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4 years ago

How do you do this????

Mathematics

2 answers:

Juli2301 [7.4K]4 years ago

6 0

H 6 Using 8^2 +b^2=10^2

Svetllana [295]4 years ago

4 0

Tot answer his question will need to use the Pythagorean Theron which involves you using a squared plus b squared Egypt’s c squared that will be the Theron need to solve this problem

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Simplify the following to standard form

irina1246 [14]

Answer:

1, 11x + 3y + 15z

2, 3x + 12y + -5z

Step-by-step explanation:

all you need to do is combined the like terms and make sure the signs (positive and negative signs) are right

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3 years ago

Please help I'm almost done with this course I just need a few more answered

iogann1982 [59]

The answer would be B.

Hope this helped :)

6 0

3 years ago

Can some one really help me please!!!! :(((

GREYUIT [131]

Answer:

for the first on its a=24 and for the second its a= 48

Step-by-step explanation:

for the first one you have to find the length which is 8 and then you find the width which is 3 then you multiply that to get 24

for the second one you find the base which is the strait line which is 8 then you find the height which is 6 then you multiply it to get 48 i hopes this helped!

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3 years ago

1/2 mi = 880 yd O True O False Help

Black_prince [1.1K]

I don’t know I am just guessing but it’s False good luck

5 0

4 years ago

Read 2 more answers

The population, P(t), of China, in billions, can be approximated by1 P(t)=1.394(1.006)t, where t is the number of years since th

vitfil [10]

Answer:

At the start of 2014, the population was growing at 8.34 million people per year.

At the start of 2015, the population was growing at 8.39 million people per year.

Step-by-step explanation:

To find how fast was the population growing at the start of 2014 and at the start of 2015 we need to take the derivative of the function with respect to t.

The derivative shows by how much the function (the population, in this case) is changing when the variable you're deriving with respect to (time) increases one unit (one year).

We know that the population, P(t), of China, in billions, can be approximated by $P(t)=1.394(1.006)^t$

To find the derivative you need to:

$\frac{d}{dt}\left(1.394\cdot \:1.006^t\right)=\\\\\mathrm{Take\:the\:constant\:out}:\quad \left(a\cdot f\right)'=a\cdot f\:'\\\\1.394\frac{d}{dt}\left(1.006^t\right)\\\\\mathrm{Apply\:the\:derivative\:exponent\:rule}:\quad \frac{d}{dx}\left(a^x\right)=a^x\ln \left(a\right)\\\\1.394\cdot \:1.006^t\ln \left(1.006\right)\\\\\frac{d}{dt}\left(1.394\cdot \:1.006^t\right)=(1.394\cdot \ln \left(1.006\right))\cdot 1.006^t$

To find the population growing at the start of 2014 we say t = 0

$P(t)' = (1.394\cdot \ln \left(1.006\right))\cdot 1.006^t\\P(0)' = (1.394\cdot \ln \left(1.006\right))\cdot 1.006^0\\P(0)' = 0.00833901 \:Billion/year$

To find the population growing at the start of 2015 we say t = 1

$P(t)' = (1.394\cdot \ln \left(1.006\right))\cdot 1.006^t\\P(1)' = (1.394\cdot \ln \left(1.006\right))\cdot 1.006^1\\P(1)' = 0.00838904 \:Billion/year$

To convert billion to million you multiple by 1000

$P(0)' = 0.00833901 \:Billion/year \cdot 1000 = 8.34 \:Million/year \\P(1)' = 0.00838904 \:Billion/year \cdot 1000 = 8.39 \:Million/year$

6 0

3 years ago