All categories

3 years ago

What is the value of x in the parallelogram below?

Mathematics

1 answer:

Inessa [10]3 years ago

4 0

Answer:

Step-by-step explanation:

The diagonals of a parallelogram bisect each other. The value of x is the same as the length of the other half of the diagonal:

x = 11

You might be interested in

Fill in the blanks a ---÷48=-1

agasfer [191]

Ans is -48.

Hope this will help u..

4 0

3 years ago

Read 2 more answers

Simplify fully12xy24xy

Darina [25.2K]

Answer:

288xy

Step-by-step explanation:

First, you multiply 12 by 24 and you get 288. Next, you can't forget the variables, so you add those to the end and get, 288xy.

5 0

3 years ago

What is the value of x in -7+x=4? A. -11 B. -3 C. 3 D. 11 I need the answer

vichka [17]

Answer:

D.11

Step-by-step explanation:

D.11

7 0

4 years ago

Read 2 more answers

What is 9.187 to the nearest tenth? And hundredth and one

Assoli18 [71]

9.187 to the nearest tenth is 9.2

The 1 in 9.187 is in the tenths place and since the number after it is over 5, you round up.

4 0

4 years ago

Read 2 more answers

Find the equation of the tangent line to the curve when x has the given value.

Leviafan [203]

Answer:

14) The equation of the tangent line to the curve $f(x)=-2-x^2$ at x = -1 is $y=2x-1$

15) The rate of learning at the end of eight hours of instruction is $w'(8) = 5 \frac{items}{hour}$

Step-by-step explanation:

14) To find the equation of a tangent line to a curve at an indicated point you must:

1. Find the first derivative of f(x)

$f(x)=-2-x^2\\\\\frac{d}{dx} f(x)=\frac{d}{dx}(-2-x^2)\\\\\mathrm{Apply\:the\:Sum/Difference\:Rule}:\quad \left(f\pm g\right)'=f\:'\pm g'\\\\\frac{d}{dx} f(x)=\frac{d}{dx}\left(-2\right)-\frac{d}{dx}\left(x^2\right)\\\\f'(x)=-2x$

2. Plug x value of the indicated point, x = -1, into f '(x) to find the slope at x.

$f'(-1)=-2(-1)=2$

3. Plug x value into f(x) to find the y coordinate of the tangent point

$f(-1)=-2-(-1)^2=-3$

4. Combine the slope from step 2 and point from step 3 using the point-slope formula to find the equation for the tangent line

$y-y_1=m(x-x_1)\\y+3=2(x+1)\\y=2x-1$

5. Graph your function and the equation of the tangent line to check the results.

15) To find the rate of learning at the end of eight hours of instruction you must:

1. Find the first derivative of f(x)

$w(t)=15\sqrt[3]{t^2} \\\\\frac{d}{dt}w= \frac{d}{dt}(15\sqrt[3]{t^2})\\\\w'(t)=15\frac{d}{dt}\left(\sqrt[3]{t^2}\right)\\\\\mathrm{Apply\:the\:chain\:rule}:\quad \frac{df\left(u\right)}{dx}=\frac{df}{du}\cdot \frac{du}{dx}\\\\f=\sqrt[3]{u},\:\:u=\left(t^2\right)\\\\w'(t)=15\frac{d}{du}\left(\sqrt[3]{u}\right)\frac{d}{dt}\left(t^2\right)\\\\w'(t)=15\cdot \frac{1}{3u^{\frac{2}{3}}}\cdot \:2t\\\\\mathrm{Substitute\:back}\:u=\left(t^2\right)$

$w'(t)=15\cdot \frac{1}{3(t^2)^{\frac{2}{3}}}\cdot \:2t\\w'(t)=\frac{10t}{\left(t^2\right)^{\frac{2}{3}}}$

2. Evaluate the derivative a t = 8

$w'(t)=\frac{10t}{\left(t^2\right)^{\frac{2}{3}}}\\\\w'(8)=\frac{10\cdot 8}{\left(8^2\right)^{\frac{2}{3}}}\\\\=\frac{80}{\left(8^2\right)^{\frac{2}{3}}}\\\\\left(8^2\right)^{\frac{2}{3}}=16\\\\=\frac{80}{16}\\\\w'(8) = 5 \frac{items}{hour}$

The rate of learning at the end of eight hours of instruction is $w'(8) = 5 \frac{items}{hour}$

7 0

4 years ago