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2 years ago

Anyone know where I can get these answers? (There’s also a back but I can’t send it)

Mathematics

1 answer:

Nutka1998 [239]2 years ago

4 0

Answer:

See below

Step-by-step explanation:

<u>Covert equations in standard form into slope-intercept form of y = mx + b</u>

x - y = 3 ⇒ y = x - 3

5x + 3y = 6 ⇒ 3y = -5x + 6 ⇒ y = -5/3x + 2

x - 2y = 2 ⇒ 2y = x - 2 ⇒ y = 1/2x - 1

3x + 2y = -8 ⇒ 2y = -3x - 8 ⇒ y = -3/2x - 4

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Are there any vertical, complimentary, or any supplementary angles here ? And label for ex. 1 supplementary

kumpel [21]

<u>Supplementary:</u> 1 and 2 & 3 and 4

Supplementary angles: A pair of angles whose sum is equal to 180°.

<u />

<u>Vertical:</u> 1 and 3 & 2 and 4

Vertical angles: A pair of opposite angles that are made by two intersecting lines.

<u>Complimentary:</u> None

Complimentary angles: A pair of angles whose sum is equal to 90°.

Hope this helps! :D

7 0

2 years ago

I need help on this problem really I would be grateful

leonid [27]

Answer:

Step-by-step explanation:

The answer is A

4 0

2 years ago

Compare. Write <,>,or =.<br> 250 lb __ 0.25 T

Llana [10]

Answer: 0.25 T is bigger

Step-by-step explanation:

0.25 T = 500 lb

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3 years ago

Determine the domain:<br><br><img src="https://tex.z-dn.net/?f=f%28x%29%20%3D%20%20%5Cfrac%7Bln%28ln%28x%20%2B%201%29%29%7D%7Be%

Xelga [282]

The denominator cannot be zero, so

$e^x - 9 = 0 \implies e^x = 9 \implies x = \ln(9)$

is not in the domain of $f(x)$ .

$\ln(x)$ is defined only for $x>0$ , and we have

$\ln(x+1) > 0 \implies e^{\ln(x+1)} > e^0 \implies x+1 > 1 \implies x>0$

so there is no issue here.

By the same token, we need to have

$x+1 > 0 \implies x > -1$

Taking all the exclusions together, we find the domain of $f(x)$ is the set

$\left\{ x \in \Bbb R \mid x > 0 \text{ and } x \neq \ln(9)\right\}$

or equivalently, the interval $(0,\ln(9))\cup(\ln(9),\infty)$ .

3 0

1 year ago

Please help me with this and thank you

Rainbow [258]

Answer

1 ⅝

working :-

_________

8 / 13 \ 1

-8

____

= 1⅝

5 0

3 years ago