Find DEA in Circle R. *

Surface area of a rectangular prism if W = 6in L = 9in and H = 2in

Delicious77 [7]

Answer:

Step-by-step explanation:

Two faces are 6” by 9”. Two faces are 6” by 2”. Two faces are 9” by 2”.

Surface are = 2*6*9 + 2*6*2 + 2*9*2 = 108 + 24 + 36 = 168 square inches

4 0

3 years ago

Q

Tema [17]

Answer:

$y = \frac{1}{2} x+7$

Step-by-step explanation:

Slope-intercept form: y = mx + b

Slope formula: $\frac{y2-y1}{x2-x1}$

Given points: (-6, 4), (6, 10)

(-6, 4) = (x1, y1)

(6, 10) = (x2, y2)

To write the equation in slope-intercept form, we need to find the slope(m) and the y-intercept(b) of the equation.

First, let's find the slope. To do this, input the given points into the slope formula:

$\frac{10-4}{6-(-6)}$

Simplify:

10 - 4 = 6

6 - (-6) = 6 + 6 = 12

$\frac{6}{12}=\frac{1}{2}$

The slope is $\frac{1}{2}$ .

To find the y-intercept, input the slope and one of the given points(in this example I'll use point (6, 10)) into the equation and solve for b:

$10 = \frac{1}{2}(6)+b$

10 = 3 + b

7 = b

The y-intercept is 7.

Now that we know the slope and the y-intercept, we can write the equation:

$y = \frac{1}{2} x+7$

4 0

3 years ago

Find the arc length of the given curve between the specified points. x = y^4/16 + 1/2y^2 from (9/16), 1) to (9/8, 2).

lutik1710 [3]

Answer:

The arc length is $\dfrac{21}{16}$

Step-by-step explanation:

Given that,

The given curve between the specified points is

$x=\dfrac{y^4}{16}+\dfrac{1}{2y^2}$

The points from $(\dfrac{9}{16},1)$ to $(\dfrac{9}{8},2)$

We need to calculate the value of $\dfrac{dx}{dy}$

Using given equation

$x=\dfrac{y^4}{16}+\dfrac{1}{2y^2}$

On differentiating w.r.to y

$\dfrac{dx}{dy}=\dfrac{d}{dy}(\dfrac{y^2}{16}+\dfrac{1}{2y^2})$

$\dfrac{dx}{dy}=\dfrac{1}{16}\dfrac{d}{dy}(y^4)+\dfrac{1}{2}\dfrac{d}{dy}(y^{-2})$

$\dfrac{dx}{dy}=\dfrac{1}{16}(4y^{3})+\dfrac{1}{2}(-2y^{-3})$

$\dfrac{dx}{dy}=\dfrac{y^3}{4}-y^{-3}$

We need to calculate the arc length

Using formula of arc length

$L=\int_{a}^{b}{\sqrt{1+(\dfrac{dx}{dy})^2}dy}$

Put the value into the formula

$L=\int_{1}^{2}{\sqrt{1+(\dfrac{y^3}{4}-y^{-3})^2}dy}$

$L=\int_{1}^{2}{\sqrt{1+(\dfrac{y^3}{4})^2+(y^{-3})^2-2\times\dfrac{y^3}{4}\times y^{-3}}dy}$

$L=\int_{1}^{2}{\sqrt{1+(\dfrac{y^3}{4})^2+(y^{-3})^2-\dfrac{1}{2}}dy}$

$L=\int_{1}^{2}{\sqrt{(\dfrac{y^3}{4})^2+(y^{-3})^2+\dfrac{1}{2}}dy}$

$L=\int_{1}^{2}{\sqrt{(\dfrac{y^3}{4}+y^{-3})^2}dy}$

$L= \int_{1}^{2}{(\dfrac{y^3}{4}+y^{-3})dy}$

$L=(\dfrac{y^{3+1}}{4\times4}+\dfrac{y^{-3+1}}{-3+1})_{1}^{2}$

$L=(\dfrac{y^4}{16}+\dfrac{y^{-2}}{-2})_{1}^{2}$

Put the limits

$L=(\dfrac{2^4}{16}+\dfrac{2^{-2}}{-2}-\dfrac{1^4}{16}-\dfrac{(1)^{-2}}{-2})$

$L=\dfrac{21}{16}$

Hence, The arc length is $\dfrac{21}{16}$

6 0

3 years ago

Below is the frequency distribution of the weights (in kg) of some sample of teen-agers.

julia-pushkina [17]

Usually, the terms “data” and “information” are used interchangeably. However, there is a subtle difference between the two.

In a nutshell, data can be a number, symbol, character, word, codes, graphs, etc. On the other hand, information is data put into context. Information is utilised by humans in some significant way (such as to make decisions, forecasts etc).

A basic example of information would be a computer. A computer uses programming scripts, formulas, or software applications to turn data into information.

Let us have a detailed look at the difference between data and information in a tabular column below

3 0

3 years ago

PLEASE FINISH QUICKLYYY!!!!!!

swat32

1) Area = leg(1) * leg(2) * .5 = 15 * 36 * .5 = 270

Perimeter = leg(1) + leg(2) + hypotenuse = 15 + 36 + 39 = 90

2) Area = leg(1) * leg(2) = 20 * 80 = 1600

3) Median (divides it in 1/2)

4) Both (divides it in 1/2 and makes a 90 degree angle w/ base)

5) Altitude (makes 90 degree angle w/ base)

6)sqrt(30^2 + 16^2) = sqrt( 900 + 256) = sqrt(1156) = 34

7) sqrt(24^2 + 18^2) = sqrt( 576 + 324) = sqrt(900) = 30

8) sqrt(40^2 + 96^2) = sqrt(1600 + 9216) = sqrt(10816) = 104

9) sqrt(150^2 - 90^2) = sqrt(22500 - 8100) = sqrt(14400) = 120

10) sqrt(35^2 - 25^2) = sqrt(1225 - 625) = sqrt(600) = 10sqrt(6), approx. 24.5

7 0

3 years ago