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3 years ago

13

Need help please -f-5(2f-3)

1 answer:

lana [24]3 years ago

3 0

-f - 5(2f - 3)

1. -5 * 2f = -10f
2. -5 * -3 = 15
3. -f - 10f + 15
4.Add like terms,
-f - 10f = -11f.
5.-11f + 15

Answer: -11f + 15

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All of the following are equivalent to 2(2 a+ b) + 8 except __________

Masteriza [31]

2 * (2a + b) + 8 = 4a + 2b + 8
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A. =
B. =
C. ≠ 4* (a + b + 2) = 4a + 4b + 8
D. =

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3 years ago

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weqwewe [10]

Answer:

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Step-by-step explanation:

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3 years ago

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What is the equation for the line with a slope of - 2/3 that passes through (0, -2)

viktelen [127]

Answer:

y= -2/3x-2

Step-by-step explanation:

Since we have a point and a slope, we can use the point slope formula

$y-y_{1} =m(x-x_{1})$

m is the slope. In this case, m is -2/3

y1 is the y coordinate of the point. In this case, it is -2

x1 is the x coordinate. In this case it is 0.

We can substitute these values into the formula

y- -2= -2/3(x-0)

Distribute -2/3

y+2 = -2/3*x + -2/3*0

y+2= -2/3x+0

Subtract 2 from both sides. We want to get y by itself. This is because we want our line to be in y=mx+b form.

y+2-2= -2/3x+0-2

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3 years ago

Write a solution in Interval Notation - (you don't have to help me on all, 1 or 2 is fine c: )

Gemiola [76]

QUESTION 1

The given inequality is

$|m|-2>0$

We group like terms to get,

$|m|>2$

This implies that,

$-m>2$ or $m>2$ .

We simplify the inequality to get,

$m$ or $m>2$ .

We can write this interval notation to get,

$(-\infty,-2)\cup (2,+\infty)$ .

QUESTION 2

$|x-4|-3\:>\:5$ .

We group like terms to get,

$|x-4|\:>\:5+3$ .

$|x-4|\:>\:8$

We split the absolute value sign to get,

$-(x-4)\:>\:8$ or $x-4\:>\:8$

This implies that,

$x-4\:$ or $x-4\:>\:8$

$x\:$ or $x\:>\:8+4$

$x\:$ or $x\:>\:12$

We can write this interval notation to get,

$(-\infty,-4)\cup (12,+\infty)$ .

QUESTION 3

The given inequality is

$|6+9x|\leq 24$

We split the absolute value sign to obtain,

$-(6+9x)\leq 24$ or $(6+9x)\leq 24$

This simplifies to

$6+9x\ge -24$ and $6+9x\leq 24$

$9x\ge -24-6$ and $9x\leq 24-6$

$9x\ge -30$ and $9x\leq 18$

$x\ge -\frac{10}{3}$ and $x\leq 2$

$-\frac{10}{3}\leq x\leq2$

We write this in interval form to get,

$[-\frac{10}{3},2]$

QUESTION 4

The given inequality is

$|1-5a|>29$

We split the absolute value sign to get,

$-(1-5a)>29$ or $1-5a>29$

This simplifies to,

$1-5a\:$ or $1-5a\:>\:29$

This implies that,

$-5a\:$ or $-5a\:>\:29-1$

$-5a\:$ or $-5a\:>\:28$

$a\:>\:6$ or $a\:$

We write this in interval notation to get,

$(-\infty,-\frac{28}{5})\cup (6,+\infty)$

7 0

2 years ago

Me ayudan plis es para hoy

sladkih [1.3K]

Answer:

si yo te ayudo pero no puedo ver bien la foto

6 0

3 years ago

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