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3 years ago

I need help can anyone help me

Mathematics

1 answer:

kolbaska11 [484]3 years ago

6 0

So for each day, Tessa ran 2.25 miles. So, multiply 2.25 miles by the number of days, which is 14, to get the total number of miles ran.

2.25 x 14 = 31.5 miles

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Kyilah bought 41.19 grams of chili and 19.415 grams of sugar. Did Kyilah buy more chili or sugar?

Sauron [17]

Answer:

Chili I think, I don't know for certain

Step-by-step explanation:

7 0

2 years ago

On Tuesday Lily spent 1/4 hour working on her science fair project.been work three times is mom on his science fair project is L

scZoUnD [109]

45 minutes because 1/4 of an hour is 15 min so 15+15+15=45

7 0

3 years ago

Write a direct or inverse variation equation for each relation.

igor_vitrenko [27]

Answer:

6 it is 6

Step-by-step explanation:

6 0

2 years ago

SOLUTION We observe that f '(x) = -1 / (1 + x2) and find the required series by integrating the power series for -1 / (1 + x2).

Ann [662]

Answer:

Required series is:

$\int{\frac{-1}{1+x^{2}} \, dx =-x+\frac{x^{3}}{3}-\frac{x^{5}}{5}+\frac{x^{7}}{7}+.....$

Step-by-step explanation:

Given that

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

We know that:

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

Comparing (1) and (2)

$f'(x)=-(tan^{-1}x)$ ---- (3)

Using power series expansion for $tan^{-1}x$

$f'(x)=-tan^{-1}x=-\int {\frac{1}{1+x^{2}} \, dx$

$= -\int{ \sum\limits^{ \infty}_{n=0} (-1)^{n}x^{2n}} \, dx$

$= -\sum{ \int\limits^{ \infty}_{n=0} (-1)^{n}x^{2n}} \, dx$

$=-[c+\sum\limits^{ \infty}_{n=0} (-1)^{n}\frac{x^{2n+1}}{2n+1}]$

$=C+\sum\limits^{ \infty}_{n=0} (-1)^{n+1}\frac{x^{2n+1}}{2n+1}$

$=C-x+\frac{x^{3}}{3}-\frac{x^{5}}{5}+\frac{x^{7}}{7}+.....$

$tan^{-1}(0)=0 \implies C=0$

Hence,

$\int{\frac{-1}{1+x^{2}} \, dx =-x+\frac{x^{3}}{3}-\frac{x^{5}}{5}+\frac{x^{7}}{7}+.....$

7 0

3 years ago

Find the error in the following sample of work, if one exists.

ch4aika [34]

Answer:

D. There is no mistake.

Step-by-step explanation:

The following lines show the process of factorization by using common factor.

Line 1:

In line 1, the equation is given and is completely fine.

$y^{2}+8xy+2y+16x=0$

The only thing missing was equate to zero, but the options below talk about correct factors only, therefore this can't be considered as a mistake and can be ignored completely.

Line 2:

In line 2, the terms are grouped, from which we can factor out common terms.

$(y^{2}+8xy)+(2y+16x)=0$

This is also fine.

Line 3:

In line 3, the common term y is taken out from group 1 and 2 from other group.

$y(y+8x)+2(y+8x)=0$

which is exactly what is given in line 3.

Line 4:

In line 4 the common factors can be seen and easily split into 2 factors.

$(y+2)(y+8x)=0$

which is exactly what is given in line 4.

Options:

A. The grouping is correct in line 2. So this option is does not hold.

B. Common factor was factored correctly from group 1. So this option does not hold.

C. Common factor was factored correctly from group 2. So this option does not hold.

D. There is no mistake. This is correct. Thus we choose this option as correct answer.

3 0

2 years ago