<img src="https://tex.z-dn.net/?f=5%20%7Bx%7D%5E3%20-%2020%20%7Bx%7D%5E%7B2%7D%20%3D%200" id="TexFormula1" title="5 {x}^3 - 20 {

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3 years ago

13

x}^{2} = 0" alt="5 {x}^3 - 20 {x}^{2} = 0" align="absmiddle" class="latex-formula">

1 answer:

adell [148]3 years ago

7 0

$5x^{2} \times (x - 4) = 0 \\ x^{2} \times (x - 4) = 0 \\ {x}^{2} = 0 \: x = 0 \\ x - 4 = 0 \: x = 4$

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Find the solution of 5 times the square root of the quantity of x plus 7 equals negative 10, and determine if it is an extraneou

PilotLPTM [1.2K]

5sqrt(x+7)=-10
Dividing both sides by 5:
5sqrt(x+7)/5=-10/5
sqrt(x+7)=-2
Squaring both sides of the equation:
[sqrt(x+7)]^2=(-2)^2
x+7=4
x+7-7=4-7
x=-3

Replacing in the original equation:
5sqrt(x+7)=-10
5sqrt(-3+7)=-10
5sqrt(4)=-10
5(2)=-10
10=-10 It's not equal, then x=-3 is an extraneous solution

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3 years ago

The sum of the measures of two of the angles formed by the intersection of two lines is 114°. Find the measures of all angles. I

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Answer:

Measures of all angles are 57°, 57°, 123°, 123°

Step-by-step explanation:

Given:

Sum of two angles = 114°

Find:

Measures of all angles

Computation:

Sum of all 4 angles = 360°

Sum of remain 2 angles = 360° - 114°

Sum of remain 2 angles = 246°

So, first two angles = 114° / 2 = 57°

Second two angles = 246° / 2 = 123°

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3 years ago

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4 years ago

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#1. bring the fraction: b / 7a^2c to a denominator of 35a^3c^3

WITCHER [35]

Part A: Term $5ac$ is used to bring denominator of $35a^{3} c^{3}$

Part B: Term $a+4$ is used to bring denominator of $16-a^{2}$

Explanation:

Part A: To bring the fraction $\frac{b}{7a^{2}c }$ to a denominator of $35a^{3} c^{3}$ , we need to multiply the denominator by the term $5ac$

The term $5ac$ is determined by $\frac{35a^{3} c^{3}}{7a^{2}c }}$ which equals $5ac$

Thus, the fraction becomes

$\frac{b}{7a^{2}c }\times\frac{5ac}{5ac} =\frac{5abc}{35a^{3} c^{3}}$

Thus, the term $5ac$ is used to bring denominator of $35a^{3} c^{3}$

Part B: To bring the fraction $\frac{a}{a-4}$ to a denominator of $16-a^{2}$ , we need to multiply the denominator by $a+4$

Thus, we have,

$\frac{a}{a-4}\times\frac{a+4}{a+4} =\frac{a(a+4)}{16-a^{2}}$

Thus, the term $a+4$ is used to bring denominator of $16-a^{2}$

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Answer:

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Step-by-step explanation:

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3 years ago

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