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3 years ago

Help me pleaseeeeeeeee

Mathematics

1 answer:

SIZIF [17.4K]3 years ago

4 0

Answer:

(f o g)(x) = 4x^4 - 24x^2 + 37

(g o f)(x) = 16x^4 + 8x^2 - 2

Step-by-step explanation:

f(x) = 4x^2 + 1

g(x) = x^2 - 3

(f o g)(x) = f(g(x)) = f(x^2 - 3) = 4(x^2 - 3)^2 + 1 =

= 4(x^4 - 6x^2 + 9) + 1

= 4x^4 - 24x^2 + 37

(g o f)(x) = g(f(x)) = g(4x^2 + 1) = (4x^2 + 1)^2 - 3 =

= 16x^4 + 8x^2 - 2

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For the following system, if you isolated x in the first equation to use the substitution method, what expression would you subs

alekssr [168]

Answer:

$x=2y+6$

Step-by-step explanation:

So we have the system:

$-x+2y=-6\\3x+y=8$

If we isolate the x-variable in the first equation:

$-x+2y=-6$

Subtract 2y from both sides:

$-x=-6-2y$

Divide both sides by -1:

$x=2y+6$

Therefore, we would substitute the above into the second equation:

$3x+y=8\\3(2y+6)+y=8$

The answer is 2y+6

Further notes:

To solve for the system, distribute:

$6y+18+y=8$

Simplify:

$7y+18=8$

Subtract:

$7y=-10$

Divide:

$y=-10/7\approx-1.4286$

Now, substitute this value back into the isolated equation:

$x=2(-10/7)+6\\x=-20/7+42/7\\x=22/7\approx3.1429$

3 0

3 years ago

Read 2 more answers

part 3. Find the coordinates of the vertices of the triangle after a reflection across the y-axis and then across the line y= 2.

garik1379 [7]

Answer:

c. S'(3, 1), T'(1, -1), U'(0, 1)

Step-by-step explanation:

Reflection across the y-axis negates the x-coordinate, so is equivalent to the transformation ...

(x, y) ⇒ (-x, y)

Reflection across the horizontal line y=c is equivalent to the transformation ...

(x, y) ⇒ (x, 2c-y)

So, the combined reflections are equivalent to the transformation ...

(x, y) ⇒ (-x, 4 -y)

Then we have ...

S(-3, 3) ⇒ S'(-(-3), 4-3) = S'(3, 1)

T(-1, 5) ⇒ T'(-(-1), 4-5) = T'(1, -1)

U(0, 3) ⇒ U'(-(0), 4-3) = U'(0, 1) . . . . matches choice C

7 0

3 years ago

Mr. Mole left his burrow and started digging his way down at a constant rate. Time (minutes) Altitude (meters) 666 -20.4−20.4min

Juli2301 [7.4K]

Answer:

$-6$ meters.

Step-by-step explanation:

We have been given Mr. Mole left his burrow and started digging his way down at a constant rate.

We are also given a table of data as:

Time (minutes) Altitude (meters)

6 -20.4

9 -27.6

12 -34.8

First of all, we will find Mr. Mole's digging rate using slope formula and given information as:

$m=\frac{y_2-y_1}{x_2-x_1}$ , where,

$y_2-y_1$ represents difference of two y-coordinates,

$x_2-x_1$ represents difference of two corresponding x-coordinates of y-coordinates.

Let $(6,-20.4)$ be $(x_1,y_1)$ and $(9,-27.6)$ be $(x_2,y_2)$ .

$m=\frac{-27.6-(-20.4)}{9-6}$

$m=\frac{-27.6+20.4}{3}$

$m=\frac{-7.2}{3}$

$m=-2.4$

Now, we will use slope-intercept form of equation to find altitude of Mr. Mole's burrow.

$y=mx+b$ , where,

m = Slope,

b = The initial value or the y-intercept.

Upon substituting $m=-2.4$ and coordinates of point $(6,-20.4)$ , we will get:

$-20.4=-2.4(6)+b$

$-20.4=-14.4+b$

$-20.4+14.4=-14.4+14.4+b$

$-6=b$

Since in our given case y-intercept represents the altitude of Mr. Mole's burrow, therefore, the altitude of Mr. Mole's burrow is $-6$ meters.

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3 years ago

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A person drives north 6 blocks, then drives west 6 blocks. the displacement is a straight line from the starting point to the fi

klemol [59]

For the answer to the question above, A person drives north 6 blocks, then drives west 6 blocks. the displacement is a straight line from the starting point to the finish in 8.49 blocks in a Northwest direction.
The Solution:
A^2+B^2=C^2

6*6+6*6 = sqrt(72)

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3 years ago

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What is the solution of the system of equations? Y=x-4 and y=-3x

nata0808 [166]

Answer:

(-7,-3)

Step-by-step explanation:

step one) solve one equation for x.

y=-3x

-3x/-3=y/-3

x=(-y/3)

step two) plug in x for the other original problem.

y=x-4

y=(-y/3)-4

step three) solve for y.

y=(-y/3)-4

y+4=(-y/3)-4+4

3*(y+4)=(-y/3)*3

3y+12=-y

(3y+12)-3y=-y-3y

12/-4=-4y/-4

y=-3

step four) plug in -3 for y and solve for x.

y=x-4

-3=x+4

-3-4=x+4-4

-7=x

I am in algebra two so you can trust my answer. if you need more help lmk in the comments. happy holidays and stay safe!

5 0

3 years ago