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Kipish [7]
3 years ago
12

|a+x|/2 − |a−x|/2 , if a=−2; x=−6 PLZ HELP!!!

Mathematics
2 answers:
Alchen [17]3 years ago
8 0

Answer:

2

Step-by-step explanation:

|a+x|/2 − |a−x|/2 , if a=−2; x=−6

Evaluate this expression

Simply plug in the numbers

| -2 + -6 | /2   -   |-2 - -6|/2

|-8| /2  -   |4|/2

4 - 2

2

ad-work [718]3 years ago
5 0

Answer:

2

Step-by-step explanation:

So you're dealing with absolute value which means that whatever is inside these symbols |  | will be positive after all the calculations inside are done.

Plug in your values

|(-2) + (-6)|/2

-2 - 6 = -8

absolute value of -8 is 8

8/2 = 4

First part is 4

|(-2) - (-6)|/2

-2 + 6 = 4

4/2 = 2

Second part is 2

Put the two values back into your original equation

4-2 = 2

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ABC= ADC, BA= 25, DA=
geniusboy [140]

Answer:

DA = 25

Step-by-step explanation:

Given that,

→ ABC = ADC ==> (1)

→ BA = 25 ==> (2)

From the given (1) and (2),

→ BA = DA

→ BA = 25

→ [ DA = 25 ]

Hence, the answer is 25.

4 0
1 year ago
What is the volume of the triangular prism? Round to the nearest tenth.
Sloan [31]

Answer:

1185 bin

Step-by-step explanation:

i rounded so i and i multiply and didved and i got that i got 1185 bin

6 0
2 years ago
Please help me..I don't understand how to solve problems like
Nuetrik [128]
Okay. So first you subtract 2x from each side. That will give you 4y= -2x+18 (you want to get y by itself). Then you divide both sides by 4. It should equal> y= 1/2x +9/2
3 0
3 years ago
Read 2 more answers
This hyperbola is centered at theorigin. Find its equation.Foci: (-2,0) and (2,0)Vertices: (-1,0) and (1,0)
beks73 [17]

SOLUTION

From the question, the center of the hyperbola is

\begin{gathered} (h,k),\text{ which is } \\ (0,0) \end{gathered}

a is the distance between the center to vertex, which is -1 or 1, and

c is the distance between the center to foci, which is -2 or 2.

b is given as

\begin{gathered} b^2=c^2-a^2 \\ b^2=2^2-1^2 \\ b=\sqrt[]{3} \end{gathered}

But equation of a hyperbola is given as

\frac{(x-h)^2}{a^2}-\frac{(y-k)^2}{b^2}=1

Substituting the values of a, b, h and k, we have

\begin{gathered} \frac{(x-0)^2}{1^2}-\frac{(y-0)^2}{\sqrt[]{3}^2}=1 \\ \frac{x^2}{1}-\frac{y^2}{3}=1 \end{gathered}

Hence the answer is

\frac{x^2}{1}-\frac{y^2}{3}=1

5 0
11 months ago
2
melisa1 [442]

Answer:

\boxed{\sf 10}

Step-by-step explanation:

The additive number of any number is the number when added to the number gives a result of zero.

So, if we add 10 to -10 we get a result of zero.

=> -10+10

=> Zero

5 0
2 years ago
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