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3 years ago

Can Someone please help me on all four of these.

Mathematics

1 answer:

bonufazy [111]3 years ago

3 0

1st one is x > 1
2nd one is x < -4
3rd one is x < 70
4th one is x > -2

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What’s the domain of the graph?

KATRIN_1 [288]

Answer:

(-10, 6)

Step-by-step explanation:

To find the domain of a function, you just look at where the x starts and the x ends. since there are points at either end you do the points. if it were to go on forever like a line then the domain would be from Infinity to a point or Infinity to Infinity and etc.

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3 years ago

Read 2 more answers

Multiply

Dmitriy789 [7]

Answer:

The product of given two fractions is

$\frac{4(x+2)}{(x-1)(x+3)}$

Therefore $\frac{2(2+3)}{x(x-1)}\times \frac{4x(x+2)}{10(x+3)}=\frac{4(x+2)}{(x-1)(x+3)}$

Step-by-step explanation:

Given expression is

$\frac{2(2+3)}{x(x-1)}\times \frac{4x(x+2)}{10(x+3)}$

To find the product of two given fractions as below :

$\frac{2(2+3)}{x(x-1)}\times \frac{4x(x+2)}{10(x+3)}=\frac{2(5)}{x(x-1)}\times \frac{2x(x+2)}{5(x+3)}$

$=\frac{10}{x(x-1)}\times \frac{2x(x+2)}{5(x+3)}$

$=\frac{20x(x+2)}{5x(x-1)(x+3)}$

$=\frac{20x^2+40x}{(5x^2-5x)(x+3)}$ (multiply each term in the factor to each term in the another factor )

$=\frac{20x^2+40x}{5x^3+15x^2-5x^2-15x}$ ( adding the like terms )

$=\frac{20x^2+40x}{5x^3+10x^2-15x}$

$=\frac{5x(4x+8)}{5x(x^2+2x-3)}$

$=\frac{4(x+2)}{(x-1)(x+3)}$

Therefore $\frac{2(2+3)}{x(x-1)}\times \frac{4x(x+2)}{10(x+3)}=\frac{4(x+2)}{(x-1)(x+3)}$

Therefore the product of given two fractions is $\frac{4(x+2)}{(x-1)(x+3)}$

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3 years ago

Carly owns a toy store. She sells teddy bears and charges $25 per bear. Her total monthly cost is $3765. How many bears does Car

Zinaida [17]

Answer:

divide 3765 by 25 to get how many bears she sold, then divide that by 3765

3 0

2 years ago

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When a number is added to six times its square, the result is twelve. Find the number.

7nadin3 [17]

Let's represent our number as x.
x is being added to 6 * x², and it equals 12
$x + 6x^{2}=12\\6x^{2}+x-12=0$
Factor:
$6x^{2}+x-12=0\\(2x+3)(3x-4)=0$
Solve for x:
$(2x+3)=0\\2x=-3\\x=\frac{-3}{2}\\\\(3x-4)=0\\3x=4\\x=\frac{4}{3}$

Therefore, x = -3/2 and 4/3.

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3 years ago

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If a point is the midpoint of a segment, then it divides the segment into two congruent segments. R is the midpoint of QS. Concl

DIA [1.3K]

Answer:

All given information leads to conclusion described on statement. ( $QR = RS$ )

Step-by-step explanation:

The strategy of problem consist in make a mathematical demonstration from given information. There are two assumptions: i) $R$ is a point of segment $QS$ and ii) $R$ is the midpoint of $QS$ . We proceed to perform the demonstration:

1) $R\in QS$ , $R$ is the midpoint of $QS$ . Given.

2) $Q - R = R -S$ Definition of midpoint.

3) $QR = RS$ Definition of segment/Result.

All given information leads to conclusion described on statement. ( $QR = RS$ )

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3 years ago