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3 years ago

Find the value of x.

Mathematics

1 answer:

vazorg [7]3 years ago

4 0

Step-by-step explanation:

Espero que te ajude de mucho

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The manager of an apartment building increases the monthly

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The answer is 513

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3 years ago

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Can you help find the quotient for 4/5 ÷ 10

pochemuha

Answer:

2/25

Step-by-step explanation:

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3 years ago

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Find the quotient of n(x) = 6x^5y^7 and m(x) = 2x^3y

likoan [24]

Answer:

a. Yes

b. Yes

c. Yes

d. Degree 8

Step-by-step explanation:

a. Yes, n(x) is a polynomial of one single term (also called monomial) because it contains variables raised to positive integers.

b. Yes, m(x) is a polynomial of also one single term (also called monomial) because it contains variables raised to positive integers.

c. The quotient of n(x) / m(x) can be reduced to a polynomial of one single term as follows:

$\frac{n(x)}{m(x)} =\frac{6\,x^5\,y^7}{2\,x^3\,y} =3\,x^2\,y^6$

which as can be seen, also contains variables raised to positive integers.

d. The degree of the polynomial resultant is the addition of the powers of all variables present (x and y) which results in: 2 + 6 = 8

Therefore the degree of this polynomial is 8.

6 0

3 years ago

The gift shop is having a sale. your friend sells half her stock of keyrings every hour. she has 24 boxes at the start of the sa

Bingel [31]

To solve for the given problem above:

Let x be the number of boxes left after 3 hours

3 * ½ X = 24

3/2 X = 24

X = 24 * 2/3

X = 16

The correct answer is 16.

3 0

2 years ago

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NO LINKS!! Use the method of substitution to solve the system. (If there's no solution, enter no solution). Part 11z

Roman55 [17]

Answer:

$(x,y)=\left(\; \boxed{-1,-8} \; \right)\quad \textsf{(smaller $x$-value)}$

$(x,y)=\left(\; \boxed{5,16} \; \right)\quad \textsf{(larger $x$-value)}$

Step-by-step explanation:

Given system of equations:

$\begin{cases}y=x^2-9\\y=4x-4\end{cases}$

To solve by the method of substitution, substitute the first equation into the second equation and rearrange so that the equation equals zero:

$\begin{aligned}x^2-9&=4x-4\\x^2-4x-9&=-4\\x^2-4x-5&=0\end{aligned}$

Factor the quadratic:

$\begin{aligned}x^2-4x-5&=0\\x^2-5x+x-5&=0\\x(x-5)+1(x-5)&=0\\(x+1)(x-5)&=0\end{aligned}$

Apply the <u>zero-product property</u> and solve for x:

$\implies x+1=0 \implies x=-1$

$\implies x-5=0 \implies x=5$

Substitute the found values of x into the <u>second equation</u> and solve for y:

$\begin{aligned}x=-1 \implies y&=4(-1)-4\\y&=-4-4\\y&=-8\end{aligned}$

$\begin{aligned}x=5 \implies y&=4(5)-4\\y&=20-4\\y&=16\end{aligned}$

Therefore, the solutions are:

$(x,y)=\left(\; \boxed{-1,-8} \; \right)\quad \textsf{(smaller $x$-value)}$

$(x,y)=\left(\; \boxed{5,16} \; \right)\quad \textsf{(larger $x$-value)}$

6 0

11 months ago

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