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3 years ago

A men earns $62 per week. how many will he earn if he works a full year 52weeks

Mathematics

2 answers:

Sever21 [200]3 years ago

5 0

62$ (earns of 1 week) multiplied with 52 weeks (how often he earns the 62$) is 3224

NARA [144]3 years ago

5 0

$3224 he will earn......

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PLEASE HELP IM ON A TIME LIMIT :(

Mice21 [21]

Answer:

hey hope this helps

<h3 /><h3>Comparing sides AB and DE </h3>

AB =

$\sqrt{ {1}^{2} + {1}^{2} }$

$= \sqrt{2}$

$= \sqrt{ {(3 - 5)}^{2} + {(1 + 1}^{2} } \\ = \sqrt{ {( - 2)}^{2} + {(2)}^{2} } \\ = \sqrt{4 + 4} \\ = \sqrt{8} \\ = 2 \sqrt{2}$

So DE = 2 × AB

and since the new triangle formed is similar to the original one, their side ratio will be same for all sides.

<u>scale factor</u> = AB/DE

= 2

It's been reflected across the Y-axis

<em>moved thru the translation of 3 units towards the right of positive x- axis </em>

for this let's compare the location of points B and D

For both the y coordinate is same while the x coordinate of B is 0 and that of D is 3

so the triangle has been shifted by 3 units across the positive x axis

7 0

2 years ago

ANSWER THIS!!!!!!!!!!

cluponka [151]

Answer:

r = sqrt( 3V/( pih))

Step-by-step explanation:

V = 1/3 pi r^2 h

Solving for r

Multiply each side by 3

3V = pi r^2h

Divide each side by pi h

3V/( pi h) = r^2

Take the square root of each side

sqrt( 3V/ (pi h) )= sqrt( r^2)

sqrt( 3V/ (pi h)) = r

4 0

2 years ago

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

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