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3 years ago

Suppose y varies directly with x. Write a direct variation equation that relates x and y.

Mathematics

1 answer:

timama [110]3 years ago

8 0

$\huge\mathfrak\purple{Answer:-}$

★ Option (A) y = -5x is the right answer.

<h3>I hope this helps! :)</h3>

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For an art project, Kelvin cut a 4-foot piece of yarn into pieces. He then measured the length of each piece of yarn to the near

oksian1 [2.3K]

Answer:

I think answers is

label the x-axis "Pieces of Yarn," and label the y-axis "Length (inches

3 0

3 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 0.002652 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

f(x)=cos(x)

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

Need math help. Thanks in advance.

Mademuasel [1]

9. D because it fits the description when it is graphed.
11. D because it has the correct vertex, which is (1.5, 4.75).
12. f(x)=-4(x-2)²+16 works with the table and provided graph!
13. 0 isn't a zero for f(x), so B is correct (the zeros are 1/2 and 4).
14. The width is about 3.831.
15. The answer is B because the equation for a parabola in this form is:
f(x) = (x - r₁)(x - r₂)

Every one of these except for 15 was solved using desmos, a graphing calculator.

Hope this helps!

7 0

3 years ago

A metalsmith is mixing two molten metals, each containing different percentages of silver. The table shows the amount of each mo

Bess [88]

Answer:

(15)(0.75) and 0.7(15 – x) + 0.9x

Step-by-step explanation:

I did that assignment.

4 0

3 years ago

Read 2 more answers

Round to the nearest tenth if neccessary 69.904 divided by 34