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spayn [35]
2 years ago
5

Write (5/3)³ as a quotient of powers.

Mathematics
1 answer:
Oxana [17]2 years ago
3 0

Answer:

The number (5/3)³ written as a quotient of powers is; 5³/3³

According to the question;

We are required to determine write the number, (5/3)³ as a quotient of powers

The number (5/3)³ can be evaluated as;

= 125/27

= 5³/3³

Step-by-step explanation:

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Now you will connect the line M and Z.

Thats it! I hope this helped!! :)




3 0
3 years ago
Read 2 more answers
A curve is given by y=(x-a)√(x-b) for x≥b, where a and b are constants, cuts the x axis at A where x=b+1. Show that the gradient
ankoles [38]

<u>Answer:</u>

A curve is given by y=(x-a)√(x-b) for x≥b. The gradient of the curve at A is 1.

<u>Solution:</u>

We need to show that the gradient of the curve at A is 1

Here given that ,

y=(x-a) \sqrt{(x-b)}  --- equation 1

Also, according to question at point A (b+1,0)

So curve at point A will, put the value of x and y

0=(b+1-a) \sqrt{(b+1-b)}

0=b+1-c --- equation 2

According to multiple rule of Differentiation,

y^{\prime}=u^{\prime} y+y^{\prime} u

so, we get

{u}^{\prime}=1

v^{\prime}=\frac{1}{2} \sqrt{(x-b)}

y^{\prime}=1 \times \sqrt{(x-b)}+(x-a) \times \frac{1}{2} \sqrt{(x-b)}

By putting value of point A and putting value of eq 2 we get

y^{\prime}=\sqrt{(b+1-b)}+(b+1-a) \times \frac{1}{2} \sqrt{(b+1-b)}

y^{\prime}=\frac{d y}{d x}=1

Hence proved that the gradient of the curve at A is 1.

7 0
2 years ago
A state park charges an entrance fee of 10$, plus 22$ for each night of camping. A family spends a total of 142$ for staying at
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7 0
3 years ago
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Solve x^2+ 17x-12=-84
kramer

Answer:

(X+9)(X+8)=0

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Step-by-step explanation:

x^2+ 17x-12=-84

x^2+17x-12+84=0

x^2+17x+72=0

x^2+(9+8)X+72=0

x^2+9x+8x+72=0

X(X+9)+8(X+9)=0

(X+9)(X+8)=0

X=-9,-8

So the value of the X is -9,-8

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Answer:

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