All categories

2 years ago

The following two parallelograms are drawn below. Find the measurement of FE or the variable “x”

Mathematics

1 answer:

labwork [276]2 years ago

4 0

Answer:

x=15 in.

Step-by-step explanation:

If the parallelograms are equal (I assume they are), then the ratios are the same. 12 is 3/4 of 16. 3/4 of 20 is 15.

You might be interested in

Which ordered pair can be removed so that the resulting graph represents a function?

Sunny_sXe [5.5K]

Answer:

(1, 3)

Step-by-step explanation:

6 0

2 years ago

Planes A & B are shown. If a new line, p, is drawn parallel to line l, which statement is true?

Alisiya [41]

The true statement is : Line p must be drawn so that it can lie on the same plane as line l

Step-by-step explanation:

From the diagram line l lies on plane A. When line p is drawn to be parallel to line l, it has to appear on plane A. This means that line p will be parallel to line l and perpendicular to line n that lies on plane B

Learn More

Line and plane :brainly.com/question/13099718

Keywords :plane, new line, parallel line, statement, true

#LearnwithBrainly

6 0

2 years ago

Adding which terms to 3x2y would result in a monomial? Check all that apply.

kkurt [141]

the answers are -12x^2 and 4x^2y. hopefully it helps and i just got it right.

8 0

2 years ago

Read 2 more answers

Find the distance between the points

Maru [420]

Answer:

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