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2 years ago

6

HELP ASAP ASAP!! NO LINKS PLEASEEEE OR U WILL BE REPORTED, I NEED AN ANSWER WITH EXPLANATION AND WORK SHOWN OR EXPLAINED, WILL G

IVE BRAINIEST!!

1 answer:

kari74 [83]2 years ago

8 0

Answer:

Method 1 of 7: Squares, Rectangles, and Parallelograms. Measure the width and height.

Step-by-step explanation:

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Step-by-step explanation:

Im not sure how to be honest.

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2 years ago

you 494 miles from home and you are driving forward home at an average of 45 mhp. How long will it take to drive home if you mai

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3 years ago

Help please, and explain(step-by-step) how the answer would be correct.

vekshin1

When a function is reflected, it must be reflected over a line

The new function is: $y = |4 -x| -8$

The equation is given as:

$y = |x + 4| - 8$

The rule of reflection over the y-axis is:

$(x,y) \to (-x,y)$

So, we have:

$y = |-x + 4| -8$

Rewrite as:

$y = |4 -x| -8$

Hence, the new function is:

$y = |4 -x| -8$

Read more about reflections at:

brainly.com/question/938117

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2 years ago

On the coloring sheet, thd a ratio<br> is equivalent to 3/10.

DerKrebs [107]

Answer:

Equivalent ratios:

3 : 10 6 : 20 9 : 30 12 : 40 15 : 50 18 : 60 21 : 70 24 : 80 27 : 90 30 : 100 33 : 110 36 : 120 39 : 130 42 : 140 45 : 150 48 : 160 51 : 170 54 : 180 57 : 190 60 : 200 63 : 210 66 : 220 69 : 230 72 : 240 75 : 250 78 : 260 81 : 270 84 : 280 87 : 290 90 : 300

<u>( Brainlyst will help my rank <3 )</u>

8 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

2 years ago