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Dominik [7]
2 years ago
9

△OMT≅△HZE. If \text{m}\angle O = 51^{\circ}m∠O=51

Mathematics
1 answer:
professor190 [17]2 years ago
6 0

so true

just do it mega easy

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Help I’m trying to finish this math tests
Simora [160]

Answer:

2...................

8 0
3 years ago
Which of the following are vertical asymptotes of the function y = 2cot(3x) + 4? Check all that apply. A.x = pi/3 B.x = +/- pi/2
Kisachek [45]
Vertical asymptotes occur when the denominator of a rational is 0, whilst not zeroing out the numerator, making the rational, undefined, in this case

\bf y=2cot(3x)+4\implies y=2\cdot \cfrac{cos(3x)}{sin(3x)}+4\impliedby \textit{if \underline{sin(3x)} turns to 0}\\\\
-------------------------------\\\\
\textit{let's check}
\\\\\\
A)\qquad \cfrac{cos(3x)}{sin\left( 3\frac{\pi }{3} \right)}\implies \cfrac{cos(3x)}{sin\left( \pi \right)}\implies \cfrac{cos(3x)}{0}\impliedby unde f ined

\bf B)\qquad \cfrac{cos(3x)}{sin\left( 3\frac{\pm\pi }{2} \right)}\implies\cfrac{cos(3x)}{sin\left( \frac{\pm3\pi }{2} \right)}\implies \cfrac{cos(3x)}{\pm 1}\implies \pm cos(3x)
\\\\\\
C)\qquad \cfrac{cos(3x)}{sin\left( 3(2\pi )\right)}\implies \cfrac{cos(3x)}{sin(6\pi )}\implies \cfrac{cos(3x)}{0}\impliedby unde f ined
\\\\\\
D)\qquad \cfrac{cos(3x)}{sin(3(0))}\implies \cfrac{cos(3x)}{sin(0)}\implies \cfrac{cos(3x)}{0}\impliedby unde f ined
6 0
3 years ago
Read 2 more answers
What is the mean of the data set?<br> 108, 305, 252, 113, 191
irinina [24]

Go to the third pic for your answer. I got problem 4 wrong but it shows the correct answer.

7 0
3 years ago
Read 2 more answers
140÷(x+9)+(x+4)² when x =1 show work​
Dmitrij [34]

Answer:

39

Step-by-step explanation:

If x=1, then it's equal to 140/(1+9, or 10)+(1+4, or 5)^2, and that is equal to 140/10+5^2, or 140/10+25, and 140/10=14, and 14+25=39

brainliest pls

4 0
2 years ago
3. Which polynomial is equal to
Nataly_w [17]

Which polynomial is equal to  (-3x^2 + 2x - 3) subtracted from  (x^3 - x^2 + 3x)?

<h3><u><em>Answer:</em></u></h3>

The polynomial equal to (-3x^2 + 2x - 3) subtracted from  (x^3 - x^2 + 3x) is x^3 + 2x^2 + x + 3

<h3><u><em>Solution:</em></u></h3>

Given that two polynomials are: (-3x^2 + 2x - 3) and (x^3 - x^2 + 3x)

We have to find the result when (-3x^2 + 2x - 3) is subtracted from  (x^3 - x^2 + 3x)

In basic arithmetic operations,

when "a" is subtracted from "b" , the result is b - a

Similarly,

When (-3x^2 + 2x - 3) is subtracted from  (x^3 - x^2 + 3x) , the result is:

\rightarrow (x^3 - x^2 + 3x) - (-3x^2 + 2x - 3)

Let us solve the above expression

<em><u>There are two simple rules to remember: </u></em>

  • When you multiply a negative number by a positive number then the product is always negative.
  • When you multiply two negative numbers or two positive numbers then the product is always positive.

So the above expression becomes:

\rightarrow (x^3 - x^2 + 3x) + 3x^2 -2x + 3

Removing the brackets we get,

\rightarrow x^3 - x^2 + 3x + 3x^2 -2x + 3

Combining the like terms,

\rightarrow x^3 -x^2 + 3x^2 + 3x - 2x + 3

\rightarrow x^3 + 2x^2 + x + 3

Thus the resulting polynomial is found

5 0
3 years ago
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