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ValentinkaMS [17]
2 years ago
12

Pls answer clearly , just the answer i dont need explenation:) please <3

Mathematics
1 answer:
lawyer [7]2 years ago
5 0

Answer:

option D

Step-by-step explanation:

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If f(x)=x+4/x , g(x)= x-2, and h(x) =4x-1, what is f(h(g(x)
xxMikexx [17]

Answer:

(16x² - 72x + 85) / (4x-9)

Step-by-step explanation:

first put g(x) into h(x)

h(g(x)) = 4(x-2) - 1 = 4x - 8 - 1 = 4x - 9, then

f(h(g(x))) = (4x - 9) + (4/4x-9) = ((4x-9)(4x-9) + 4)/(4x-9) =

(16x² - 72x + 81 + 4)/(4x-9) = (16x² - 72x + 85)/(4x-9)

If you are missing parentheses around (x+4)/x, then the answer would be different.

3 0
3 years ago
Read 2 more answers
WILL GIVE BRAINLST <br><br> HAVE A GOOD DAY
Sedbober [7]
1st one 17/100 2nd one 0.25


Because 0.17 would be like 17% out of 100% so the fraction simplified is 17/100

2. Since the area of a square is length×length
But according to the question we are asked to find the length and the area is given so we will have to solve this
L×L= area
L^2=0.25
Square root both sides
L=√0.25
L=0.5
Therefore the length of the square is 0.5
6 0
3 years ago
Round 458,673 to the nearest hundred thousand
Viefleur [7K]
500,000 is that number rounded.
4 0
3 years ago
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Need help geometry ^^^
sdas [7]
The term corresponding angles means that the angles are in the same relative position. if you look at A, r and w are in the same positions if we.were.to overlap the 2 intersections. So they are corresponding angles
8 0
4 years ago
Find the equation of the plane passing through the points
erica [24]

Extract the normal vectors from the given planes:

-2x+y+z+2=0\implies\vec n_1=(-2,1,1)

x+y-3z+1=0\implies\vec n_2=(1,1,-3)

(which are unique up to their signs, meaning either \vec n_1 or -\vec n_1 are valid choices for the normal vector)

The third plane must be perpendicular to both these given planes, which means it would be parallel to both \vec n_1 and \vec n_2, which in turn means its own normal vector \vec n_3 should be perpendicular to both \vec n_1 and \vec n_2.

Enter the cross product:

\vec n_3=\vec n_1\times\vec n_2=(-4,-5,-3)

or (4, 5, 3), which also works.

The given plane passes through (-1, 1, 4), so its equation is

(x+1,y-1,z-4)\cdot\vec n_3=0

Simplify:

(x+1,y-1,z-4)\cdot(4,5,3)=0

4(x+1)+5(y-1)+3(z-4)=0

\boxed{4x+5y+3z=13}

3 0
3 years ago
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