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kifflom [539]
3 years ago
11

What is the height of the triangle 35 and 819

Mathematics
1 answer:
kakasveta [241]3 years ago
4 0
I don’t know what answers you have but I think it’s 23.4
You might be interested in
Solve |3k-2|=2|k+12|
Stella [2.4K]

Answer:

k = 26   or   k = -\dfrac{22}{5}

Step-by-step explanation:

|3k - 2| = 2|k + 12|

\dfrac{|3k - 2|}{|k + 12|} = 2

|\dfrac{3k - 2}{k + 12}| = 2

3k - 2 = 2(k + 12)   or   3k - 2 = -2(k + 12)

3k - 2 = 2k + 24   or   3k - 2 = -2k - 24

k = 26   or   5k = -22

k = 26   or   k = -\dfrac{22}{5}

7 0
3 years ago
The solution to the inequality 5x - 4 > 21 is represented by which number line?
andrew-mc [135]

Answer:

Step-by-step explanation:

Start with the given inequality and solve it for x.

First, add 4 to both sides:  5x > 25

Next, divide both sides by 5:  x > 5

The solution (set) is x > 5.

7 0
3 years ago
Find the domain of the function y = 3 tan(23x)
solmaris [256]

Answer:

\mathbb{R} \backslash \displaystyle \left\lbrace \left. \frac{1}{23}\, \left(k\, \pi + \frac{\pi}{2}\right)  \; \right| k \in \mathbb{Z}  \right\rbrace.

In other words, the x in f(x) = 3\, \tan(23\, x) could be any real number as long as x \ne \displaystyle \frac{1}{23}\, \left(k\, \pi + \frac{\pi}{2}\right) for all integer k (including negative integers.)

Step-by-step explanation:

The tangent function y = \tan(x) has a real value for real inputs x as long as the input x \ne \displaystyle k\, \pi + \frac{\pi}{2} for all integer k.

Hence, the domain of the original tangent function is \mathbb{R} \backslash \displaystyle \left\lbrace \left. \left(k\, \pi + \frac{\pi}{2}\right)  \; \right| k \in \mathbb{Z}  \right\rbrace.

On the other hand, in the function f(x) = 3\, \tan(23\, x), the input to the tangent function is replaced with (23\, x).

The transformed tangent function \tan(23\, x) would have a real value as long as its input (23\, x) ensures that 23\, x\ne \displaystyle k\, \pi + \frac{\pi}{2} for all integer k.

In other words, \tan(23\, x) would have a real value as long as x\ne \displaystyle \frac{1}{23} \, \left(k\, \pi + \frac{\pi}{2}\right).

Accordingly, the domain of f(x) = 3\, \tan(23\, x) would be \mathbb{R} \backslash \displaystyle \left\lbrace \left. \frac{1}{23}\, \left(k\, \pi + \frac{\pi}{2}\right)  \; \right| k \in \mathbb{Z}  \right\rbrace.

4 0
3 years ago
Darrel divided 575 by 14 by using partial quotients, what is the quotient?
Usimov [2.4K]
143r3 (143 remainder 3)
3 0
3 years ago
Can some pls help me find the slope of this line i really really need help​
Damm [24]

Answer:

Slope = Rise over Run

Rise = 2

Run = 3

Slope = 2/3

Let me know  if this helps!

4 0
3 years ago
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