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Furkat [3]
2 years ago
15

How many 4/5are in 1

Mathematics
2 answers:
OverLord2011 [107]2 years ago
6 0

Answer:

If your gonna ask it on groups its one group

natita [175]2 years ago
3 0
There are one 4/5 in 1 cus (1=5/5)
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100 POINTS AND BRAINLIST NEEED ASAPPP:Use the picture of the function below to answer the following questions, Horizontal Asympt
Leni [432]

Answer:

Concept: Graph Analysis

  1. Horizontal Asy: 1
  2. Vertical Asy: 4
  3. Domain: -inf to inf U -inft to inf
  4. X intercept: 7
  5. Y intercept: 2
  6. End behavior: Continous to real numbers, hence negative and positive infinity
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6 0
3 years ago
Find the least common denominator for these two rational expressions -5/s -1/2s
Ronch [10]
<span>-5/s -1/2s
.The least common denominator is 2, since 2 is divisible by 1 and by 2

The answer will be : </span><span>-5/s -1/2s = -10/2s - 1/2s = -11s/2</span>
6 0
3 years ago
Mia's office recycled a total of 9 pounds of paper over 3 weeks. After 7 weeks, how many pounds of paper will Mia's office have
alukav5142 [94]

Answer:

63

will you mark me brainllest?

Step-by-step explanation:

7 0
2 years ago
Read 2 more answers
What is the value of X?
Citrus2011 [14]
(X+12) + 140 = 180

X + 12 + 140 = 180

X + 152 = 180

X = 180 - 152

X = 28
4 0
2 years ago
How do you do this question?
Sergeeva-Olga [200]

Answer:

V = 2000r³/3

Step-by-step explanation:

We know that the base is a circular disk, so it creates a circle on the xy plane. It would be in the form x² + y² = r². In other words x² + y² = (5r)². Let's isolate y in this equation now:

x² + y² = (5r)²,

x² + y² = 25r²,

y² = 25r² - x²,

y = √25r² - x² ---- (1)

Now remember that parallel cross sections perpendicular to the base are squares. Therefore Area = length^2. The length will then be = 2√25r² - x² --- (2). Now we can evaluate the integral from -5r to 5r, of [ 2√25r² - x² ]² dx.

\int _{-5r}^{5r}\:\left[\:2\sqrt{\left(25r^2\:-\:x^2\right)}\:\right]\:^2\:dx\\=\int _{-5r}^{5r}4\left(25r^2-x^2\right)dx\\\\= 4\cdot \int _{-5r}^{5r}25r^2-x^2dx\\\\= 4\left(\int _{-5r}^{5r}25r^2dx-\int _{-5r}^{5r}x^2dx\right)\\\\= 4\left(250r^3-\frac{250r^3}{3}\right)\\\\= 4\cdot \frac{500r^3}{3}\\\\= \frac{2000r^3}{3}

As you can see, your exact solution would be, V = 2000r³/3. Hope that helps!

3 0
2 years ago
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