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3 years ago

What is the location of point C? (-7, -2) (-2, -7) (-2, 7) (-7, 2)

Mathematics

2 answers:

rewona [7]3 years ago

5 0

Answer:

(-2,-7)

Step-by-step explanation:

x comes before Y

Lilit [14]3 years ago

5 0

Answer: (-2, -7).

Step-by-step explanation:

Find -2 on the x axis, then down that line to find point C at -7 on the y axis.

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Please help: what is 2+4+4+10? I will five brainliest

telo118 [61]

Answer:

It is 20

Step-by-step explanation:

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5 0

3 years ago

What is the factorization of 729^15+1000?

igomit [66]

Answer:

The factorization of $729x^{15} +1000$ is $(9x^{5} +10)(81x^{10} -90x^{5} +100)$

Step-by-step explanation:

This is a case of factorization by sum and difference of cubes, this type of factorization applies only in binomials of the form $(a^{3} +b^{3} )$ or $(a^{3} -b^{3})$ . It is easy to recognize because the coefficients of the terms are perfect cube numbers (which means numbers that have exact cubic root, such as 1, 8, 27, 64, 125, 216, 343, 512, 729, 1000, etc.) and the exponents of the letters a and b are multiples of three (such as 3, 6, 9, 12, 15, 18, etc.).

Let's solve the factorization of $729x^{15} +1000$ by using the sum and difference of cubes factorization.

1.) We calculate the cubic root of each term in the equation $729x^{15} +1000$ , and the exponent of the letter x is divided by 3.

$\sqrt[3]{729x^{15}} =9x^{5}$

$1000=10^{3}$ then $\sqrt[3]{10^{3}} =10$

So, we got that

$729x^{15} +1000=(9x^{5})^{3} + (10)^{3}$ which has the form of $(a^{3} +b^{3} )$ which means is a sum of cubes.

Sum of cubes

$(a^{3} +b^{3} )=(a+b)(a^{2} -ab+b^{2})$

with $a= 9x^{5}$ y $b=10$

2.) Solving the sum of cubes.

$(9x^{5})^{3} + (10)^{3}=(9x^{5} +10)((9x^{5})^{2}-(9x^{5})(10)+10^{2} )$

$(9x^{5})^{3} + (10)^{3}=(9x^{5} +10)(81x^{10}-90x^{5}+100)$

8 0

3 years ago

find the formula for a rational function that satisfies all of the following. its x-intercept is (-3,0). it has a vertical asymp

Norma-Jean [14]

Let that be

$\\ \rm\Rightarrow y=\dfrac{a(f(x))}{g(x)}$

Two vertical asymptotes at -1 and 0

$\\ \rm\Rightarrow y=\dfrac{a(f(x)}{x(x+1)}$

If we simply

$\\ \rm\Rightarrow y=\dfrac{a(f(x)}{x^2+1}$

Denominator has degree 2
Numerator should have degree as 2 and coefficient as 3 inorder to get horizontal asymptote y=3 means the quadratic equation should contain 3x²
But there should be a x intercept at -3 so one zeros should be -3

Find a equation