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aleksandrvk [35]
3 years ago
9

Which expression is equivalent to StartFraction RootIndex 7 StartRoot x squared EndRoot Over RootIndex 5 StartRoot y cubed EndRo

ot? Assume y not-equals 0.
(x Superscript two-sevenths Baseline) (y Superscript negative three-fifths Baseline)
(x Superscript two-sevenths Baseline) (y Superscript five-thirds Baseline)
(x Superscript two-sevenths Baseline) (y Superscript three-fifths Baseline)
(x Superscript seven-halves Baseline) (y Superscript negative five-thirds Baseline)
Mathematics
2 answers:
Lera25 [3.4K]3 years ago
8 0

Answer:

Option A.

Step-by-step explanation:

The given expression is

\dfrac{\sqrt[7]{x^2}}{\sqrt[5]{y^3}}

where, y\neq 0.

We need to find the expression which is equivalent to the given expression.

The given expression can be rewritten as

\dfrac{(x^2)^{\frac{1}{7}}}{(y^3)^{\frac{1}{5}}}      [\because \sqrt[n]{x}=x^{\frac{1}{n}}]

\dfrac{x^{\frac{2}{7}}}{y^{\frac{3}{5}}}      [\because (a^m)^n=a^{mn}]

x^{\frac{2}{7}}y^{-\frac{3}{5}}      [\because a^{-n}=\dfrac{1}{a^n}]

Therefore, the correct option is A.

Contact [7]3 years ago
8 0

Answer:

it is A on edgenuity

Step-by-step explanation:

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Use slopes and y-intercepts to determine if the lines 10x+3y=−3 and 5x−4y=−3 are parallel.
kvasek [131]

Answer:

They are not parallel

Step-by-step explanation:

original equation

10x + 3y = -3

subtract 10x

3y = -10x - 3

divide by 3

y = -10/3x - 1

original equation

5x - 4y = -3

subtract 5x

-4y = -5x-3

divide by -4

y = 5/4x + 3/4

the slopes are not equal to each other (5/4x and -10/3x) so they are not parallel

4 0
2 years ago
The answer ASAP I don’t understand this at all
RSB [31]

Answer:

T = 530N + 250

Step-by-step explanation:

For the first plan, Heather will deposit $250 and then save $135 per month.

So, that is 250 + (135 x N) where N is the number of months she saves the $135.

So if it is for 3months,

We will have:

t¹ = 135N + 250

= 250 + (135 x 3)

= 250 + 405

= $655

For the second plan, there is no initial deposit, but she will save $395 per month.

That is 395 x N

t² = 395N

For 3months, we have 395 x 3 = $1185

Therefore the total for both plans in 3months = 655 + 1185 = $1840

Equation relating T to N

T = t¹ + t²

T = (135N + 250) + 395N

T = 135N + 250 + 395N

T = 530N + 250

5 0
3 years ago
Find the distance between the two points rounding to the nearest tenth (if necessary).
Virty [35]

\bf ~~~~~~~~~~~~\textit{distance between 2 points} \\\\ (\stackrel{x_1}{3}~,~\stackrel{y_1}{6})\qquad (\stackrel{x_2}{1}~,~\stackrel{y_2}{9})\qquad \qquad d = \sqrt{( x_2- x_1)^2 + ( y_2- y_1)^2} \\\\\\ d=\sqrt{(1-3)^2+(9-6)^2}\implies d=\sqrt{(-2)^2+3^2} \\\\\\ d=\sqrt{13}\implies d\approx 3.6056\implies \stackrel{\textit{rounded up}}{d=3.6}

4 0
3 years ago
6 1/2lb=___oz I need to know now or I will get it wrong
Semmy [17]
It i 104 oz 

Hope This Helps
4 0
3 years ago
Read 2 more answers
According to a survey, high school girls average 100 text messages daily (The Boston Globe, April 21, 2010). Assume the populati
Ghella [55]

Answer:

Step-by-step explanation:

Hello!

The variable of interest is:

X: number of daily text messages a high school girl sends.

This variable has a population standard deviation of 20 text messages.

A sample of 50 high school girls is taken.

The is no information about the variable distribution, but since the sample is large enough, n ≥ 30, you can apply the Central Limit Theorem and approximate the distribution of the sample mean to normal:

X[bar]≈N(μ;δ²/n)

This way you can use an approximation of the standard normal to calculate the asked probabilities of the sample mean of daily text messages of high school girls:

Z=(X[bar]-μ)/(δ/√n)≈ N(0;1)

a.

P(X[bar]<95) = P(Z<(95-100)/(20/√50))= P(Z<-1.77)= 0.03836

b.

P(95≤X[bar]≤105)= P(X[bar]≤105)-P(X[bar]≤95)

P(Z≤(105-100)/(20/√50))-P(Z≤(95-100)/(20/√50))= P(Z≤1.77)-P(Z≤-1.77)= 0.96164-0.03836= 0.92328

I hope you have a SUPER day!

3 0
3 years ago
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