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zhuklara [117]
3 years ago
10

Autum school holding a voleturing challeng. Students are encouraged to volunteer for at least 1 1/2 hours per week. Volenteers a

bou the same amount each week. During a 3- week period she volunteers for 6 3/4 hours. Autumn wants to compare her volunteers rate with the challenge rate
Mathematics
2 answers:
Pani-rosa [81]3 years ago
8 0

Given :

Challenge given is to volunteer for about 1 1/2 hours per week .

Autum did at a rate of 6 3/4 hours per week .

To Find :

The comparison ( ratio )  her volunteers rate with the challenge rate .

Solution :

Simplifying challenge rate in in normal fraction , 1\dfrac{1}{2}=\dfrac{2+1}{2}=\dfrac{3}{2} .

Also , simplifying her rate , 6\dfrac{3}{4}=\dfrac{(6\times 4)+3}{4}=\dfrac{27}{4} .

Now , ratio of his work by challenge is :

R=\dfrac{\dfrac{27}{4}}{\dfrac{3}{2}}\\\\\\R=\dfrac{27\times 2}{4\times 3}\\\\R=\dfrac{9}{2}

Therefore , the ratio is 9/2 or 4 1/2 .

Hence , this is the required solution .

Anit [1.1K]3 years ago
6 0

Answer:

its 2 1/4

Step-by-step explanation:

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Lunna [17]

Answer:no

the answer is c

Step-by-step explanation:

3 0
2 years ago
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If line segment BH = 3, line segment BF = 10, line segment GH= 2, find HD 15 11 10.5 ?
pav-90 [236]

HD = 10.5

Step-by-step explanation:

Given BH = 3, GH = 2, BF = 10

Step 1: To find HF:

HF = BF – BH

HF = 10 – 3

HF = 7

Step 2: To find HD:

We know that if two chords intersects inside a circle, the product of the lengths of the segments of one chord equals the product of the lengths of the segments of the other chord.

⇒ GH × HD = BH × HF

⇒ 2 × HD = 3 × 7

⇒ HD = 10.5

Hence, the value of HD = 10.5.

5 0
3 years ago
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Does anyone know any 7th grade math? cus i really need help with this.
sladkih [1.3K]

Answer:

B

Step-by-step explanation:

All we have to do is plug in the values! 1/2(4)(10+5) will give us the area. Now we simplify!

2(15) = A

30=A

Our answer is B!

8 0
3 years ago
in a coordinate plane the midpoint of AB is (4,-3) and A has coordinate (1,5). if B has coordinate (x,y) find the value x+y​
Aloiza [94]

Answer:

Please check the explanation.

Step-by-step explanation:

The midpoint (a, b) of line joining points (x₁, y₁) and (x₂, y₂)

a = x₁ + x₂ / 2

b = y₁ + y₂ / 2

Given that the midpoint of AB is (4, -3).

i.e. (a, b) = (4, -3)

Given that A has coordinate (1, 5).

i.e. (x₁, y₁) = (1, 5)

We have to determine the coordinates of B.

i.e. (x₂, y₂) = B

Thus,

4 = (1 + x₂)/2

(1 + x₂) = 4 × 2

1 + x₂ = 8

x₂ = 7

and

-3 = (5 + y₂)/2

(5 + y₂) = -3 × 2

5 + y₂ = -6

y₂ = -11

so (x₂, y₂) = (7, -3) = B

Thus, the coordinates of B = (x₂, y₂)  = (7, -3)

Therefore,

x₂ + y₂ = 7 + (-3)

          = 7 - 3

          = 4

Hence, the value of x₂ + y₂ = 4

4 0
3 years ago
Find the area lying outside r=6cos(theta) and inside r=3+3cos(theta)
Sloan [31]
The area bounded by the 2 parabolas is A(θ) = 1/2∫(r₂²- r₁²).dθ between limits θ = a,b... 

<span>the limits are solution to 3cosθ = 1+cosθ the points of intersection of curves. </span>
<span>2cosθ = 1 => θ = ±π/3 </span>

<span>A(θ) = 1/2∫(r₂²- r₁²).dθ = 1/2∫(3cosθ)² - (1+cosθ)².dθ </span>
<span>= 1/2∫(3cosθ)².dθ - 1/2∫(1+cosθ)².dθ </span>
<span>= 9/8[2θ + sin(2θ)] - 1/8[6θ + 8sinθ +sin(2θ)] .. </span>
<span>.............where I have used ∫(cosθ)².dθ=1/4[2θ + sin(2θ)] </span>
<span>= 3θ/2 +sin(2θ) - sin(θ) </span>

<span>Area = A(π/3) - A(-π/3) </span>
<span>= 3π/6 + sin(2π/3) -sin(π/3) - (-3π/6) - sin(-2π/3) + sin(-π/3) </span>
<span>= π.</span>
7 0
3 years ago
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