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3 years ago

1.Which coordinates are not of the vertices of the feasible region for the system of inequalities?

1 answer:

3 0

2. I think the answer is 24

subracting the last 2 inequalities we get

6x + 7y <= 42
3x + 2y <= 18 subtract:_-
3x + 5y <= 24

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Hello can someone help me simplify 4) and 5)? Thank you and please include steps :)

Leni [432]

Alrighty

remember some rules
$(ab)^c=(a^c)(b^c)$
and
$x^{-m}=\frac{1}{x^m}$
and
$x^0=1$ for all real values of x
and
$(a^b)^c=a^{bc}$
and
$(\frac{a}{b})^c=\frac{a^c}{b^c}$
and
(a/b)/(c/d)=(ad)/(bc)
and
$(a^b)(a^c)=a^{b+c}$
and
$\frac{a^m}{a^n}=a^{m-n}$
and don't forget pemdas
example: $-x^m=-1(x^m)$ but $(-x)^m=(-1)^m(x^m)$

so

4.
$(\frac{c^{-2}}{2})^{-2}=$

$\frac{(c^{-2})^{-2}}{2^{-2}}=$

$\frac{c^4}{\frac{1}{2^2}}=$

$\frac{c^4}{\frac{1}{4}}=$

$4c^4$

5.

$\frac{(-a)^4bc^5}{-a^2b^{-3}c^0}=$

$\frac{(-1)^4(a)^4bc^5}{-1(a^2)(\frac{1}{b^3}(1)}=$

$\frac{(1)a^4bc^5}{\frac{(-1)(a^2)}{b^3}}=$

$\frac{(a^4bc^5)b^3}{(-1)(a^2)}=$

$\frac{-a^4b^4c^5}{a^2}=$

$-a^2b^4c^5$

3 0

3 years ago

Pleaseee help me with number eight,, I don’t know how to do it at all.

iogann1982 [59]

1,200 because 4x 3 is 12 and 40 x 30=1,200

5 0

3 years ago

How do I do this Cp geometry

ryzh [129]

Whats your geometry problem? i dont understand the question

5 0

3 years ago

(3x+7)(3x-7) if necessary combine like terms

34kurt

Answer:

9x^2 -49

Step-by-step explanation:

We recognize that this is of the form

(a+b)(a-b) = a^2 - b^2

3x = a and 7 =b

(3x)^2 - 7^2

9x^2 -49

3 0

4 years ago

Expand this expression: 0.3(9x + 2)<br> Plz help

Vikentia [17]

Step-by-step explanation:

0.3(9x + 2) [Then we multiply to expand]

= 2.7x + 0.6 (Ans)

4 0

3 years ago