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3 years ago

8

Which table of ordered pairs, when plotted, will form a straight line? Select two answers.

1 answer:

lukranit [14]3 years ago

6 0

Answer:

B

Step-by-step explanation:

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15 points. Need help with this math question

kodGreya [7K]

Answer:

0.0098 < 0.01935 < 0.02 < 0.14999 < 0.1589

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3 years ago

Read 2 more answers

What is the unknown fraction?

vaieri [72.5K]

Answer:

38/100

Step-by-step explanation:

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3 years ago

A donut machine makes small glaze donuts the diameter of each doughnut is 8 cm. What is the circumference of each doughnut creat

kotegsom [21]

Answer:

50.29 cm

Step-by-step explanation:

Given:

Diameter of each doughnut = 8 cm

To find: circumference of each doughnut

Solution:

Circumference refers to the boundary of a geometric figure.

A doughnut has the shape of a circle.

Let r represents the radius of the circle.

Diameter of each doughnut = 8 cm

Radius of each doughnut (r) = Diameter/2

$=\frac{8}{2}\\ =4 \,\,cm$

Circumference of each doughnut $=\pi r^2\\$

Put $r=4\,\,,\pi=\frac{22}{7}$

So, circumference = $=\frac{22}{7} (4)^2$

$=\frac{352}{7} \,\,cm\\=50.29\,\,cm$

4 0

3 years ago

Is (-3,-5) a solution of the graphed inequality?<br> Choose 1 answer,<br> Yes<br> B<br> No

ArbitrLikvidat [17]

There’s not graph so if you could provide a graph

7 0

3 years ago

Find the taylor polynomial t3(x) for the function f centered at the number

inysia [295]

Answer:

$t_3(x)=\frac{7\pi}{4}+\frac{7}{2}(x-1)-\frac{7}{4}(x-1)^2+\frac{7}{12}(x-1)^3$

Step-by-step explanation:

We are given that

$f(x)=7tan^{-1}(x)$

a=1

$T_n(x)=\sum_{r=0}^{n}\frac{f^r(a)(x-a)^r}{r!}$

Substitute n=3 and a=1

$t_3(x)=f(1)+f'(1)(x-1)+\frac{f''(1)(x-1)^2}{2!}+\frac{f'''(1)(x-1)^3}{3!}$

$f(x)=7tan^{-1}(x)$

$f(1)=7tan^{-1}(1)=7\times \frac{\pi}{4}=\frac{7\pi}{4}$

Where $tan^{-1}(1)=\frac{\pi}{4}$

$f'(x)=\frac{7}{1+x^2}$

Using the formula

$\frac{d(tan^{-1}(x))}{dx}=\frac{1}{1+x^2}$

$f'(1)=\frac{7}{2}$

$f''(x)=\frac{-14x}{(1+x^2)^2}$

$f''(1)=-\frac{7}{2}$

$f''(x)=-14x(x^2+1)^{-2}$

$f'''(x)=-14((x^2+1)^{-2}-4x^2(x^2+1)^{-3}})$

By using the formula

$(uv)'=u'v+v'u$

$f'''(x)=-14(\frac{x^2+1-4x^2}{(1+x^2)^3}$

$f'''(x)=(-14)\frac{-3x^2+1}{(1+x^2)^3}$

$f'''(1)=-14(\frac{-3(1)+1}{2^3})=\frac{7}{2}$

Substitute the values

$t_3(x)=\frac{7\pi}{4}+\frac{7}{2}(x-1)-\frac{7}{4}(x-1)^2+\frac{7}{2\times 3\times 2\times 1}(x-1)^3$

$t_3(x)=\frac{7\pi}{4}+\frac{7}{2}(x-1)-\frac{7}{4}(x-1)^2+\frac{7}{12}(x-1)^3$

7 0

3 years ago