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crimeas [40]
3 years ago
5

Please answer this correctly

Mathematics
2 answers:
Elden [556K]3 years ago
7 0

Answer:

2 minutes and 50 seconds

I think thats the answer, sorry if it's wrong ;;

zysi [14]3 years ago
3 0

Answer:

it's 2.31 mins

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Which equivalent equation is the result of combining like terms in the given equation? 2r + 55 = 6r + 7 – 2r
nikklg [1K]
I hope I help you out
6 0
3 years ago
eff can paint 28 square feet every 30 minutes. Angela can paint 50 square feet every hour. Who can paint faster, and by how many
lisabon 2012 [21]
Guessing “eff” is Jeff ;)
28 sq feet/30 min = 56 sq feet/60 minutes

Angela
50 sq feet/60 min

So Jeff paints 6 square feet more per hour.
4 0
3 years ago
Two mechanics worked on a car. The first mechanic worked for 10 hours, and the second mechanic worked for 5 hours. Together they
almond37 [142]

Answer:

102 an hour

Step-by-step explanation:

all you do is add all of them together

5 0
3 years ago
10 pt
lord [1]

Answer:

C. 434π

Step-by-step explanation:

Given:

Radius (r) = 7 in.

Height (h) = 24 in.

Required:

Surface area of the cylinder

Solution:

S.A = 2πrh + 2πr²

Plug in the values

S.A = 2*π*7*24 + 2*π*7²

S.A = 336π + 98π

S.A = 434π

5 0
3 years ago
Use the mid-point rule with n = 4 to approximate the area of the region bounded by y = x3 and y = x. (10 points)
USPshnik [31]
See the graph attached.

The midpoint rule states that you can calculate the area under a curve by using the formula:
M_{n} = \frac{b - a}{2} [ f(\frac{x_{0} + x_{1} }{2}) +  f(\frac{x_{1} + x_{2} }{2}) + ... +  f(\frac{x_{n-1} + x_{n} }{2})]

In your case:
a = 0
b = 1
n = 4
x₀ = 0
x₁ = 1/4
x₂ = 1/2
x₃ = 3/4
x₄ = 1

Therefore, you'll have:
M_{4} = \frac{1 - 0}{4} [ f(\frac{0 +  \frac{1}{4} }{2}) +  f(\frac{ \frac{1}{4} + \frac{1}{2} }{2}) +  f(\frac{\frac{1}{2} + \frac{3}{4} }{2}) + f(\frac{\frac{3}{4} + 1} {2})]
M_{4} = \frac{1}{4} [ f(\frac{1}{8}) +  f(\frac{3}{8}) +  f(\frac{5}{8}) + f(\frac{7}{8})]

Now, to evaluate your f(x), you need to look at the graph and notice that:
f(x) = x - x³

Therefore:
M_{4} = \frac{1}{4} [(\frac{1}{8} - (\frac{1}{8})^{3}) + (\frac{3}{8} - (\frac{3}{8})^{3}) + (\frac{5}{8} - (\frac{5}{8})^{3}) + (\frac{7}{8} - (\frac{7}{8})^{3})]

M_{4} = \frac{1}{4} [(\frac{1}{8} - \frac{1}{512}) + (\frac{3}{8} - \frac{27}{512}) + (\frac{5}{8} - \frac{125}{512}) + (\frac{7}{8} - \frac{343}{512})]

M₄ = 1/4 · (2 - 478/512)
     = 0.2666

Hence, the <span>area of the region bounded by y = x³ and y = x</span> is approximately 0.267 square units.

6 0
3 years ago
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