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2 years ago

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The segment JL has a definite length of [Drop Down 1]. Segments JK and KL have both definite lengths of [Drop Down 2] and [Drop

Down 3], respectively. Both segments are [Drop Down 4]. JL=5x-9, JK=3x and KL=1x+4

1 answer:

saul85 [17]2 years ago

6 0

Answer:

$JL=56$ $JK=39$ $KL=17$

Step-by-step explanation:

<em>Your question is not well presented/formatted; however, I can deduce that the question requires that you calculate the following lengths;</em>

JK, KL and JL

Given that

$JL=5x-9$

$JK=3x$

$KL=1x+4$

The relationship between the given parameters is:

$JL = JK + KL$

Substitute 5x - 9 for JL; 3x for JK and 1x + 4 for KL

$5x - 9 = 3x + 1x + 4$

Collect Like Terms

$5x - 3x - 1x = 9 + 4$

$x = 13$

Now that the value of x has been solved, we can easily get the numeric values of JK, KL and JL by substituting 13 for x

So;

$JL=5x-9$

$JL=5*13-9$

$JL=65 - 9$

$JL=56$

$JK=3x$

$JK=3 * 13$

$JK=39$

$KL=1x+4$

$KL=1 * 13+4$

$KL=13+4$

$KL=17$

Hence;

$JL=56$ $JK=39$ $KL=17$

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Step-by-step explanation:

The options are missing; However, I'll simplify the given expression.

Given

$\frac{\sqrt[3]{32x^3y^6}}{\sqrt[3]{2x^9y^2} }$

Required

Write Equivalent Expression

To solve this expression, we'll make use of laws of indices throughout.

From laws of indices $\sqrt[n]{a} = a^{\frac{1}{n}}$

So,

$\frac{\sqrt[3]{32x^3y^6}}{\sqrt[3]{2x^9y^2} }$ gives

$\frac{(32x^3y^6)^{\frac{1}{3}}}{(2x^9y^2)^\frac{1}{3}}$

Also from laws of indices

$(ab)^n = a^nb^n$

So, the above expression can be further simplified to

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Substitute $2^5$ for 32

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This law can be applied to the expression above;

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