystyle\frac{x^3}{3!}+f^{(4)}(0)\displaystyle\frac{x^4}{4!}+f^{(5)}(0)\displaystyle\frac{x^5}{5!}+R_6(x)

with

\bf R_6(x)=f^{(6)}(c)\displaystyle\frac{x^6}{6!}

for some c in the interval (-x, x)

In the particular case f

<em>f(x)=cos(x) </em>

<em> </em>

we have

\bf f'(x)=-sin(x)\\f''(x)=-cos(x)\\f^{(3)}(x)=sin(x)\\f^{(4)}(x)=cos(x)\\f^{(5)}(x)=-sin(x)\\f^{(6)}(x)=-cos(x)

therefore

\bf f'(x)=-sin(0)=0\\f''(0)=-cos(0)=-1\\f^{(3)}(0)=sin(0)=0\\f^{(4)}(0)=cos(0)=1\\f^{(5)}(0)=-sin(0)=0

and the polynomial approximation of T5(x) of cos(x) would be

\bf cos(x)\approx1-\displaystyle\frac{x^2}{2}+\displaystyle\frac{x^4}{4!}=\\\\=1-\displaystyle\frac{x^2}{2}+\displaystyle\frac{x^4}{24}

In order to find all the values of x for which this approximation is within 0.002652 of the right answer, we notice that

\bf R_6(x)=-cos(c)\displaystyle\frac{x^6}{6!}

for some c in (-x,x). So

\bf |R_6(x)|\leq|\displaystyle\frac{x^6}{6!}|=\displaystyle\frac{|x|^6}{6!}

and we must find the values of x for which

\bf \displaystyle\frac{|x|^6}{6!}\leq0.002652

Working this inequality out, we find

\bf \displaystyle\frac{|x|^6}{6!}\leq0.002652\Rightarrow |x|^6\leq1.90944\Rightarrow\\\\\Rightarrow |x|\leq\sqrt[6]{1.90944}\Rightarrow |x|\leq1.113826815

Therefore the polynomial is an approximation with an error less than or equals to 0.002652 for x in the interval

[-1.113826815, 1.113826815]

8 0
3 years ago
PLS ANSWER ASAP <br> simplify (-8^3)^2
Tanya [424]

Answer:

(-8^3)^2

=-8^2*3

=-8^6

=1/8^6

=1/262144

Step-by-step explanation:

7 0
3 years ago
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alexgriva [62]

Answer:

47,100.00

Step-by-step explanation:

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3 years ago
Please solve -2 = -x + 4 AND PLEASE PROVIDE WORK I WANNA KNOW HOW
frosja888 [35]

Answer:

Step-by-step explanation: x = 6

Subtract 4 from -2 to get -6. Then divide -6 by -x to get x = 6

8 0
3 years ago
Read 2 more answers
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