Find the area of the shaded sector. Round your

SOVA2 [1]

Hi!
To find this math problem your going to need to use a^2 + b^2 = c^2 (^2 means squared)

So if we look at the shape , they entire shape makes for triangles . (I'm going to be using the upper right triangle to solve this problem just so your following me)
So one side of the "triangle" is 6 and that's like the bottom of the triangle . And your hypotenuse is 10.

So using that formula , plug in 6 and 10 so we can find the other missing side
a^2 + 6^2 = 10^2
A^2 + 36 = 100
100-36 = 64
SQAURE ROOT of 64 = 8

Your missing side length is 8
8 + 8 = 16

DB= 16. ( you have to add them together to get the full length)

I hope this helps you :)

3 0

4 years ago

Solve for y 8x + y = 32

oksian1 [2.3K]

Answer:

Subtract 8x8x from both sides of the equation.

y=32−8x

8 0

3 years ago

Wilton collected $8.50 in dimes and quarters while cleaning his father's truck. If the number of quarters is 3 times the number

dusya [7]

Answer:

Wilton found 10 dimes while cleaning his father's truck.

Step-by-step explanation:

Given that:

Total amount collected = $8.50 = 8.50*100 = 850 cents

Let,

x be the number of dimes

y be the number of quarters

According to given statement,

10x+25y=850 Eqn 1

y = 3x Eqn 2

Putting y=3x in Eqn 1

$10x+25(3x)=850\\10x+75x=850\\85x=850$

Dividing both sides by 85

$\frac{85x}{85}=\frac{850}{85}\\x=10$

As x represents dimes, therefore, number of dimes is 10.

Hence,

Wilton found 10 dimes while cleaning his father's truck.

8 0

3 years ago

Hunter and his workout partner arelifting weights together, doing manysets of each exercise. On a certainexercise, Hunter is usi

Soloha48 [4]

9514 1404 393

Answer:

1 set completed
21 kg

Step-by-step explanation:

The difference in mass of the bars is 19-15 = 4 kg. The difference in mass of the added weight is 6-2 = 4 kg. The difference in total mass (bar +added) is reduced by the difference in added mass after each set. The number of sets required to reduce the difference to zero is ...

(4 kg)/(4 kg/set) = 1 set

After 1 set, Hunter will increase the mass to 19 kg + 2 kg = 21 kg.

After 1 set, his partner will increase the mass to 15 kg +6 kg = 21 kg.

Both will be lifting 21 kg for their 2nd set, after they have completed 1 set.

_____

Additional comment

You can write equations for the mass being lifted by Hunter (h) and his partner (p) after s sets:

h = 19 +2s

p = 15 +6s

The masses will be equal when ...

h = p

19 +2s = 15 +6s

19 = 15 +4s . . . . . . . subtract 2s

4 = 4s . . . . . . . . . . . subtract 15

Note that the number on the left side of the equation is 19-15, the difference of bar masses. The coefficient on the right side is 6-2, the difference in the added masses. To find the number of sets (s), we divide the equation by the coefficient of s:

4/4 = s = 1

In the above solution we skipped directly to this final division in order to find the number of sets until the lifted masses were equal.

5 0

3 years ago

Read 2 more answers

35 – 10 ÷ 5 + [(5 + 3) • 4] what is the answer to this evaluated

alexandr1967 [171]

35−10/5+(5+3)(4)

=35−2+(5+3)(4)

=33+(5+3)(4)

=33+(8)(4)

=33+32

=65

8 0

3 years ago

Read 2 more answers