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2 years ago

9

PLEASE HELP PLEASE HELP PLEASE

1 answer:

TEA [102]2 years ago

5 0

Answer:

1:1

Step-by-step explanation:

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Answer:

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Step-by-step explanation:

If you divide 240 by 4 you should get $60 a day. If I did my math correctly the answers should be right.

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3 years ago

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3 years ago

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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 <em>sum and difference of cubes</em>, 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 <u><em>perfect cube numbers</em></u> (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 <em>sum and difference of cubes </em>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 <em>sum of cubes.</em>

<em>Sum of cubes</em>

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

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

If you are solving the equation by factoring, which of the following equations would you use the zero product property on?

Artist 52 [7]

Answer:

(x + 5)(x + 10) = 0

Step-by-step explanation:

When you solve for the equation: $\frac{1}{x-4}-\frac{14}{x+2}+1=0$ , you get that the answer is $x=10,\:x=5$ .

<em>Hope this helped!!</em>

8 0

3 years ago

Write a quadratic function in standard form with zeros 1 and -10

Licemer1 [7]

Answer:

f(x)=x^2+9x-10

Step-by-step explanation:

<u>Standard Form of Quadratic Function</u>

The standard form of a quadratic function is:

$f(x)=ax^2+bx+c$

where a,b, and c are constants.

The factored form of a quadratic equation is:

$f(x)=a(x-\alpha)(x-\beta)$

Where $\alpha$ and $\beta$ are the roots or zeros of f, and a is constant.

We know the zeros of the function are 1 and -10. The function is:

$f(x)=a(x-1)(x-(-10))$

$f(x)=a(x-1)(x+10)$

Operating:

$f(x)=a(x^2+10x-x-10)$

Joining like terms:

$f(x)=a(x^2+9x-10)$

Since we are not given any more restrictions, we can choose the value of a=1, thus. the required function is:

$\boxed{f(x)=x^2+9x-10}$

6 0

3 years ago