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Marat540 [252]
3 years ago
15

A number x is greater than 0. write the word sentence as an inequality

Mathematics
1 answer:
IgorLugansk [536]3 years ago
8 0

9514 1404 393

Answer:

  x > 0

Step-by-step explanation:

The only thing that needs translation to symbols is "greater than". Your "word sentence" already uses the symbols x and 0. The symbol for greater than is >. Your inequality is ...

  x > 0

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HELPP PLISS
Otrada [13]

Answer:

C. Tan

Step-by-step explanation:

This is because you are given the measure of the opposite angle and the adjacent side.

7 0
3 years ago
F(x) = <img src="https://tex.z-dn.net/?f=%5Csqrt%7Bx%2B7%7D%20-%5Csqrt%7Bx%5E2%2B2x-15%7D" id="TexFormula1" title="\sqrt{x+7} -\
elena-s [515]

Answer:

x >= -7  ................(1a)

x >= 3   ...............(2a1)

Step-by-step explanation:

f(x) =  \sqrt{x+7}-\sqrt{x^2+2x-15}  .............(0)

find the domain.

To find the (real) domain, we need to ensure that each term remains a real number.

which means the following conditions must be met

x+7 >= 0  .....................(1)

AND

x^2+2x-15 >= 0 ..........(2)

To satisfy (1),  x >= -7  .....................(1a) by transposition of (1)

To satisfy (2), we need first to find the roots of (2)

factor (2)

(x+5)(x-3) >= 0

This implis

(x+5) >= 0 AND (x-3) >= 0.....................(2a)

OR

(x+5) <= 0 AND (x-3) <= 0 ...................(2b)

(2a) is satisfied with x >= 3   ...............(2a1)

(2b) is satisfied with x <= -5 ................(2b1)

Combine the conditions (1a), (2a1), and (2b1),

x >= -7  ................(1a)

AND

(

x >= 3   ...............(2a1)

OR

x <= -5 ................(2b1)

)

which expands to

(1a) and (2a1)   OR  (1a) and (2b1)

( x >= -7 and x >= 3 )  OR ( x >= -7 and x <= -5 )

Simplifying, we have

x >= 3  OR ( -7 <= x <= -5 )

Referring to attached figure, the domain is indicated in dark (purple), the red-brown and white regions satisfiy only one of the two conditions.

3 0
3 years ago
Read 2 more answers
Angle 0 lies in the second quadrant and sin 0 = 3/5 cos= coy =
Elza [17]
Consider the right triangle ABC with legs AB=4, AC=3 and hypotenuse BC=5. Angle B has
 \sin B=\frac{opposite}{hypotenuse} = \frac{3}{5} and
 \cos B = \frac{adjacent}{hypotenuse} = \frac{4}{5}.

Since O lies in second quadrant \cos O\ \textless \ 0 and 
\cos O= -\cos B =- \frac{4}{5}.
Answer: \cos O=- \frac{4}{5}. 


3 0
4 years ago
What is the area of the regular hexagon shown?
andreyandreev [35.5K]

Answer:

584.57

Step-by-step explanation:

5 0
3 years ago
Solve by substitution. Show each step of your work.<br><br> 2x + y = 7<br><br> 3x + 5y = 14
Monica [59]
Alright...simple...showing all steps.. ;)

You have the equations...

2x+y=7
and
3x+5y=14

To be able to even solve for any of the variables...multiply the equations by...2...and..3...

2x+y=7----*3--> 6x+3y=21
and
3x+5y=14-----*2--->6x+10y=28

Thus,

6x+3y=21
-
6x+10y=28

=========

-7y=-7

y=1

Now, plug y back into any of the original equations....we'll use 2x+y=7 in this case....

2x+(1)=7

2x+1=7
      -1  -1

2x=6

x=3

Thus, the point of intersection for these two equations is (3,1)
8 0
4 years ago
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