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3 years ago

Lines a and b are parallel. Find the measure of each angle. Angle 8 is 113 degrees

Mathematics

1 answer:

pentagon [3]3 years ago

3 0

Answer:

angles 1 4 and 5 are all equal to 113 and angles 2 3 6 and 7 equal 67

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The figure contains of two quarter circles and a square what is the perimeter of this figure

sineoko [7]

Answer: 114.24 in

Step-by-step explanation:

To solve this problem you must apply the following proccedure:

1. You can see two quarter circles. You know that the radius of the circle is 16 inches. Then, the lenght a each quarter of of the circumference is:

$\frac{C}{4}=\frac{(3.14*16in*2)}{2}=25.12in$

3. Therefore, the perimeter is:

$P=25.12in+16in+16in+16in+16in\\P=114.24in$

The answer is the second option.

4 0

4 years ago

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I need help figuring this out. (Please)

zavuch27 [327]

You can try on photo math it works for me its free on app store

7 0

4 years ago

find a positive angle less than one revolution around the unit circle that is co-terminal with the given angle -15pi/17

natta225 [31]

ANSWER
$=\frac{19}{17} \pi$

EXPLANATION

To find a positive angle that is coterminal with
$- \frac{15}{17} \pi$

We add multiples of 2π until we get a positive angle that is less than one revolution,

We add to obtain,

$- \frac{15}{17} \pi + 2\pi$

This simplifies to,

$= \frac{19}{17} \pi$

This is the first positive angle that is coterminal with
$- \frac{15}{17} \pi$

and is less than one revolution.

7 0

4 years ago

A rectangular lamina of uniform density is situated with opposite corners at (0,0) and (15,4). calculate its radii of gyration a

Sphinxa [80]

First, you have to find the moment of inertia along the x and y axes. Constant density is denoted as k.

$I_{x}= \int\limits^15_0\int\limits^4_0 {k y^{2} } \, dx dy= \frac{1}{3}k (15-4)^4=4880.33k$

$I_{y}= \int\limits^15_0\int\limits^4_0 {k x^{2} } \, dx dy= \frac{1}{3}k (15-4)^4=4880.33k$

Then, the radii of gyration for

x = √[I_x/m]
y = [I_y/m]

where m = k(15-4)² = 121k. Then,

x = y = [4880.33k/121k] = 40.33

I hope I was able to help you. Have a good day.

7 0

3 years ago

Find the quadratic function that passes through the points: (-1,6), (1,4), and (2,9).

Masja [62]

Answer:

The quadratic function that passes through given points is y = 2 x² - x + 3 .

Step-by-step explanation:

The given quadratic function as

y = a x² + b x + c

The equation passes through the points ( - 1 , 6 ) , ( 1 , 4 ) and ( 2, 9 )

As The points passes through equation then

At points ( - 1 , 6 )

6 = a (1)² + b ×( - 1 ) + c

Or, a - b + c = 6 .....A

Again At points ( 1 , 4 )

4 = a (1)² + b × 1 + c

Or, a + b + c = 4 .......B

Similarly At points ( 2 , 9 )

9 = a (2)² + b × 2 + c

Or, 4 a +2 b + c = 9 ....,,,C

Now solving equation A and B

( a - b + c ) + ( a + b + c ) = 6 + 4

Or, a + c = $\frac{10}{2}$

I.e a + c = 5 ......D

Similarly Solving equation B and C

( 4 a +2 b + c ) - 2 × ( a + b + c ) = 9 - 2 × 4

Or, ( 4 a - 2 a + 2 b - 2 b + c - 2 c ) = 9 - 8

Or, ( 2 a - c ) = 1 .....E

Solving D and E

( a + c ) + ( 2 a - c ) = 5 + 1

Or, 3 a = 6

∴ a = $\frac{6}{3}$

I.e a = 2

Put the value of a in Eq D

So , a + c = 5

Or, c = 5 - a

∴ c = 5 - 2 = 3

I.e c = 3

Put The value of a and c in Eq A

a - b + c = 6

Or, b = a + c - 6

Or . b = 2 + 3 - 6

∴ , b = 5 - 6

I.e b = - 1

Now, Putting the values of a , b , c in the given quadratic equation

I.e y = a x² + b x + c

Or, y = 2 x² + ( - 1 ) x + 3

∴ The quadratic eq is y = 2 x² - x + 3

Hence The quadratic function that passes through given points is y = 2 x² - x + 3 . Answer

8 0

3 years ago