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n200080 [17]
3 years ago
14

1. Erika has 4 ¾ gallons of Power Aid to serve her fitness club members. If each member gets ⅔ of a gallon of Power Aid, how man

y bottles does Erika need to fill up?
Mathematics
1 answer:
Natasha_Volkova [10]3 years ago
5 0
If u divide 4 3/4 by 2/3 you will get 7 1/8
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WILL MARK BRAINLIEST!!
Rudik [331]

Answer:

C = (A*P - 8.4Y -330T + 200I) / 100

Step-by-step explanation:

P = (8.4Y + 330T + 100C -200I ) / A

now we have to calculate completed passes C for given P, Y, T, I, A

A*P = 8.4Y + 330T -200I +100C

100C = A*P - 8.4Y - 330T + 200I

C = (A*P - 8.4Y -330T + 200I) / 100

I just solved the equation for C

5 0
3 years ago
Can someone please explain?
geniusboy [140]

This is an isosceles triangle. The angles at the base are congruent (they have the same measures).

Look at the picture.

We know: The sum of the measures of angles in a triangle is equal 180°.

Therefore we have the equation:

2x-10+2x-10+3x+25=180       <em>combine like terms</em>

(2x+2x+3x)+(-10-10+25)=180

7x+5=180           <em>subtract 5 from both sides</em>

7x=175              <em>divide both sides by 7</em>

x=25

m\angle A=2x-10\to m\angle A=2(25)-10=50-10=40\\\\m\angle C=m\angle A=40\\\\m\angle B=3x+25\to m\angle B=3(25)+25=75+25=100^o

<h3>Answer: D. m∠C = 40</h3>

6 0
3 years ago
What is the arc measure of an arc with length 4.189 cm and radius equal to 3 cm.'?
Mazyrski [523]
Given that the arc length is 4.189 cm and the radius is 3 cm, the size of the arc will found as follows;
C=theta/360 πd
suppose:
size  of arc=theta=x
d=3*2=6 cm
hence;
4.189=x/360*π*6
4.189=0.0524x
x=4.189/0.0524
x=80.004°
The size of the arc length is 80.004°
6 0
3 years ago
Which transformation is a direct isometry?
Scilla [17]
In mathematics, an isometry is a distance-preserving injective map between metric spaces. A composition of two opposite isometries is a direct <span>isometry.
Hope it helped!!</span>
3 0
3 years ago
Let f(x) = -5e^(-x/3)<br><br> Find f^(6) (1)
Luba_88 [7]

Let's take the first derivative:

f'(x) = (-5e^{-x/3})' = -5 \times (e^{-x/3})' = -5e^{-x/3} \times \left(-\dfrac{1}{3}\right) = \dfrac{5}{3}e^{-x/3}.

Notice that we can write this as:

f'(x) = -\dfrac{f(x)}{3}.

By taking the derivative of both sides n times, we get:

f^{(n+1)}(x) = -\dfrac{f^{(n)}}{3}.

This means that each time you take a derivative, a factor of -\dfrac{1}{3} will appear. So we conclude that:

f^{(n)}(x) = \left(-\dfrac{1}{3}\right)^nf(x).

Taking n=6 and x=1, we get:

f^{(6)}(1) = \left(-\dfrac{1}{3}\right)^6 f(1) = \dfrac{-5e^{-1/3}}{729} = -\dfrac{5}{729}e^{-1/3}.

So we finally get:

\boxed{-\dfrac{5}{729}e^{-1/3}}.

6 0
3 years ago
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