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Nitella [24]
3 years ago
6

The question is in the picture!! pls help

Mathematics
1 answer:
evablogger [386]3 years ago
7 0

Answer:

I HELPEDDDDDDDDD :D

Step-by-step explanation:

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"find the cost per ounce of a sunscreen made from 140 oz of lotion that cost $3.45 per ounce and 90 oz of lotion that cost $14.0
Olegator [25]
Sunscreen approx. 40.6 cents an oz 
Lotion 6.4 cents an oz
5 0
4 years ago
Please explain the answer to 4367 ÷ 0.004
boyakko [2]
4367 ÷ 0.004 is 1091750. To solve this, you can use long division, by setting up the equation. Make sure to move the decimal. Then, divide the first digit, then the first two digits, and so on.
7 0
4 years ago
Angelica calculated the distance between the two points shown on the graph below. (1st picture)
Alex Ar [27]
Use the distance formula to find the value of the side lengths.
d=√((x1-x2)²+(y1-y2)²

d of side AC is 6

d of side CB is 10 

Angela's use of the Pythagorean Theorem of 10²+6²+c² is incorrect; she put the right values in the wrong spots, the formula needed is:
6²+10²=c²

Option C- Angelica's side lengths were too long.
7 0
3 years ago
Read 2 more answers
Charles wants to take out a 5 year loan on a car that costs $15,000. He is trying to make a rough estimate to decide if he can a
marin [14]

Answer:


No, This is not reasonable estimate.  

Step-by-step explanation:


Charles wants to take out a 5 year loan on car that cost $15000.  

He make rough estimate to make payment. He can afford to pay up to $200 per month.  

First we find total number of months in 5 years.  

Total months = 5x12= 60 months

Total estimate to pay = 60\times 200 =$12000

we can see cost of a car $15,000 is greater than his estimation $12,000.  

Therefore, Charles estimation will not work.  

No, This is not reasonable estimate.  


5 0
3 years ago
evaluate the line integral ∫cf⋅dr, where f(x,y,z)=5xi−yj+zk and c is given by the vector function r(t)=⟨sint,cost,t⟩, 0≤t≤3π/2.
meriva

We have

\displaystyle \int_C \vec f \cdot d\vec r = \int_0^{\frac{3\pi}2} \vec f(\vec r(t)) \cdot \dfrac{d\vec r}{dt} \, dt

and

\vec f(\vec r(t)) = 5\sin(t) \, \vec\imath - \cos(t) \, \vec\jmath + t \, \vec k

\vec r(t) = \sin(t)\,\vec\imath + \cos(t)\,\vec\jmath + t\,\vec k \implies \dfrac{d\vec r}{dt} = \cos(t) \, \vec\imath - \sin(t) \, \vec\jmath + \vec k

so the line integral is equilvalent to

\displaystyle \int_C \vec f \cdot d\vec r = \int_0^{\frac{3\pi}2} (5\sin(t) \cos(t) + \sin(t)\cos(t) + t) \, dt

\displaystyle \int_C \vec f \cdot d\vec r = \int_0^{\frac{3\pi}2} (6\sin(t) \cos(t) + t) \, dt

\displaystyle \int_C \vec f \cdot d\vec r = \int_0^{\frac{3\pi}2} (3\sin(2t) + t) \, dt

\displaystyle \int_C \vec f \cdot d\vec r = \left(-\frac32 \cos(2t) + \frac12 t^2\right) \bigg_0^{\frac{3\pi}2}

\displaystyle \int_C \vec f \cdot d\vec r = \left(\frac32 + \frac{9\pi^2}8\right) - \left(-\frac32\right) = \boxed{3 + \frac{9\pi^2}8}

7 0
2 years ago
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