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3 years ago

Find the measure of

Mathematics

1 answer:

MatroZZZ [7]3 years ago

8 0

Angles D and G are 80 degrees

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If a wedding ring is dropped from a tall building, the distance it has fallen in feet is given by 16t2, where t is the time in s

Aneli [31]

It would be 40 seconds.

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3 years ago

(+41)x(+16)<br> pleasderr

larisa [96]

Answer:

$656$

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3 years ago

Write an equation in slope-intercept form that goes through (12, 4) and (20,8).

spayn [35]

Answer:

Equation in slope-intercept form that goes through (12, 4) and (20,8) is: $y = \frac{1}{2}x-2$

Step-by-step explanation:

Given two points are:

$(x_1,y_1) = (12,4)\\(x_2,y_2) = (20,8)$

Slope intercept form of line is given as:

$y = mx+b$

Here m is the slope of the line and b is the y-intercept.

Slope of a line is calculated by the formula:

$m = \frac{y_2-y_1}{x_2-x_1}$

Putting the values

$m = \frac{8-4}{20-12}\\m = \frac{4}{8}\\m=\frac{1}{2}$

Putting the value of slope in slope-intercept form we get

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

To find the value of b, any one point will be put in the equation

Putting the first point (12,4) in the equation

$4 = \frac{1}{2}(12) + b\\4 = 6+b\\b = 4-6\\b = -2$

Putting the value of b

$y = \frac{1}{2}x-2$

Hence,

Equation in slope-intercept form that goes through (12, 4) and (20,8) is: $y = \frac{1}{2}x-2$

4 0

3 years ago

119 kilometers per hour equals how many miles per hour

kakasveta [241]

You divide the number of kilometers by 0.621371 to get the number of miles per hour. I hope this helped!

8 0

3 years ago

A circular swimming pool has a diameter of 40 meters. The depth of the pool is constant along west-east lines and increases line

Nataliya [291]

Answer:

Volume is $2000\pi\ m^{3}$

Solution:

As per the question:

Diameter, d = 40 m

Radius, r = 20 m

Now,

From north to south, we consider this vertical distance as 'y' and height, h varies linearly as a function of y:

iff

h(y) = cy + d

Then

when y = 1 m

h(- 20) = 1 m

1 = c.(- 20) + d = - 20c + d (1)

when y = 9 m

h(20) = 9 m

9 = c.20 + d = 20c + d (2)

Adding eqn (1) and (2)

d = 5 m

Using d = 5 in eqn (2), we get:

$c = \frac{1}{5}$

Therefore,

$h(y) = \frac{1}{5}y + 5$

Now, the Volume of the pool is given by:

$V = \int h(y)dA$

where

A = $r\theta$

$A = rdr\ d\theta$

Thus

$V = \int (\frac{1}{5}y + 5)dA$

$V = \int_{0}^{2\pi}\int_{0}^{20} (\frac{1}{5}rsin\theta + 5) rdr\ d\theta$

$V = \int_{0}^{2\pi}\int_{0}^{20} (\frac{1}{5}r^{2}sin\theta + 5r}) dr\ d\theta$

$V = \int_{0}^{2\pi} (\frac{1}{15}20^{3}sin\theta + 1000) d\theta$

$V = [- 533.33cos\theta + 1000\theta]_{0}^{2\pi}$

$V = 0 + 2\pi \times 1000 = 2000\pi\ m^{3}$

7 0

3 years ago