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3 years ago

6

eleven times one number is 9 more than 8 times another number. the first number minus four is half the second number. find the n

umbers.

1 answer:

andreev551 [17]3 years ago

6 0

Answer:

haha...the first number is 55 & the second number is 102.

if am right, pls sende more of this question or even more complicated quest of this type.

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What five-digit positive integer with an 8 in the ten-thousands place is the cube of an integer?

Sloan [31]

Answer:

44

Step-by-step explanation:

44^3 = 85,184

five digit positive

eight in the ten thousand place

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3 years ago

What numbers complete this equation?

Morgarella [4.7K]

Answer:

5 and 10

Step-by-step explanation:

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3 years ago

Mr. Stewart said that he would pay $750 for a bedroom suite. The suite Mrs. Stewart wants costs $250 less than twice as much mon

xxMikexx [17]

Answer:

$1,250

Step-by-step explanation:

Mr. Stewart will spend $750

Mrs. Steward wants $250 less than twice that

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brainliest? please and thank you.

3 0

3 years ago

Let <img src="https://tex.z-dn.net/?f=r%28x%29%20%3D%20%5Cfrac%7B8x-x%5E%7B2%7D%20%7D%7Bx%5E%7B4%7D-64x%5E%7B2%7D%7D" id="TexFor

Reil [10]

Answer:

<em>x=8</em>

Step-by-step explanation:

<u>Discontinuity of a Function</u>

We can find some functions whose graphs cannot be plotted in one stroke. It can be a hole or a vertical asymptote or a jump. To find a possible hole in a rational function, we must set both numerator and denominator to 0 independently. If a common point is found, it's a candidate for a hole if the function could eventually be redefined as continuous.

Let's find the zeros of the numerator

$8x-x^2=0$

Factoring

$x(8-x)=0$

We find two solutions: x=0, x=8

Let's find the zeros of the denominator

$x^4-64x^2=0$

Factoring

$x^2(x-8)(x+8)=0$

We find three roots: x=0, x=8, x=-8

There are two common points where the function can have holes, those are

$x=0,\ x=8$

We are not sure if those values are holes or not until we find the limits

$\displaystyle \lim\limits_{x \rightarrow 8}\frac{x(8-x)}{x^2(x-8)(x+8)}$

Simplifying

$\displaystyle =\lim\limits_{x \rightarrow 8}-\frac{1}{x(x+8)}$

$\displaystyle =-\frac{1}{128}$

Since the limit exists, the function can be redefined to cover up the hole. Now let's find the limit in x=0

$\displaystyle \lim\limits_{x \rightarrow 0}\frac{x(8-x)}{x^2(x-8)(x+8)}$

Simplifying

$\displaystyle =\lim\limits_{x \rightarrow 0}-\frac{1}{x(x+8)}$

$\displaystyle =-\frac{1}{0}=-\infty$

The limit does not exist and goes to infinity, it's not a hole, thus the only hole occurs when x=8

5 0

3 years ago

Put the following equation of a line into slope-intercept form, simplifying all fractions 2x-2y=14

hoa [83]

Answer:

y = x - 7

Step-by-step explanation:

8 0

3 years ago

Read 2 more answers