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2 years ago

Evaluate 4(a2 + 2b) - 2b

Mathematics

1 answer:

iogann1982 [59]2 years ago

5 0

<span>4(a^2 + 2b) - 2b
= 4a^2 + 8b - 2b
=</span> 4a^2 + 6b

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If 2 tacos and 5 drinks cost $20, And three tacos and five drinks cost $25 how much does a taco cost

san4es73 [151]

A taco costs $5
This is because there was no difference in the cost except for $5. The only item added was one taco. Drinks cost $2 each.

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2 years ago

1) A 1.5 kg apple pie is sitting on the kitchen counter . If it has 13.44 J of gravitational potential energy, how tall is the c

vampirchik [111]

Answer: 0.91 m

Explanation:

We know that,

P.E. = m g h

Where,

P.E = Potential energy

m = Mass of the object

g = acceleration due to gravity (9.8 m/s²)

It is given that, m = 1.5 kg

P.E. = 13.44 J

⇒ 13.44 = 1.5 kg × 9.8 m/s² × h

⇒ h = 0.91 m

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Explanation:

P.E. = m g h

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height, h = 1.8 m ( given)

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Hence, the energy possessed by the suitcase sitting on the counter is 216 J.

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2 years ago

Niveen makes muffins and sells them for $0.65 each. To make 12 muffins, she must spend $5.25 on muffin mix, $0.75 on vegetable o

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Answer:

selling 36 muffins at a discount rate of 10% off the original price

Step-by-step explanation:

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1 year ago

The German autobahns are the nationally coordinated motorway system in Germany and have no general speed limit, but the advisory

melamori03 [73]

Answer:

The required equation is: $s=0.0508v^2$

The stopping distance of your car if you were moving at 95 mph is 458.47 feet.

Step-by-step explanation:

Consider the provided information.

Part (A)

Let us consider "s" is the distance "v" is the velocity and "k" is the constant of variation.

It is given that the stopping distance of a car varies directly as the square of the velocity of car when the brakes are applied

This can be written as:

$s=kv^2$ ......(1)

It is given that A car moving at 85 mph can stop in 367 feet.

Substitute v=85 and s=367 in above formula and solve for k.

$367=k(85)^2$

$367=7225k$

$\frac{367}{7225}=k$

$0.0508=k$

Now to find the equation that relates the stopping distance to the velocity substitute the value of k in equation 1.

$s=0.0508v^2$

Hence, the required equation is: $s=0.0508v^2$

Part (B)

Calculate the stopping distance of your car if you were moving at 95 mph.

Substitute the value of v=95 in $s=0.0508v^2$

$s=0.0508(95)^2$

$s=0.0508\times 9025$

$s=458.47$

Hence, the stopping distance of your car if you were moving at 95 mph is 458.47 feet.

4 0

2 years ago

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5a is ur answer!!!!!

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2 years ago