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Arada [10]
3 years ago
15

Who can help me on this ?

Mathematics
2 answers:
Anna [14]3 years ago
5 0

Answer: 3

Step-by-step explanation:

Marat540 [252]3 years ago
3 0

Answer:

b= -4

Step-by-step explanation:

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I need to know what's the answer
FrozenT [24]
Hi friend,

Answer of 4th u need nah as u ticked on it-:

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8*x*x*y*z*z*z*z/16*x*y*y*z
=x*z^3/2y

Hope it helps..........



3 0
3 years ago
How do you solve this linear equation 3x+2=x+2x+4 ?
Svetach [21]
3x+2=x+2x+4
     -2              -2
3x=x+2x+4-2
 3x=3x+4-2
 -3x  -3x
x= 4-2
x=2
7 0
3 years ago
Read 2 more answers
The functions f and g are defined as follows.
g100num [7]

Answer:

f(-5)=-16\\g(2)=-11

Step-by-step explanation:

By definition, a relation is a function if and only if each input value have one and only one output value.

The input values are the x-values and the output values are the y-values.

Given the function f(x):

f(x)=3x-1

You need to substitute x=-5 into this function:

f(-5)=3(-5)-1

And now you must evaluate in order to find the corresponding output value.

You get:

f(-5)=-15-1\\\\f(-5)=-16

The function g(x) is:

g(x)=-2x^2-3

Then, you need to substitute x=2 in the function:

g(2)=-2(2)^2-3

And  finally you must evaluate in order to find the corresponding output value. This is:

g(2)=-2(4)-3\\\\g(2)=-8-3\\\\g(2)=-11

8 0
3 years ago
Which of the following is not an advantage of using credit cards?
kvv77 [185]

Answer:a

Step-by-step explanation:

6 0
3 years ago
Someone help me please
RoseWind [281]

9514 1404 393

Answer:

  • A = (0, 1)
  • B = (3, -2)
  • area = 4.5 square units

Step-by-step explanation:

Rewriting the equations to make x the subject, we have ...

  x = y² -1 . . . . . [eq1]

  x = 1 - y . . . . . .[eq2]

At the points of intersection, the difference will be zero.

  y² -1 -(1 -y) = 0

  y² +y -2 = 0

  (y -1)(y +2) = 0

The y-coordinates of points A and B are 1 and -2.

The corresponding x-coordinates are ...

  x = 1 -{1, -2} = {1 -1, 1+2} = {0, 3}

Then A = (0, 1) and B = (3, -2).

__

A differential of area can be written ...

  (x2 -x1)dy = ((1 -y) -(y² -1))dy = (2 -y -y²)dy

Integrating this over the interval y = [-2, 1] gives the area.

  \displaystyle A=\int_{-2}^1(2-y-y^2)\,dy=\left.(2y-\dfrac{1}{2}y^2-\dfrac{1}{3}y^3)\right|_{-2}^1\\\\=\left(2-\dfrac{1}{2}-\dfrac{1}{3}\right)-\left(2(-2)-\dfrac{(-2)^2}{2}-\dfrac{(-2)^3}{3}\right)=\dfrac{7}{6}+4+2-\dfrac{8}{3}\\\\=\boxed{4.5}

The area of the shaded region is 4.5 square units.

5 0
3 years ago
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