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Alika [10]
3 years ago
9

Use synthetic division to divide (x^ 3 + x^ 2 – 40x – 4) ÷ (x – 6)

Mathematics
2 answers:
Marina CMI [18]3 years ago
7 0
The answer is C.

Hope this helps!!

mafiozo [28]3 years ago
4 0

For this case we must build a quotient that, when multiplied by the divisor, eliminates the terms of the dividend until reaching the remainder.

It must be fulfilled that:

Dividend = Quotient * Divider + Remainder

Looking at the attached image we have that the quotient is given by:

x ^ 2 + 7x + 2

Answer:

Option C

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Write an equation for the quadratic graphed below
uranmaximum [27]

Answer:

Step-by-step explanation:

LOL The graph doesn’t match the y intercept :)

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If we have point (0,-3) we have a quadratic of

y=ax^2+bx-3 we are given points (-1,0) and (2,0) so

a-b-3=0 and 4a+2b-3=0

4a+2b-3+2(a-b-3)=0

4a+2b-3+2a-2b-6=0

6a-9=0

6a=9

a=1.5, since a-b=3

1.5-b=3

b=-1.5

y=1.5x^2-1.5x-3

6 0
3 years ago
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Molodets [167]

Answer:

8.8%

Step-by-step explanation:

you have ( $150 / $1695 ) * 100 = 8.8

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3 years ago
5. Determine the 2d shape that would be created if the 3d shape were sliced as shown.
geniusboy [140]
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3 years ago
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An engineer is designing a large steel pad to be installed on the deck of an aircraft carrier. Its total volume will be 18 yd^3.
Shtirlitz [24]

Answer:

The cost of the material for the full-sized pad is;

d) 222,750.00

Step-by-step explanation:

The parameters of the steel pad and the scale model are;

The volume of the deck = 18 yd³

The dimensions of the model of the same material = 6 feet by 4.5 feet and 4.5 inches thick

The weight of the sample, m₂ = 75 lbs

The cost of the steel, P = $55/lb

The dimensions of the sample in yards are;

6 feet = 2 yards, 4.5 feet = 1.5 yards, and 4 inches = 0.\overline 1 yards

The volume of the sample, V₂ = 2 yd. × 1.5 yd. × 0.\overline 1 yd. =  (1/3) yd.³

The density of the sample material, ρ = m₂/V₂

∴ The density of the sample material, ρ = 75 lbs/((1/3)yd.³) = 225 lbs/yd³

Therefore, the density of the material of the pad, ρ = 225 lbs/yd³

The mass of the steel  pad, m₁ = ρ × V₁

∴ The mass of the steel  pad, m₁ = 225 lbs/yd³ × 18 yd³ = 4,050 lbs

The cost of the material for the full-sized steel pad, C = m₁ × P

∴ C = 4,050 lbs × $55/lb = $222,750

6 0
3 years ago
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