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3 years ago

Write a quadratic equation to represent a function with the vertex (-4, 1)

Mathematics

1 answer:

pishuonlain [190]3 years ago

6 0

Answer:

y=a(x+4)^2 +1, for a real number a hat is different of 0.

Step-by-step explanation:

If we know only the vertex, there are many quandratic equations.

* Result: Vertex (m,n), the quadratic equation is y= a(x-m)^2 + n for a real number a that is different of 0.

Here, the vertex (-4,1), we have the equation: y=a(x+4)^2 +1, for a real number a hat is different of 0.

Hope you understand.

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3 years ago

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3 years ago

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2 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.

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

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

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3 years ago

Find the surface area of the rectangular and triangular prisms

harina [27]

Is there a picture to this????

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3 years ago