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3 years ago

5

Mr.Jamison brought 15 tubs of paint for the art classes he teaches for his Elementary kids. He paid $63.90 for the paint.If each

tube of paint costs the same amount, what was the price of one tube?

1 answer:

xeze [42]3 years ago

6 0

Answer:

$4.26

Step-by-step explanation:

63.90 divide by 15 = $4.26

the price of one tube is $4.26

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Find the measure of each angle (help me please)

adelina 88 [10]

Answer:

<1=39 degrees

<2=141 degrees

Step-by-step explanation:

Detailed ans is in attachment

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3 years ago

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Please explain how to do this

11Alexandr11 [23.1K]

Answer:

A = 5t +7000

Step-by-step explanation:

Tim's rate of increasing altitude is 5 ft/min, so his change in altitude in t minutes is 5t.

If he started at 7000 ft, his altitude as a function of time is the sum of his initial altitude and his change in altitude:

A = 5t +7000

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4 years ago

A local grocery store is giving out

Marat540 [252]

Answer:

The original price range of soda is $10 to $39.4

Step-by-step explanation:

Given as :

The discount coupon applied on grocery = 5%

The cost of soda after discount coupon ranges = $0.50 to $1.97

Let the original price range of soda = $A

Now, According to question

discount on original price = the cost after discount

So, initial original price = x

5% of x = $0.50

0.05 × x = $0.50

∴ x = $\dfrac{0.50}{0.05}$

i.e x = $10

Again final original price = y

5% of y = $1.97

0.05 × y = $1.97

∴ y = $\dfrac{1.97}{0.05}$

i.e y = $39.4

So, The original price range of soda = $A = $10 to $39.4

Hence, The original price range of soda is $10 to $39.4 Answer

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4 years ago

What percentage of 130 liters is 6.5 liters?

DochEvi [55]

Answer:

2%

Correct Answer : 2%

Required Percentage = [ (130/(6.5 * 1000)) * 100]% = 2%.

Step-by-step explanation:

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3 years ago

Please help me to prove this!

Sophie [7]

Answer: see proof below

<u>Step-by-step explanation:</u>

Given: A + B = C → A = C - B

→ B = C - A

Use the Double Angle Identity: cos 2A = 2 cos² A - 1

→ (cos 2A + 1)/2 = cos² A

Use Sum to Product Identity: cos A + cos B = 2 cos [(A + B)/2] · 2 cos [(A - B)/2]

Use Even/Odd Identity: cos (-A) = cos (A)

<u>Proof LHS → RHS:</u>

LHS: cos² A + cos² B + cos² C

$\text{Double Angle:}\qquad \dfrac{\cos 2A+1}{2}+\dfrac{\cos 2B+1}{2}+\cos^2 C\\\\\\.\qquad \qquad \qquad =\dfrac{1}{2}\bigg(2+\cos 2A+\cos 2B\bigg)+\cos^2 C\\\\\\.\qquad \qquad \qquad =1+\dfrac{1}{2}\bigg(\cos 2A+\cos 2B\bigg)+\cos^2 C$

$\text{Sum to Product:}\quad 1+\dfrac{1}{2}\bigg[2\cos \bigg(\dfrac{2A+2B}{2}\bigg)\cdot \cos \bigg(\dfrac{2A-2B}{2}\bigg)\bigg]+\cos^2 C\\\\\\.\qquad \qquad \qquad =1+\cos (A+B)\cdot \cos (A-B)+\cos^2 C$

$\text{Given:}\qquad \qquad 1+\cos C\cdot \cos (A-B)+\cos^2C$

$\text{Factor:}\qquad \qquad 1+\cos C[\cos (A-B)+\cos C]$

$\text{Sum to Product:}\quad 1+\cos C\bigg[2\cos \bigg(\dfrac{A-B+C}{2}\bigg)\cdot \cos \bigg(\dfrac{A-B-C}{2}\bigg)\bigg]\\\\\\.\qquad \qquad \qquad =1+2\cos C\cdot \cos \bigg(\dfrac{A+(C-B)}{2}\bigg)\cdot \cos \bigg(\dfrac{-B-(C-A)}{2}\bigg)$

$\text{Given:}\qquad \qquad =1+2\cos C\cdot \cos \bigg(\dfrac{A+A}{2}\bigg)\cdot \cos \bigg(\dfrac{-B-B}{2}\bigg)\\\\\\.\qquad \qquad \qquad =1+2\cos C \cdot \cos A\cdot \cos (-B)$

$\text{Even/Odd:}\qquad \qquad 1+2\cos C \cdot \cos A\cdot \cos B\\\\\\.\qquad \qquad \qquad \quad =1+2\cos A \cdot \cos B\cdot \cos C$

LHS = RHS: 1 + 2 cos A · cos B · cos C = 1 + 2 cos A · cos B · cos C $\checkmark$

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3 years ago