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k0ka [10]
3 years ago
8

Gizela teaches 5 yoga classes that are each 3/4 hour. Use the drop down menu to complete the equation below to find the number o

f hours Gizela teaches yoga
5 classes _______ 3/4 hour per class = / hours ​
Mathematics
1 answer:
Assoli18 [71]3 years ago
4 0

Answer:

3.75 hours (3 hours and 45 minutes)

Step-by-step explanation:

5 times 3/4 = 3.75

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A circle has an area of 314.1592. What is the diameter? (round to the nearest whole number)
ad-work [718]

Answer:

20

Step-by-step explanation:

The area of a circle is equal to pi*radius^2

We can then write this as

314.1592=π*r^2

We then divide pi to get

100=r^2

take the square root and you find that the radius equals around 10

The diameter is always equal to the radius times two which means that the diameter is 20

hope this helps

8 0
3 years ago
Solve the equation<br> 0.2(d-6)=0.3d+5-3+0.1d for d
WINSTONCH [101]

Answer:

d= -16

Step-by-step explanation:

To answer this question you'll have to first simplify the equation to

0.2d+-1.2=0.3d+5-3+0.1d

then combining like terms you get

0.2d-1.2=(0.3d+0.1d)+(5+-3)

0.2d-1.2=0.4d+2

then subtract 0.4d from both sides leaving you with

-0.2d-1.2=2

then add 1.2 to both sides

-0.2d=3.2

after you divide both sides by -0.2

leaving you with the answer

D= -16

7 0
2 years ago
Read 2 more answers
Please Help Fast Algebra 1
Ostrovityanka [42]

Answer:

it's answer is C. 1

Step-by-step explanation:

Given

f(x) = -2x + 7

f(3) = - 2 * 3 + 7

= - 6 + 7

= 1

4 0
3 years ago
Read 2 more answers
Help me PLEASEEEEEEEEEEEEE
steposvetlana [31]

Answer: d

Step-by-step explanation:

its d

4 0
3 years ago
Can someone help me
marshall27 [118]

Answer:

16682.7 cm³

Step-by-step explanation:

The total volume is the volume of the cone plus half the volume of a sphere:

V = ⅓ π r² h + ½ (4/3 π r³)

V = ⅓ π r² h + ⅔ π r³

V = ⅓ π r² (h + 2r)

Given that r = 4.15 cm and h = 10.2 cm:

V = ⅓ π (4.15)² (10.2 + 2×4.15)

V ≈ 333.65

The volume needed for 50 cones is therefore:

50V ≈ 16682.7 cm³

7 0
3 years ago
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