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Dominik [7]
3 years ago
8

How do I solve this 6-3=□-2

Mathematics
1 answer:
Sonja [21]3 years ago
8 0
Well, first what is 6-3=3 so now whats 3-2? 3-2=1 
6-3=3-2
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The weight of water is 62 1/2 lb per cubic foot water that weighs 300 lb will fill how many cubic feet
Grace [21]
4.8 cubic feet.    300/62.5=4.8
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3 years ago
Given the formula A =pieab solve for a.
Romashka [77]

Answer:

a = A/πb

Step-by-step explanation:

To solve this subject of the formulae given that A = πab.

<u>solution</u>

A = πab

the next step is to look for a unique way to get rid of the variables disturbing "a" from standing alone. this variables are π and b, we need to detach dem from a

A = πab

divide both sides by πb

A/πb = πab/πb

A/πb = a

a = A/πb

therefore  the value of a in the fomular A = πab is evaluated to be  a = A/πb

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3 years ago
Use the Two-Transversal Proportionality Corollary to find the value of x.
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I hope this helps you




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5 0
3 years ago
Read 2 more answers
9. What is the distance along the x axis from order pairs -8, 6 and 4,6
Blababa [14]

Answer:

12 units

Step-by-step explanation:

got it right on edg

5 0
3 years ago
Measure the lengths of the sides of ∆ABC in GeoGebra, and compute the sine and the cosine of ∠A and ∠B. Verify your calculations
marusya05 [52]

Answer:

Sin \angle A =0.80

Cos \angle A=0.60

Sin \angle B =0.60

Cos \angle B=0.80

Step-by-step explanation:

Given

I will answer this question using the attached triangle

Solving (a): Sine and Cosine A

In trigonometry:

Sin \theta =\frac{Opposite}{Hypotenuse} and

Cos \theta =\frac{Adjacent}{Hypotenuse}

So:

Sin \angle A =\frac{BC}{BA}

Substitute values for BC and BA

Sin \angle A =\frac{8cm}{10cm}

Sin \angle A =\frac{8}{10}

Sin \angle A =0.80

Cos \angle A=\frac{AC}{BA}

Substitute values for AC and BA

Cos \angle A=\frac{6cm}{10cm}

Cos \angle A=\frac{6}{10}

Cos \angle A=0.60

Solving (b): Sine and Cosine B

In trigonometry:

Sin \theta =\frac{Opposite}{Hypotenuse} and

Cos \theta =\frac{Adjacent}{Hypotenuse}

So:

Sin \angle B =\frac{AC}{BA}

Substitute values for AC and BA

Sin \angle B =\frac{6cm}{10cm}

Sin \angle B =\frac{6}{10}

Sin \angle B =0.60

Cos \angle B=\frac{BC}{BA}

Substitute values for BC and BA

Cos \angle B=\frac{8cm}{10cm}

Cos \angle B=\frac{8}{10}

Cos \angle B=0.80

Using a calculator:

A = 53^{\circ}

So:

Sin(53^{\circ}) =0.7986

Sin(53^{\circ}) =0.80 -- approximated

Cos(53^{\circ}) = 0.6018

Cos(53^{\circ}) = 0.60 -- approximated

B = 37^{\circ}

So:

Sin(37^{\circ}) = 0.6018

Sin(37^{\circ}) = 0.60 --- approximated

Cos(37^{\circ}) = 0.7986

Cos(37^{\circ}) = 0.80 --- approximated

8 0
3 years ago
Read 2 more answers
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