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anygoal [31]
3 years ago
6

Another company offers a rate of 0.05per minute how would you find the unit rate per hour

Mathematics
1 answer:
Dominik [7]3 years ago
7 0
If you want to figure out how much it is an hour you need to convert 1 hour into minutes. 1 hour = 60 minutes After you've figured out your conversion multiply the rate and units. .05(60) When you multiply that you should get 3 as your answer.
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ANSWER CORRECTLY AND YOU WILL GET BRAINLIEST WORTH IT
AveGali [126]

64

80               80

thats how you havw to put it

8 0
2 years ago
Read 2 more answers
the length of an arc of a circle is 269⁢πcentimeters and the measure of the corresponding central angle is 65∘. what is the leng
scZoUnD [109]

The length of the circle's radius = 744.92 cm

Given the length of arc of a circle, arc length = 269⁢π cm

Central angle of a circle is the angle made between the radius through the arc length at the center of the circle.

The corresponding central angle = 65°

To find the corresponding central angle in radians = 65° x π/180 = 13π/36 radians

We have, arc length of a circle = radius x central angle

Therefore, radius of the circle = arc length / central angle

                                                  =  269⁢π /(13π/36)

                                                  = 744.92 cm

Learn more about arc length of a circle at brainly.com/question/28108430

#SPJ4

6 0
2 years ago
What is the derivative of x times squaareo rot of x+ 6?
Dafna1 [17]
Hey there, hope I can help!

\mathrm{Apply\:the\:Product\:Rule}: \left(f\cdot g\right)^'=f^'\cdot g+f\cdot g^'
f=x,\:g=\sqrt{x+6} \ \textgreater \  \frac{d}{dx}\left(x\right)\sqrt{x+6}+\frac{d}{dx}\left(\sqrt{x+6}\right)x \ \textgreater \  \frac{d}{dx}\left(x\right) \ \textgreater \  1

\frac{d}{dx}\left(\sqrt{x+6}\right) \ \textgreater \  \mathrm{Apply\:the\:chain\:rule}: \frac{df\left(u\right)}{dx}=\frac{df}{du}\cdot \frac{du}{dx} \ \textgreater \  =\sqrt{u},\:\:u=x+6
\frac{d}{du}\left(\sqrt{u}\right)\frac{d}{dx}\left(x+6\right)

\frac{d}{du}\left(\sqrt{u}\right) \ \textgreater \  \mathrm{Apply\:radical\:rule}: \sqrt{a}=a^{\frac{1}{2}} \ \textgreater \  \frac{d}{du}\left(u^{\frac{1}{2}}\right)
\mathrm{Apply\:the\:Power\:Rule}: \frac{d}{dx}\left(x^a\right)=a\cdot x^{a-1} \ \textgreater \  \frac{1}{2}u^{\frac{1}{2}-1} \ \textgreater \  Simplify \ \textgreater \  \frac{1}{2\sqrt{u}}

\frac{d}{dx}\left(x+6\right) \ \textgreater \  \mathrm{Apply\:the\:Sum/Difference\:Rule}: \left(f\pm g\right)^'=f^'\pm g^'
\frac{d}{dx}\left(x\right)+\frac{d}{dx}\left(6\right)

\frac{d}{dx}\left(x\right) \ \textgreater \  1
\frac{d}{dx}\left(6\right) \ \textgreater \  0

\frac{1}{2\sqrt{u}}\cdot \:1 \ \textgreater \  \mathrm{Substitute\:back}\:u=x+6 \ \textgreater \  \frac{1}{2\sqrt{x+6}}\cdot \:1 \ \textgreater \  Simplify \ \textgreater \  \frac{1}{2\sqrt{x+6}}

1\cdot \sqrt{x+6}+\frac{1}{2\sqrt{x+6}}x \ \textgreater \  Simplify

1\cdot \sqrt{x+6} \ \textgreater \  \sqrt{x+6}
\frac{1}{2\sqrt{x+6}}x \ \textgreater \  \frac{x}{2\sqrt{x+6}}
\sqrt{x+6}+\frac{x}{2\sqrt{x+6}}

\mathrm{Convert\:element\:to\:fraction}: \sqrt{x+6}=\frac{\sqrt{x+6}}{1} \ \textgreater \  \frac{x}{2\sqrt{x+6}}+\frac{\sqrt{x+6}}{1}

Find the LCD
2\sqrt{x+6} \ \textgreater \  \mathrm{Adjust\:Fractions\:based\:on\:the\:LCD} \ \textgreater \  \frac{x}{2\sqrt{x+6}}+\frac{\sqrt{x+6}\cdot \:2\sqrt{x+6}}{2\sqrt{x+6}}

Since\:the\:denominators\:are\:equal,\:combine\:the\:fractions
\frac{a}{c}\pm \frac{b}{c}=\frac{a\pm \:b}{c} \ \textgreater \  \frac{x+2\sqrt{x+6}\sqrt{x+6}}{2\sqrt{x+6}}

x+2\sqrt{x+6}\sqrt{x+6} \ \textgreater \  \mathrm{Apply\:exponent\:rule}: \:a^b\cdot \:a^c=a^{b+c}
\sqrt{x+6}\sqrt{x+6}=\:\left(x+6\right)^{\frac{1}{2}+\frac{1}{2}}=\:\left(x+6\right)^1=\:x+6 \ \textgreater \  x+2\left(x+6\right)
\frac{x+2\left(x+6\right)}{2\sqrt{x+6}}

x+2\left(x+6\right) \ \textgreater \  2\left(x+6\right) \ \textgreater \  2\cdot \:x+2\cdot \:6 \ \textgreater \  2x+12 \ \textgreater \  x+2x+12
3x+12

Therefore the derivative of the given equation is
\frac{3x+12}{2\sqrt{x+6}}

Hope this helps!
8 0
3 years ago
Tom and his best friend are going to join a gym. Tom saw that Platinum gym has a sale with a one time fee of $90 and a monthly f
worty [1.4K]

Answer:

<u>Part 1:</u>

For Platinum Gym:

90 + 30x

For Super Fit Gym:

200 + 20x

<u>Part 2:</u>  $270

<u>Part 3:</u>  $320

<u>Part 4:</u>  11 months

<u>Part 5:</u>  See explanation below

Step-by-step explanation:

<u>Part 1:</u>

Let "x" be the number of months:

For Platinum Gym:

90 + 30x

For Super Fit Gym:

200 + 20x

<u>Part 2:</u>

We put x = 6 in platinum gym's equation and get our answer.

90 + 30x

90 + 30(6)

90 + 180

=$270

<u>Part 3:</u>

We put x = 6 into super fit's equation and get our answer.

200 + 20x

200 + 20(6)

200 + 120

=$320

<u>Part 4:</u>

To find the number of months for both gyms to cost same, we need to equate both equations and solve for x:

90 + 30x = 200 + 20x

10x = 110

x = 11

So 11 months

<u>Part 5:</u>

We know for 11 months, they will cost same. Let's check for 10 months and 12 months.

In 10 months:

Platinum = 90 + 30(10) = 390

Super Fit = 200 + 20(10) = 400

In 12 months:

Platinum = 90 + 30(12) = 450

Super Fit = 200 + 20(12) = 440

Thus, we can see that Platinum Gym is a better deal if you want to get membership for months less than 11 and Super Fit is a better deal if you want to get membership for months greater than 11.

4 0
3 years ago
Can someone help me with the homework
r-ruslan [8.4K]
Start with:
How many inches are in 15 feet?
15 feet = ? inches
6 0
3 years ago
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