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3 years ago

How do I simplify (6x-10x^2+3)-(5x^2+3x-4) ???

Mathematics

1 answer:

Neporo4naja [7]3 years ago

3 0

Answer:

Step-by-step explanation:

6x-10x^2+3-(5x^2+3x-4)

when there is a - in front of an expression in (), change the sign of each term in the expression

6x-10x^2+3-5x^2-3x+4

now collect like terms

3x-10x^2+3-5x^2+4

3x-15x^2+3+4

3x-15x^2+7

now use the commutative property to reorder the terms

-15x^2+3x+7

=-15x^2+3x+7

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What is the vertical intercept for this function? A.(0,1) B. (1,0) C.(0,0)

balandron [24]

I am not that sure but I am guessing it’s B.

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3 years ago

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19. Mr. Green teaches band, choir, and math. This year, he has 57 students that take at least one of (3 points)

kvasek [131]

Answer:

The answer is "17 students with Mr. Green take exactly two courses".

Step-by-step explanation:

Following are the calculation to this question:

Alone band 11

Alone chorus 17

Choir and band alone 4

7 Maths alone and the band

Choir and Math alone 6

Chorus and Math Band 3

mathematics alone 9 +11+4+10+6+17 + M = 57

2 classes = 4+7+6 = 17

So, there are 17 students with Mr. Green take exactly two courses.

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3 years ago

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fgiga [73]

Answer:

4x +7y

Step-by-step explanation:

(8x +10y) +(-4x -3y) = x(8 -4) +y(10 -3) = 4x +7y

Combine like terms. Pay attention to signs.

5 0

3 years ago

T/3 > Solve > 15. Graph the solution 3 The solution is

MissTica

Answer:

$T > 45$

Step-by-step explanation:

Given

$\frac{T}{3} > 15$

Required

Graph the solution

$\frac{T}{3} > 15$

Multiply both sides by 3

$3 * \frac{T}{3} > 15*3$

$T > 15*3$

$T > 45$

See attachment for graph (Assume T is on the x-axis)

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3 years ago

What is the maximum vertical distance between the line y = x + 20 and the parabola y = x2 for −4 ≤ x ≤ 5?

Fofino [41]

The vertical distance can be found by subtracting parabola from the given line.

Given line equation is y= x+20

Equation of parabola is y= x^2

Subtract parabola from line equation

so y = $x+20 -x^2$

y = $-x^2 + x + 20$

Now we take derivative to find out the maximum that is the vertex

y' = -2x + 1

Now we set derivative =0 and solve for x

0 = -2x+1

2x = 1

$x=\frac{1}{2}$

Now we plug in x values in $y= -x^2 + x + 20$

$y = (\frac{1}{2}^2) + \frac{1}{2} + 20$

Take common denominator

$y= \frac{81}{4}$

So our maximum vertical distance is $\frac{81}{4}$ at $x= \frac{1}{2}$

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3 years ago

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