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Vadim26 [7]
3 years ago
12

An artist mixes yellow and blue paint to make the perfect shade of green. She uses 2 ounces of blue paint for every 7 ounces of

yellow
Mathematics
1 answer:
My name is Ann [436]3 years ago
4 0

Answer:

what do we need to figure  out

Step-by-step explanation:

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Each month, Miss Patrick spends $60 on transportation to work, and earns $24.50 per hour.Each month, Mr. Shah spends $150 on lab
Semmy [17]

Answer:

3 hours for miss patrick 5 hours for mr shah

Step-by-step explanation:

5 0
3 years ago
Read 2 more answers
What must you do to the first inequality to get to the second inequality? A. N 4 <5;n<20 b. − n 4 <5;n>−20 A. A. Add
adell [148]

A. You have two inequalities:

  • \dfrac{n}{4}
  • n

If you multiply the first inequality by positive number 4 (multiplying inequality by positive number doesn't change the inequality sign), then you get the inequality

4\cdot \dfrac{n}{4}

B. You have two inequalities:

  • -\dfrac{n}{4}
  • n>-20.

If you multiply the first inequality by negative number -4 (multiplying inequality by negative number changes the inequality sign), then you get the inequality

-4\cdot \left(-\dfrac{n}{4}\right)>-4\cdot 5,\\ \\n>-20.

Answer: option D.

5 0
3 years ago
1. The sign outside Mrs. Washington’s catering company is in the shape of a trapezoid, as shown below.
valentina_108 [34]
The perimeter of the sign is 11 ft
The area of the sign is 7.5ft^2

Hope this helps :)
3 0
3 years ago
Read 2 more answers
Help!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
ella [17]

Answer: 20%??? I don't know Honestly

Step-by-step explanation:

6 0
3 years ago
PLZ HELP!!! Use limits to evaluate the integral.
Marrrta [24]

Split up the interval [0, 2] into <em>n</em> equally spaced subintervals:

\left[0,\dfrac2n\right],\left[\dfrac2n,\dfrac4n\right],\left[\dfrac4n,\dfrac6n\right],\ldots,\left[\dfrac{2(n-1)}n,2\right]

Let's use the right endpoints as our sampling points; they are given by the arithmetic sequence,

r_i=\dfrac{2i}n

where 1\le i\le n. Each interval has length \Delta x_i=\frac{2-0}n=\frac2n.

At these sampling points, the function takes on values of

f(r_i)=7{r_i}^3=7\left(\dfrac{2i}n\right)^3=\dfrac{56i^3}{n^3}

We approximate the integral with the Riemann sum:

\displaystyle\sum_{i=1}^nf(r_i)\Delta x_i=\frac{112}n\sum_{i=1}^ni^3

Recall that

\displaystyle\sum_{i=1}^ni^3=\frac{n^2(n+1)^2}4

so that the sum reduces to

\displaystyle\sum_{i=1}^nf(r_i)\Delta x_i=\frac{28n^2(n+1)^2}{n^4}

Take the limit as <em>n</em> approaches infinity, and the Riemann sum converges to the value of the integral:

\displaystyle\int_0^27x^3\,\mathrm dx=\lim_{n\to\infty}\frac{28n^2(n+1)^2}{n^4}=\boxed{28}

Just to check:

\displaystyle\int_0^27x^3\,\mathrm dx=\frac{7x^4}4\bigg|_0^2=\frac{7\cdot2^4}4=28

4 0
3 years ago
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