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2 years ago

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Andrew is searching for a cup and discovers that he has 20 plates for every 5 pairs of cups. if he has 40 plates, how many pairs

of cups does he have?

1 answer:

irakobra [83]2 years ago

3 0

You get 40 by multiplying 20 by 2 so all you have to do is multiply 5 by 2 also and you get 10. So the answer is 10 pairs of cups.

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Ludmilka [50]

The solution of your problem is shown on the picture below.

3 0

2 years ago

Answer C and D please

DENIUS [597]

Answer:

C. solution:

3(3c-2) = 5(2c-1)

or, 9c-6 = 10c-5

or, -6+5 =<em> </em>10c-9c

or, -1 = 1c

Hence, c = -1.

D. solution:

5(5-2a) = 4(6-a)

or, 25-10a = 24 - 4a

or, 25-24 = -4a+10a

or, 1 = 6a

or, 1/6 = a

Hence, a = 1/6.

6 0

2 years ago

(PLEASE HELP!!)

Flauer [41]

Answer:

b

Step-by-step explanation:

5 0

2 years ago

Read 2 more answers

Suppose that the functions g and h are defined for all real numbers r as follows. gx) -4x +5 h (x) = 6x write the expressions fo

aleksandr82 [10.1K]

Answer: Our required values would be -10x+5, 2x+5 and -25.

Step-by-step explanation:

Since we have given that

g(x) = -4x+5

and

h(x) = 6x

We need to find (g-h)(x) and (g+h)(x).

So, (g-h)(x) is given by

$g(x)-h(x)\\\\=-4x+5-6x\\\\=-10x+5$

and (g+h)(x) is given by

$g(x)+h(x)\\\\=-4x+5+6x\\\\=2x+5$

and (g-h)(3) is given by

$-10(3)+5\\\\=-30+5\\\\=-25$

Hence, our required values would be -10x+5, 2x+5 and -25.

5 0

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

3 years ago