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9966 [12]
3 years ago
7

What is the value of -2 1/6 - 1 1/4 + 1 3/4

Mathematics
1 answer:
madam [21]3 years ago
4 0
-1.66666666667 u could just use a calculator lol
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What are the solutions to the quadratic equation 2x^2-8x-24=0
Vinvika [58]

2x^2 - 8x - 24

First, we can factor a 2 out of this expression to simplify it.

2(x^2 - 4x - 12)

Now, we can try factoring this two ways: by using the quadratic formula, or by using the AC method.

We're gonna try using the AC method first.

List factors of -12.

1 * -12

-1 * 12

2 * -6

-2 * 6 (these digits satisfy the criteria.)

Split the middle term.

2(x^2 - 2x + 6x - 12)

Factor by grouping.

2(x(x - 2) + 6(x - 2)

Rearrange terms.

<h3><u>(2)(x + 6)(x - 2) is the fully factored form of the given polynomial.</u></h3>
7 0
3 years ago
Read 2 more answers
Which algebraic expression is a polynomial? 3m2n – StartFraction 2 m Over n EndFraction + StartFraction 1 Over n EndFraction Sta
SOVA2 [1]

Answer:

The Answer is D

Step-by-step explanation:

4 0
3 years ago
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Charlie made $198 for 11 hours of work. At the same rate, how many hours would he have to work to make $126? ​
goldenfox [79]

Answer:

Charlie would have to work 7 hours to make $126.

Step-by-step explanation:

198 ÷ 11 = 18

$18 - 1 hour

126 ÷ 18 = 7

4 0
3 years ago
Find the solution to the initial value problem
Karo-lina-s [1.5K]

Answer:

y = (11x + 13)e^(-4x-4)

Step-by-step explanation:

Given y'' + 8y' + 16 = 0

The auxiliary equation to the differential equation is:

m² + 8m + 16 = 0

Factorizing this, we have

(m + 4)² = 0

m = -4 twice

The complimentary solution is

y_c = (C1 + C2x)e^(-4x)

Using the initial conditions

y(-1) = 2

2 = (C1 -C2) e^4

C1 - C2 = 2e^(-4).................................(1)

y'(-1) = 3

y'_c = -4(C1 + C2x)e^(-4x) + C2e^(-4x)

3 = -4(C1 - C2)e^4 + C2e^4

-4C1 + 5C2 = 3e^(-4)..............................(2)

Solving (1) and (2) simultaneously, we have

From (1)

C1 = 2e^(-4) + C2

Using this in (2)

-4[2e^(-4) + C2] + 5C2 = 3e^(-4)

C2 = 11e^(-4)

C1 = 2e^(-4) + 11e^(-4)

= 13e^(-4)

The general solution is now

y = [13e^(-4) + 11xe^(-4)]e^(-4x)

= (11x + 13)e^(-4x-4)

3 0
3 years ago
A region R in the xy-plane is given. Find equations for a transformation T that maps a rectangular region S in the uv-plane onto
ruslelena [56]
X(u, v) = (2(v - c) / (d - c) + 1)cos(pi * (u - a) / (2b - 2a))
y(u, v) = (2(v - c) / (d - c) + 1)sin(pi * (u - a) / (2b - 2a))

As v ranges from c to d, 2(v - c) / (d - c) + 1 will range from 1 to 3, which is the perfect range for the radius. As u ranges from a to b, pi * (u - a) / (2b - 2a) will range from 0 to pi/2, which is the perfect range for the angle. So, this maps the rectangle to R.
6 0
3 years ago
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