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nalin [4]
2 years ago
11

Wish someone can do this for me.

Mathematics
1 answer:
Makovka662 [10]2 years ago
5 0

Given : Two inequality is given to us . The inequality is v + 8 ≤ -4 and v - 6 ≥ 10 .

To Find : To write those two inequality as a compound inequality with integers .

Solution: First inequality given to us is v + 8 ≤ -4 . So let's simplify it ;

⇒ v + 8 ≤ -4 .

⇒ v ≤ -4 - 8.

⇒ v ≤ -12 .

Now , on simplifying the second inequality ,

⇒ v - 6 ≥ 10 .

⇒ v ≥ 10 + 6.

⇒ v ≥ 16 .

Hence the required answer will be :

\Large{\boxed{\red{\bf \blue{\dag} v\leqslant -12 \:\:or\:\:v\geqslant 16}}}

First one implies that v is less than or equal to -12 whereas the second one implies that v is greater than or equal to 16 .

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Which statement accurately describes how adding a number, n, to the function f(x)=sin(x) affects its graph?
Hitman42 [59]

The <em>correct answer</em> is:


A) There is a vertical shift of n units.


Explanation:


When a parent function is transformed, each type of transformation results ina different change in the graph.


The parent function in this question is f(x) = sin(x).


Adding n units to the end of this function would result in f(x) = sin(x) + n.


When a function is transformed by adding a constant to the end of a function results in a vertical shift by that number of units. In this case, it would be a vertical shift n units.

7 0
3 years ago
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Find what x =_______​
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Answer:

x=18

Step-by-step explanation:

ok, so you mulityply base and hieght then divide by 2

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your welcome

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2 years ago
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VMariaS [17]

Answer:

10:40

Step-by-step explanation:

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Answer =10:40

5 0
3 years ago
Will mark brainlist if it is right
Pepsi [2]

Answer:

2/3 < y < 6

Step-by-step explanation:

So what we want to do is get the equation to become just y surrounded by the gereater than signs. So we have 3y-4 and we first add 4 to all numbers. We now 2 < 3y < 18. WE can divide everything by 3 to get 2/3 < y < 6.

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3 years ago
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mel-nik [20]
\log_{(x-1)} 9=2

the domain:
x-1 >0 \ \land \ x-1 \not=1 \\&#10;x>1 \ \land \ x \not= 2 \\&#10;x \in (1; 2) \cup (2;+\infty)

the equation:
\log_{(x-1)}9=2 \\&#10;(x-1)^2=9 \\&#10;\sqrt{(x-1)^2}=\sqrt{9} \\&#10;|x-1|=3 \\&#10;x-1=3 \ \lor \ x-1=-3 \\&#10;x=4 \ \lor \ x=-2

4 is in the domain
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The answer:
x=4

***
Also, the answer in 13 is {8}. -3 is not in the domain. Replace x with -3 and you'll see:
x=\sqrt{5x+24} \\&#10;-3=\sqrt{5 \times (-3)+24} \\&#10;-3=\sqrt{-15+24} \\&#10;-3=\sqrt{9} \\&#10;-3=3
It's not true so -3 isn't a solution to this equation.
4 0
3 years ago
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