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jarptica [38.1K]
3 years ago
12

What is -20.3-14.34?

Mathematics
1 answer:
djyliett [7]3 years ago
8 0

Answer:

-34.64

Step-by-step explanation:

-20.3-14.34

-34.64

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NEED HELP ASAP ILL MAKE BRAINLIEST PLEASE HELP
nordsb [41]

Answer:

33% (B)

Step-by-step explanation

It increased by 150, and 150 is 1/3 of 450, so 33% Hope this helped! Good luck!

6 0
3 years ago
What is the solution to -55+q=7<br><br> A. a=-62<br><br> B. q=62<br><br> C. q=42<br><br> D. q=-42
Scorpion4ik [409]

Answer:

C, as q = 62.

Step-by-step explanation:

When you have an equation, your goal is to get the letter you are solving for alone. To do this, you employ a simple rule: what you do to one side of the equals sign, you must do the other.

To isolate q in -55 + q = 7, you must add 55 to the left side. q is now alone. However, because we added 55 to the left side, we must also do it to the right! 7 + 55 = 62, so the new right side is 62. Hence, we get to this:

q = 62

The answer is now in plain sight!

4 0
3 years ago
Read 2 more answers
What is x if the box is 70in to the 3rd power?
koban [17]

Step-by-step explanation:

I doint know the answer but have a great day today or tomorrow

8 0
3 years ago
Select the functions that have a value of 0.
Sphinxa [80]

Answer:

<h2>cos90° = 0°, tan0° = 0°,</h2><h2>cos(-90°) = 0°, cot270° = 0°</h2>

Step-by-step explanation:

k\in\mathbb{Z}\\\\\sin x=0\iff x=180^ok\\\\\cos x=0\iff x=90^o+180^ok\\\\\tan x=0\iff x=180^ok\\\\\cot x=0\iff x=90^o+180^ok\\\\\csc x\neq0\ \text{for}\ x\in\mathbb{R}-\{180^ok\}\\========================\\\\\sin270^o\neq0\\\\\cos90^o=\cos(90^o+180^o\cdot0)=0\\\\\tan0^o=\tan(180^o\cdot0)=0\\\\\csc(-180^o)\neq0\\\\\cos(-90^o)=\cos(90^o-180^o)=\cos(90^o+180^o\cdot(-1))=0\\\\\cot270^o=\cot(90^o+180^o)=0

3 0
3 years ago
If f x equals x -3 + g x equals 1 - x 2 - 9 find ( f og) (x and give any restrictions on its domain
evablogger [386]
\begin{gathered} f(x)=x-3 \\ g(x)=\frac{1}{x^2-9} \\ (\text{fog)(x)}=\frac{1}{x^2-9}-3 \end{gathered}

the denominator cannot be zero, because the division by zero is not defined, therefore:

\begin{gathered} x^2-9=0 \\ \text{Solving for x:} \\ x^2=9 \\ \sqrt[]{x^2}=\sqrt[]{9} \\ x=\pm3 \end{gathered}

Therefore the domain of (f o g)(x) is:

D\colon(-\infty,-3)\cup(-3,3)\cup(3,\infty)

3 0
1 year ago
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