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ludmilkaskok [199]
3 years ago
11

Please help…………………………….

Mathematics
2 answers:
Inessa05 [86]3 years ago
5 0

if i'm correct its 7x

Alla [95]3 years ago
3 0

Answer:

Domen is [ -2 , infinitely) or can be written as

-2 <u><</u> x < infinitely in inequality form.

Step-by-step explanation:

if (x+2) is negative than function will be imaginary so to be real function (x+2) should be positive

hence answer is : Domen = -2 < x < infinitely or can be written as [ -2 , infinitely )

You might be interested in
Find 2x2 − 2z4 + y2 − x2 + z4 if x = −4, y = 3, and z = 2.
maw [93]
If you put in the substitutions it would be 2*2-(2*2*4)+(3*2)-(-4*2)+(2*4).
Simplified further by multiplying it would be 2*2-8+6--8+8
the negative 8 could be simplified further since the negatives cancel out so you'll have 2*2-8+6+8+8.
Then using the order of operations you would multiply the 2's together first to get 4 so you have 4-8+6+8+8. 
After that it's simple addition giving you an answer of -26.
I'm not sure if you're looking for the final answer or just the equation with substitutions but there's both.

3 0
3 years ago
What is this shape called?
shepuryov [24]

Answer:

This is a triangular prism

Step-by-step explanation:

6 0
3 years ago
If cos(θ) = 6/8 and θ is in the IV quadrant, then fine:
svetoff [14.1K]

Answer:

a) 1

b) \frac{4}{3}

c) = 1

Step-by-step explanation:

We are given the following in the question:

\cos \theta = \dfrac{6}{8}

θ is in the IV quadrant.

\sin^2 \theta + \cos^2 \theta = 1\\\\\sin \theta = \sqrt{1-\dfrac{36}{64}} = -\dfrac{2\sqrt7}{8}\\\\\tan \theta = \dfrac{\sin \theta}{\cos \theta} = -\dfrac{2\sqrt7}{6}\\\\\csc \theta = \dfrac{1}{\sin \theta} = -\dfrac{8}{2\sqrt7}

Evaluate the following:

a)

\tan \theta\times \cot \theta =\tan \theta\times\dfrac{1}{\tan \theta} = 1

b)

\csc \theta\times \tan \theta\\\\= -\dfrac{8}{2\sqrt7}\times -\dfrac{2\sqrt7}{6} = \dfrac{4}{3}

c)

\sin^2 \theta + \cos^2 \theta = 1\\\text{using the trignometric identity}

5 0
4 years ago
Consider the universal set R,R, define the interval A=[−7,1],A=[−7,1], interval B=(−1,5),B=(−1,5), and CC be the negative real n
AURORKA [14]
Draw a diagram representing the real number line, and the sets A  and B.

According to the diagram:

<span>1. A∪B= [-7, 5)

2. </span><span>2. A∩B∩C= (-1, 0)

</span>3. A∩(B∪C)= [-7, 1] <span>∩ (-INFINITY, 5) = [-7, 1]

4. </span>4. A^c ∪ B^c∪ C^c= (not A) ∪ (not B) ∪ (not C) = 

(-infinity, -7) ∪ (1, infinity)   ∪  (-infinity, -1] ∪ [5, infinity)   ∪ [0, infinity)

= (-infinity, -1] ∪ (1, infinity ) ∪ [0, infinity) = (-infinity, -1] ∪ [0, infinity) 

5. (C−A)∪(B−C)= "in C but not in A" union "in B but not in C"

=(-infinity, -7) ∪ [0, 5) 

8 0
4 years ago
Two sides of a triangle have lengths 8 and 17. What must be true about the length of the third side?
ololo11 [35]

Answer:

c. less than 25

Step-by-step explanation:

The third side can not be greater than the sum of the other two.

5 0
3 years ago
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