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dexar [7]
2 years ago
8

Plot the inequality x < 0​

Mathematics
1 answer:
Sholpan [36]2 years ago
7 0

Step-by-step explanation:

This is saying " Every number possible thats less than, but not 0" so you show that

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What is the answer to the equation: 8+10
alina1380 [7]

Answer:

18, the answer to 8+10 is 18, like you learn in 1st grade

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garik1379 [7]

Answer:

Step-by-step explanation:

This problem is just really long- it's trying to confuse you with complicated wording. Here's it demystified:

a)

t = 0 degrees Celsius

d = 1500 meters

Find the temperature, T.

b)

d = 300 meters

T = 26 degrees Celsius

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NOTE: The temperatures are different! T is the final temperature and t is the ground temperature.

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3 years ago
Is (-3, -9) a solution of the equation y = 3x ?
makkiz [27]

Answer:

Yes.

Step-by-step explanation:

-3 * 3 = -9

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Which graph represents the solution set of this inequality?
Maru [420]

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the answerrrr is cccccc

Step-by-step explanation:

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6 0
3 years ago
Solve the following equation by completing the square. 3x^2-3x-5=13
mr Goodwill [35]

we'll start off by grouping some

\bf 3x^2-3x-5=13\implies (3x^2-3x)-5=13\implies 3(x^2-x)-5=13 \\\\\\ 3(x^2-x)=18\implies (x^2-x)=\cfrac{18}{3}\implies (x^2-x)=6\implies (x^2-x+~?^2)=6

so we have a missing guy at the end in order to get the a perfect square trinomial from that group, hmmm, what is it anyway?

well, let's recall that a perfect square trinomial is

\bf \qquad \textit{perfect square trinomial} \\\\ (a\pm b)^2\implies a^2\pm \stackrel{\stackrel{\text{\small 2}\cdot \sqrt{\textit{\small a}^2}\cdot \sqrt{\textit{\small b}^2}}{\downarrow }}{2ab} + b^2

so we know that the middle term in the trinomial, is really 2 times the other two without the exponent, well, in our case, the middle term is just "x", well is really -x, but we'll add the minus later, we only use the positive coefficient and variable, so we'll use "x" to find the last term.

\bf \stackrel{\textit{middle term}}{2(x)(?)}=\stackrel{\textit{middle term}}{x}\implies ?=\cfrac{x}{2x}\implies ?=\cfrac{1}{2}

so, there's our fellow, however, let's recall that all we're doing is borrowing from our very good friend Mr Zero, 0, so if we add (1/2)², we also have to subtract (1/2)²

\bf \left( x^2 -x +\left[ \cfrac{1}{2} \right]^2-\left[ \cfrac{1}{2} \right]^2 \right)=6\implies \left( x^2 -x +\left[ \cfrac{1}{2} \right]^2 \right)-\left[ \cfrac{1}{2} \right]^2=6 \\\\\\ \left(x-\cfrac{1}{2} \right)^2=6+\cfrac{1}{4}\implies \left(x-\cfrac{1}{2} \right)^2=\cfrac{25}{4}\implies x-\cfrac{1}{2}=\sqrt{\cfrac{25}{4}} \\\\\\ x-\cfrac{1}{2}=\cfrac{\sqrt{25}}{\sqrt{4}}\implies x-\cfrac{1}{2}=\cfrac{5}{2}\implies x=\cfrac{5}{2}+\cfrac{1}{2}\implies x=\cfrac{6}{2}\implies \boxed{x=3}

6 0
3 years ago
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