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Nitella [24]
3 years ago
10

Select the equation of the line parallel to the equation 2x + 4y = -5 that passes through the point (-4, -8).

Mathematics
1 answer:
shtirl [24]3 years ago
7 0

Answer:

D.

Step-by-step explanation:

First, put your original equation in slope-intercept form to find your slope.

y=mx+b\\2x+4y=-5\\4y=-2x-5\\y=-\frac{1}{2} x-\frac{5}{4}

Now that you have your slope (-\frac{1}{2}) and a point, you can use point-slope form to find your y-intercept.

y-y1=m(x-x1)\\y-(-8)=-\frac{1}{2} (x-(-4))\\y+8=-\frac{1}{2} (x+4)\\y+8=-\frac{1}{2} x-2\\y=-\frac{1}{2} x-10

Your answer choices are all in Ax+By=C form, so lets convert our equation into that form.

y=-\frac{1}{2} x-10\\1/2x+y=-10\\x+2y=-20

If we multiply all of our terms by 2, we can get answer choice D.

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What is 2 1/3 + 1 2/3?
SpyIntel [72]
It totals out to 4. I/3 plus 2/3 would equal 1 whole number, then add 1+2+1, you will get 4!
5 0
3 years ago
Read 2 more answers
On a farm the number of cows and the number of sheep are in the tatio 6:5 the number of sheep and the number of pigs are in the
Rzqust [24]

Answer:

There are 70 sheep in the farm.

Step-by-step explanation:

Consider the provided information.

Let c represents the cows, s represents the sheep and p represents the pigs.

The number of cows and the number of sheep are in the ratio 6:5.

\dfrac{c}{s} =\dfrac{6}{5}

c =\dfrac{6s}{5}

The number of sheep and the number of pigs are in the ratio 2:1.

\dfrac{s}{p} =\dfrac{2}{1}

p =\dfrac{s}{2}

The total number is 189.

c+s+p=189

Substitute the value of c and p in above.

\dfrac{6s}{5}+s+\dfrac{s}{2}=189

\dfrac{12s+10s+5s}{10}=189

27s=1890

s=70

Hence, there are 70 sheep in the farm.

8 0
3 years ago
Simplify the expression below
Brilliant_brown [7]

Answer:

− 4 x + 20

Step-by-step explanation:

3 0
3 years ago
Find the angle 0 between the vectors.
Nikitich [7]

Answer:

If θ is the angle between two vectors u and v, then cosθ = (u·v

Step-by-step explanation:

6 0
3 years ago
Find constants a and b such that the function y = a sin(x) + b cos(x) satisfies the differential equation y'' + y' − 5y = sin(x)
vichka [17]

Answers:

a = -6/37

b = -1/37

============================================================

Explanation:

Let's start things off by computing the derivatives we'll need

y = a\sin(x) + b\cos(x)\\\\y' = a\cos(x) - b\sin(x)\\\\y'' = -a\sin(x) - b\cos(x)\\\\

Apply substitution to get

y'' + y' - 5y = \sin(x)\\\\\left(-a\sin(x) - b\cos(x)\right) + \left(a\cos(x) - b\sin(x)\right) - 5\left(a\sin(x) + b\cos(x)\right) = \sin(x)\\\\-a\sin(x) - b\cos(x) + a\cos(x) - b\sin(x) - 5a\sin(x) - 5b\cos(x) = \sin(x)\\\\\left(-a\sin(x) - b\sin(x) - 5a\sin(x)\right)  + \left(- b\cos(x) + a\cos(x) - 5b\cos(x)\right) = \sin(x)\\\\\left(-a - b - 5a\right)\sin(x)  + \left(- b + a - 5b\right)\cos(x) = \sin(x)\\\\\left(-6a - b\right)\sin(x)  + \left(a - 6b\right)\cos(x) = \sin(x)\\\\

I've factored things in such a way that we have something in the form Msin(x) + Ncos(x), where M and N are coefficients based on the constants a,b.

The right hand side is simply sin(x). So we want that cos(x) term to go away. To do so, we need the coefficient (a-6b) in front of that cosine to be zero

a-6b = 0

a = 6b

At the same time, we want the (-6a-b)sin(x) term to have its coefficient be 1. That way we simplify the left hand side to sin(x)

-6a  -b = 1

-6(6b) - b = 1 .... plug in a = 6b

-36b - b = 1

-37b = 1

b = -1/37

Use this to find 'a'

a = 6b

a = 6(-1/37)

a = -6/37

8 0
3 years ago
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