All categories

3 years ago

I need help with this question...

Mathematics

1 answer:

romanna [79]3 years ago

8 0

Answer:

57 inches

Step-by-step explanation:

It rained 5*6.5+7*3.5=32.5+34.5=57 inches in the past 12 days

You might be interested in

6. Name the polygon. Write whether it is regular<br> or not regular

slamgirl [31]

Answer:

this shape is a pentagon. this is a regular shape because all the sides are even and all the angles on the inside are also even.

Step-by-step explanation:

pls mark brainliest!

7 0

3 years ago

Find two points on the line to be graphed 2y+16=0

yarga [219]

Y=-8 so any x with the coordinate -8 for y i guess.

4 0

3 years ago

One number is 5 times a first number. A third number is 100 more than the first number. If the sum of the three numbers is 779,

AlladinOne [14]

Answer:

Step-by-step explanation:

x - the first number

One number is 5 times a first number:

5x - "one" number (the second number)

A third number is 100 more than the first number:

x + 100 - the third number

x + 5x + x+100 - the sum of the three numbers

x + 5x + x+100 = 779

-100 -100

7x = 679

÷7 ÷7

x = 97

5x = 5•97 = 485

x+100 = 97 + 100 = 197

Check:

97 + 485 + 197 = 779

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

Simply 3/4 - (6 - 3/4s)

Viktor [21]

Answer:

-21/4 + 3s/4

Step-by-step explanation:

6 0

3 years ago

Read 2 more answers