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ioda
3 years ago
13

Bring the fraction: b/7a^2c to a denominator of 35a^3c^3

Mathematics
1 answer:
Rina8888 [55]3 years ago
7 0

Answer:

5abc^2/35a^3c^3

Step-by-step explanation:

To bring the fraction: b/7a^2c to a denominator of 35a^3c^3, find the dividend when the 35a^3c^3 is divided by 7a^2c

=35a^3c^3/ 7a^2c

Recall that

a^x/a^y = a^x-y

Hence

35a^3c^3/ 7a^2c = 5a^3-2c^3-1

= 5ac^2

Now multiply the numerator and denominator by the result

b/7a^2c = (b * 5ac^2)/(7a^2c * 5ac^2)

Recall that

a^x * a^y = a^x+y

Hence

b/7a^2c = (b * 5ac^2)/(7a^2c * 5ac^2) = 5abc^2/35a^3c^3

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40.8 gallons of paint among 8 containers how much paint is in each container
LenKa [72]

5.1 gallons of paint are in each container.


4 0
3 years ago
Read 2 more answers
Tennis elbow is thought to be aggravated by the impact experienced when hitting the ball. The article "Forces on the Hand in the
Gnoma [55]

Answer:

Step-by-step explanation:

Hello!

The objective is to study whether there is a greater force after impacting on one- handed backhand drive in advanced tennis players than in intermediate tennis players.

Sample 1: Advanced tennis players

X₁: Force (N) on the hand just after impact on a one- handed backhand drive for an advanced tennis player.

n₁= 6

X[bar]₁= 40.29 N

S₁= 11.29

Sample 2: Intermediate players

X₂: Force (N) on the hand just after impact on a one- handed backhand drive for an intermediate tennis player.

n₂= 8

X[bar]₂= 21.40

S₂= 8.30

Assuming that both variables have a normal distribution and both population variances are equal, to compare these two populations is best to do so trough their population means using a t-test for independent samples.

If the force is greater for the advanced players than for the intermediate players, then you'd expect the population mean for the advanced players to be greater than the population mean for the intermediate players:

H₀: μ₁ ≤ μ₂

H₁: μ₁ > μ₂

α: 0.05

t= \frac{(X_[bar]_1-X[bar]_2)-(Mu_1-Mu_2)}{Sa\sqrt{\frac{1}{n_1} +\frac{1}{n_2} } } ~~t_{n_1+n_2-2}

Sa= \sqrt{\frac{(n_1-1)S_1^2+(n_2-1)S_2^2}{n_1+n_2-2} } = \sqrt{\frac{5*127.51+7*68.92}{6+8-2} }= 9.66

t_{H_0}= \frac{(40.29-21.40)-0}{9.66\sqrt{\frac{1}{6} +\frac{1}{8} } } = 3.62

Using the p-value approach, the decision rule is

If p-value ≤ α, reject the null hypothesis

If p-value > α, do not reject the null hypothesis

The p-value for this test is 0.00024, it is less than the level of significance, so the decision is to reject the null hypothesis.

This means that at a 5% significance level you can conclude that the average force experienced on the hand after a one-handed backhand drive for advanced players is greater than the average force experienced on the hand after a one-handed backhand drive for intermediate players.

I hope this helps!

3 0
3 years ago
Least to the greatest number of 103,718 and 121,356 and 113,635
muminat
103,718 113,635 121,356
7 0
3 years ago
Read 2 more answers
What is the sum of the measures of interior angles ofna regular polygon if each exterior angle measures 72°?
WARRIOR [948]
We know that

The exterior angle measures of a polygon must add up to 360°, 

therefore

 360°/72°=5 exterior angles

so

5 exterior angles, would mean 5 interior angles and therefore  5 sides to the polygon

The exterior angle and the interior angle of any polygon are supplementary and must add up to 180°.

exterior angle+interior angle=180°

interior angle=180°-exterior angle

interior angle=180°−72°-------> 108°
Each interior angle is 108°

The sum of the interior angles would be 

5*108°=540°

the answer is

540°

5 0
2 years ago
Which of the following expressions is a factor of the polynomial x 2 +3/2x-1
Elodia [21]

Answer:

\large \boxed{\sf \ \ \ (x+2)(x-\dfrac{1}{2}) \ \ \ }

Step-by-step explanation:

Hello,

let's solve

x^2+\dfrac{3}{2}x-1=0\\\\ 2x^2+3x-2=0 \ \text{multiply by 2}\\\\\\

\Delta=b^2-4ac=9+4*2*2=9+16=25

There are two solutions

x_1=\dfrac{-3-\sqrt{25}}{4}\\\\x_1=\dfrac{-3-5}{4}\\\\x_1=\dfrac{-8}{4}\\\\\boxed{x_1=-2}

And

x_2=\dfrac{-3+\sqrt{25}}{4}\\\\x_2=\dfrac{-3+5}{4}\\\\x_2=\dfrac{2}{4}\\\\\boxed{x_2=\dfrac{1}{2}}

So we can write

x^2+\dfrac{3}{2}x-1\\\\=(x-x_1)(x-x_2)\\\\=(x+2)(x-\dfrac{1}{2})

Hope this helps.

Do not hesitate if you need further explanation.

Thank you

5 0
3 years ago
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