Which monomials are factors of 9x^4

Beatrice is using a cylinder shape container to hold 52 in³ of liquid. The height of the container is 4 inches tall. What is the

asambeis [7]

Cylinder Volume = PI * radius ^2 * height

52 = PI * radius ^2 * 4

radius ^2 = 52 / (PI * 4)

radius ^2 = 52 / 12.5663706144

radius ^ 2 = 4.1380285204

radius = 2.0342144726

OR a good approximation of the radius is 2 inches

4 0

3 years ago

Read 2 more answers

A fair spinner has 11 equal sections: 4 red, 4 blue and 3 green.

Darina [25.2K]

Answer:

24/121

Step-by-step explanation:

The probability of an event happening is the ratio of the number of possible outcome to the number of total outcome.

The probability of not getting green then green

= rg or bg

Where r is the probability of getting red, g is the probability of green and b is the probability of blue.

= 4/11 * 3/11 + 4/11 * 3/11

= 12/121 + 12/121

= 24/121

4 0

3 years ago

Ten times the sum of half a number and 6 is 8

Genrish500 [490]

" · " - times

" + " - sum

n - a number

0.5n - half a number

10 · (0.5n + 6) = 8 |use distributive property

(10)(0.5n) + (10)(6) = 8

5n + 60 = 8 |substract 60 from both sides

5n = -52 |divide both sides by 5

n = -10.4

5 0

3 years ago

77 meter=_____dm ? <br><br> Can someone help ?

MariettaO [177]

Answer:

Step-by-step explanation:

1 meter = 10 decimeter. Deci means ten.

77 meters = 770 dm

8 0

3 years ago

\sqrt{53+20\sqrt{7}} can be written in the form a+b\sqrt{c}, where a, b, and c are integers and c has no factors which is a perf

choli [55]

<h3>Answer: 14</h3>

=====================================================

Explanation:

Since the range of $y = \sqrt{x}$ is $y \ge 0$ , this means that $\sqrt{53+20\sqrt{7}}$ cannot be negative.

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Let's set the given nested radical expression equal to the form a+b*sqrt(c), and then square both sides to see what happens

$\sqrt{53+20\sqrt{7}} = a+b\sqrt{c}\\\\\left(\sqrt{53+20\sqrt{7}}\right)^2 = \left(a+b\sqrt{c}\right)^2\\\\53+20\sqrt{7} = a^2+2ab\sqrt{c}+b^2c\\\\53+20\sqrt{7} = (a^2+b^2c)+2ab\sqrt{c}\\\\$

Equating the corresponding parts shows that

$53=a^2+b^2c \text{ and } 20\sqrt{7} = 2ab\sqrt{c}\\\\$

Therefore c = 7 must be the case to allow the radicands to match up.

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Since c = 7, we can update the first equation like so

$53=a^2+b^2c\\\\53=a^2+b^2*7\\\\53=a^2+7b^2\\\\$

We'll come back to this later.

-----------------------------------

Focus on $20 = 2ab\sqrt{7}$

Note how 20 and 2ab are coefficients outside the root. So 20 = 2ab leads to ab = 10.

Here are all the ways to multiply to 10 where a,b are integers

1*10 = 10
-1*(-10) = 10
2*5 = 10
-2*(-5) = 10

So we could have these possibilities.

a = 1, b = 10
a = 10, b = 1
a = -10, b = -1
a = -1, b = -10
a = 2, b = 5
a = 5, b = 2
a = -5, b = -2
a = -2, b = -5

If you try all of those a,b combinations with $53 = a^2 + 7b^2$ , then you should find that a = 5 and b = 2 works. I'm skipping a bunch of steps here. All you basically do is plug and chug.

Due to symmetry, we could have a = -5 and b = -2 as another solution to $53 = a^2 + 7b^2$ . However, if a < 0 and b < 0, then $a+b\sqrt{c}$ would be negative which contradicts the fact that $\sqrt{53+20\sqrt{7}}$ is positive. So we'll ignore any case where a,b are negative.

Therefore, the only combination that works is a = 5 and b = 2

-----------------------------------

Overall, we found that

a = 5
b = 2
c = 7

And therefore,

$\sqrt{53+20\sqrt{7}} = a+b\sqrt{c}\\\\\sqrt{53+20\sqrt{7}} = 5+2\sqrt{7}\\\\$

You can confirm this with your calculator by taking advantage of the idea "If x = y, then x-y = 0". In other words, if you subtract the left and right sides, you should get 0 as a result (the order of subtraction doesn't matter in this case). Don't forget to use parenthesis correctly, or you may not get 0. It might help to think of it as the form (x) - (y) = 0.

After we confirmed we have the correct a,b,c values, we simply add them up to get a+b+c = 5+2+7 = 14.

7 0

3 years ago