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Dimas [21]
3 years ago
15

Solve the problem n-6=21

Mathematics
2 answers:
yuradex [85]3 years ago
7 0

Answer:

n=27

Step-by-step explanation:

sammy [17]3 years ago
4 0

Answer:

27

Step-by-step explanation:

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Type the missing numbers in this sequence:<br> -101, -105,<br> -109,<br> -117
PIT_PIT [208]

Answer:

The answer is - 113 that is the correct answer skipping in three's and the first one is - 97

6 0
2 years ago
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For f(x) = 4x +1 and g(x) = x2 - 5, find (g/f) (x).
Softa [21]

Answer:

\left(g/f\right)\left(x\right)=\frac{x}{4}-\frac{1}{16}-\frac{79}{16\left(4x+1\right)}

Step-by-step explanation:

f(x)=4x+1

g\left(x\right)=x^2\:-\:5

As

(g/f)(x) = g(x) / f(x)

            =\:\frac{x^2\:-\:5}{4x+1}\:\:\:\:\:\:

            =\frac{x}{4}+\frac{-\frac{x}{4}-5}{4x+1}

              =\frac{x}{4}-\frac{1}{16}+\frac{-\frac{79}{16}}{4x+1}

               =\frac{x}{4}-\frac{1}{16}-\frac{79}{16\left(4x+1\right)}

Therefore,

\left(g/f\right)\left(x\right)=\frac{x}{4}-\frac{1}{16}-\frac{79}{16\left(4x+1\right)}

8 0
3 years ago
A translation moves A(-3,2) to a'(0,0). find b', the image of b(5,4)) under the same translation
statuscvo [17]
The image of b, b', would be (8,2)

hope this helps!
7 0
3 years ago
I WILL MARK AS BRAINLIEST! Graph each coordinate pair on the graph and then indicate which quadrant or axis the point lies on.
Kitty [74]

1. 3rd quadrant

2. 3rd quadrant

3. 1st quadrant

4. 1st quadrant

5. 4th quadrant

6. 2nd quadrant

7. 4th quadrant

8. 3rd quadrant

7 0
3 years ago
What changes would you make in your description of point, line, and plane?
weeeeeb [17]

Answer:

Here's a quick sketch of how to calculate the distance from a point P=(x1,y1,z1)

P

=

(

x

1

,

y

1

,

z

1

)

to a plane determined by normal vector N=(A,B,C)

N

=

(

A

,

B

,

C

)

and point Q=(x0,y0,z0)

Q

=

(

x

0

,

y

0

,

z

0

)

. The equation for the plane determined by N

N

and Q

Q

is A(x−x0)+B(y−y0)+C(z−z0)=0

A

(

x

−

x

0

)

+

B

(

y

−

y

0

)

+

C

(

z

−

z

0

)

=

0

, which we could write as Ax+By+Cz+D=0

A

x

+

B

y

+

C

z

+

D

=

0

, where D=−Ax0−By0−Cz0

D

=

−

A

x

0

−

B

y

0

−

C

z

0

.

This applet demonstrates the setup of the problem and the method we will use to derive a formula for the distance from the plane to the point P

P

.

Step-by-step explanation:

5 0
3 years ago
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