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riadik2000 [5.3K]

3 years ago

15

Rewrite the expression 13d - (-9d - 4) without parentheses

2 answers:

dimaraw [331]3 years ago

4 0

13d-(-9d-4)
Transfer minus sign to all numbers in parenthesis. = 13d +9d +4 =

Answer: 22d+4

Arturiano [62]3 years ago

3 0

Answer:

13d- -9d-4

Step-by-step explanation:

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There are 42 students in a class.

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Is the ordered pair (2,1) the solution the the equation y=1-x

AleksAgata [21]

(x,y)
(2,1)
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4 0

3 years ago

The largest set of x values satisfying

velikii [3]

Answer:

If $p\ge -\dfrac{1}{3},$ then $x>\dfrac{2}{3}+p$ and $mn=\dfrac{2}{3}+p=\left(\dfrac{2}{3}+p\right)\cdot 1,\ \ m+n=\dfrac{5}{3}+p$

Step-by-step explanation:

Solve two inequalities for x.

1. $2,018x-p$

Separate terms with x and without x into two sides:

$2,018x-2,020x$

Multiply by -1:

$2x>-2p\\ \\x>-p$

2. $7x+3p$

Separate terms with x and without x into two sides:

$7x-10x$

Multiply by -1:

$3x>2+3p\\ \\x>\dfrac{2}{3}+p$

Find the largest set of x values satisfying both inequalities:

$x>-p\\ \\x>\dfrac{2}{3}+p$

If

$-p>\dfrac{2}{3}+p\\ \\-2p>\dfrac{2}{3}\\ \\2p$

then $x>-p$ and $mn=-p=p\cdot (-1),\ m+n=p-1.$ In this case both m and n are negative.

If $p>-\dfrac{1}{3},$ then $x>\dfrac{2}{3}+p$ and $mn=\dfrac{2}{3}+p=\left(\dfrac{2}{3}+p\right)\cdot 1,\ \ m+n=\dfrac{5}{3}+p$

If $p=-\dfrac{1}{3},$ then $x>\dfrac{1}{3}$ and $mn=\dfrac{1}{3}=\dfrac{1}{3}\cdot 1,\ \ m+n=\dfrac{4}{3}$

7 0

3 years ago