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zlopas [31]
2 years ago
13

Flying against the wind, an airplane travels 4340 kilometers in 7 hours. Flying with the wind, the same plane travels 3120 kilom

eters in 3 hours. What is the
rate of the plane in still air and what is the rate of the wind?
Mathematics
1 answer:
Over [174]2 years ago
3 0

The rate of the wind is given to be 210 the rate of the plane in mid air is given as 830

<h3>Howe to solve for the rate of the plane and wind</h3>

In 7 hours the kilometer traveled =  4340

P + W = 4340/7

= 620

P-W = 3120/3 hours

= 1040

2p = 1040 + 620

= 1660

p =830

W = 1040 - 830

= 210

Hence the rate of the wind is given to be 210 the rate of the plane in mid air is given as 830

Read more on rate here

brainly.com/question/24304697

#SPJ1

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Match the numerical expressions to their simplest forms.
Aloiza [94]

Answer:

(a^6b^1^2)^\frac{1}{3} = a^2b^4

\frac{(a^5b^3)^\frac{1}{2}}{(ab)^-^\frac{1}{2}} = a^3b^2

(\frac{a^5}{a^-^3b^-^4})^\frac{1}{4} = a^2b

(\frac{a^3}{ab^-^6})^\frac{1}{2} = ab^3

Step-by-step explanation:

Simplify each of the expressions:

1

(a^6b^1^2)^\frac{1}{3}

Distribute the exponent. Multiply the exponent of the term outside of the parenthesis by the exponents of the variable.

(a^6b^1^2)^\frac{1}{3}

a^6^*^\frac{1}{3}b^1^2^*^\frac{1}{3}

Simplify,

a^2b^4

2

Use a similar technique to solve this problem. Remember, a fractional exponent is the same as a radical, if the denominator is (2), then the operation is taking the square root of the number.

\frac{(a^5b^3)^\frac{1}{2}}{(ab)^-^\frac{1}{2}}

Rewrite as square roots:

\frac{\sqrt{a^5b^3}}{\sqrt{(ab)}^-^1}

A negative exponent indicates one needs to take the reciprocal of the number. Apply this here:

\frac{\sqrt{a^5b^3}}{\frac{1}{\sqrt{ab}}}

Simplify,

\sqrt{a^5b^3}*\sqrt{ab}

Since both numbers are under a radical, one can rewrite them such that they are under the same radical,

\sqrt{a^5b^3*ab}

Simplify,

\sqrt{a^6b^4}

Since this operation is taking the square root, divide the exponents in half to do this operation:

a^3b^2

3

(\frac{a^5}{a^-^3b^-^4})^\frac{1}{4}

Simplify, to simplify the expression in the numerator and the denominator, the base must be the same. Remember, the base is the number that is being raised to the exponent. One subtracts the exponent of the number in the denominator from the exponent of the like base in the numerator. This only works if all terms in both the numerator and the denominator have the operation of multiplication between them:

(\frac{a^8}{b^-^4})^\frac{1}{4}

Bring the negative exponent to the numerator. Change the sign of the exponent and rewrite it in the numerator,

(a^8b^4)^\frac{1}{4}

This expression to the power of the one forth. This is the same as taking the quartic root of the expression. Rewrite the expression with such,

\sqrt[4]{a^8b^4}

SImplify, divide the exponents by (4) to simulate taking the quartic root,

a^2b

4

(\frac{a^3}{ab^-^6})^\frac{1}{2}

Using all of the rules mentioned above, simplify the fraction. The only operation happening between the numbers in both the numerator and the denominator is multiplication. Therefore, one can subtract the exponents of the terms with the like base. The term in the denomaintor can be rewritten in the numerator with its exponent times negative (1).

(a^3^-^1b^(^-^6^*^(^-^1^)^))^\frac{1}{2}

(a^2b^6)^\frac{1}{2}

Rewrite to the half-power as a square root,

\sqrt{a^2b^6}

Simplify, divide all of the exponents by (2),

ab^3

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3 years ago
Solve the following equation.<br> Round to the nearest tenth.<br> 8x -1 = -9 + 2x
motikmotik

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3 years ago
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Alexxandr [17]
<h3>Determining the solution.</h3>

The solution(s) for the provided graph are (0, 6) and (6, 0)

<h3>What is a solution of a graph?</h3>

The solution of the <u>system of equations</u> is the <u>intersection</u> of the two equations.

<h3>Reasoning.</h3>

In the image provided, we can see two solutions (i<u>ntersection points)</u> that are intersected by a line and a circle.

<u>Those two solutions are:</u>

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