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4 years ago

The number 450 is increased by 75% is ?

Mathematics

1 answer:

Vadim26 [7]4 years ago

4 0

Answer:

450 increased by 75% = 787.5

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Talib is trying to find the inverse of the function to the right. His work appears beneath it. Is his work correct? Explain your

aev [14]

The answer is is 3 hope it helps

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3 years ago

The surface area of a cube is 54cm2. what is the area of one face of the cube? What is the length of one edge of the cube?

xenn [34]

$Area\ of\ cube:\\\\A=6a^2\\\\54=6a^2\ \ \ \ \ \ \ |:6\\\\a^2=9\ -\ area\ of\ the\ one\ face\ of\ the\ cube\\\\a=3\ -\ length\ of\ one\ edge$

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3 years ago

Can someone help me solve this please?

svetlana [45]

I THINK THE ANSWER WOULD BE 2A\a+b=h

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3 years ago

Each of the ten letters in ``MATHCOUNTS'' is written on its own square tile and placed in a bag. What is the probability that a

PIT_PIT [208]

step 1

First, see how many letters in MATHCOUNTS are also in TEACH.

M no

A yes

T yes

H yes

C yes

O no

U no

N no

T yes

S no

That is 5 out of the 10 letters.

<span>the probability=</span>5/10-------> 1/2
<span>
the answer is
1/2
</span>

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3 years ago

Air pressure may be represented as a function of height (in meters) above the

denis-greek [22]

Answer: OPTION C.

Step-by-step explanation:

<h3> The complete exercise is: "Air pressure may be represented as a function of height (in meters) above the surface of the Earth, as shown below:</h3><h3> $P(h) =P_0e^{-0.00012h}$ </h3><h3> In this function $P_o$ is the air pressure at the surface of the earth, and $h$ is the height above the surface of the Earth, measured in meters. At what height will the air pressure equal 50% of the air pressure at the surface of the Earth"</h3><h3></h3>

Given the following function:

$P(h) =P_0e^{-0.00012h}$

In order to calculate at what height the air pressure will be equal 50% of the air pressure at the surface of the Earth, you can follow these steps:

1. You need to substitute $P(h)=0.5P_o$ into the function:

$0.5P_o=P_0e^{-0.00012h}$

2. Finally, you must solve for $h$ .

Remember the following property of logarithms:

$ln(b)^a=a*ln(b)\\\\ln(e)=1$

Then, you get this result:

$0.5P_o=\frac{P_o}{e^{0.00012h}}\\\\(0.5P_o)(e^{0.00012h})=P_o\\\\e^{0.00012h}=\frac{P_o}{0.5P_o}\\\\ln(e)^{0.00012h}=ln(2)\\\\0.00012h*1=ln(2)\\\\h=\frac{ln(2)}{0.00012}\\\\h=5576.2$

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4 years ago