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3 years ago

PLEASE HELP ASAP FIRST TO ANSWER CORRECTLY GETS BRAINLYIST (question in the pic)

Mathematics

1 answer:

mamaluj [8]3 years ago

8 0

Answer:

$346.50.

Step-by-step explanation:

Imagine that the 9*9 square on the left is a part of the floor then:

The area = area of a trapezium - 9^2.

So the area of the dining area = 1/2 * (31 + 21) * 12 - 9^2

= 231 ft^2.

So they should plan to spend 231 * 1.5 = $346.50.

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I’m factoring trinomials,help plz?find gcf first

Luden [163]

Factor 8 out of 16x^2 - 24x +8

8 is the GFC

Factor the following:
16 x^2 - 24 x + 8

Factor 8 out of 16 x^2 - 24 x + 8:
8 (2 x^2 - 3 x + 1)
Factor the quadratic 2 x^2 - 3 x + 1.
The coefficient of x^2 is 2 and the constant term is 1.
The product of 2 and 1 is 2.
The factors of 2 which sum to -3 are -1 and -2.
So 2 x^2 - 3 x + 1 = 2 x^2 - 2 x - x + 1 = -(2 x - 1) + x (2 x - 1):
8 x (2 x - 1) - (2 x - 1)

Factor 2 x - 1 from x (2 x - 1) - (2 x - 1):
<span>Answer: 8 (2 x - 1) (x - 1)</span>

8 0

3 years ago

Find dy/dx for 4 - xy = y^3

storchak [24]

Answer:

$\frac{dy}{dx}=-\frac{y}{3y^2+x}$

Step-by-step explanation:

4-xy=y^3

dy/dx=?

$\frac{d(4-xy)}{dx}=\frac{d(y^3)}{dx}\\ \frac{d(4)}{dx}-\frac{d(xy)}{dx}=3y^{3-1}\frac{dy}{dx}\\ 0-(\frac{dx}{dx}y+x\frac{dy}{dx})=3y^2\frac{dy}{dx}\\ -(1y+x\frac{dy}{dx})=3y^2\frac{dy}{dx}\\ -(y+x\frac{dy}{dx})=3y^2\frac{dy}{dx}\\ -y-x\frac{dy}{dx}=3y^2\frac{dy}{dx}$

Solving for dy/dx: Addind x dy/dx both sides of the equation:

$-y-x\frac{dy}{dx}+x\frac{dy}{dx}=3y^2\frac{dy}{dx}+x\frac{dy}{dx} \\ -y=3y^2\frac{dy}{dx}+x\frac{dy}{dx}$

Common factor dy/dx on the right side of the equation:

$-y=(3y^2+x)\frac{dy}{dx}$

Dividing both sides of the equation by 3y^2+x:

$\frac{-y}{3y^2+x}=\frac{(3y^2+x)}{3y^2+x}\frac{dy}{dx}\\ -\frac{y}{3y^2+x}=\frac{dy}{dx}\\ \frac{dy}{dx}=-\frac{y}{3y^2+x}$

7 0

3 years ago

I pull the throttle in my racing plane at a = 12.0 m/s2. i was originally flying at v = 100. m/s. where am i when t = 2.0, t = 5

Misha Larkins [42]

The distance covered by plane when t = 2s will be 176m and when t = 5s will be 350m and when t = 10s will be 400m found using equation of motion.

We have,

Acceleration of plane i.e. a = 12 m/s²

And,

Velocity of plane i.e. v = 100 m/s

And,

t₁ = 2s

t₂ = 5s

t₃ = 10s

So,

Now,

Using the equation of motion,

i.e.

S = vt - $\frac{1}{2}$ at²

Here,

S = Distance,

v = initial velocity,

a = acceleration

t = time taken

Now,

For t₁ = 2s,

Putting values in above equation we get,

S = (100 * 2) - ( $\frac{1}{2}$ * 12 * 2²)

On solving we get,

S = 200 - 24 = 176 m,

So,

Plane will be at 176m distance.

Now,

For t₂ = 5s,

Putting values in above equation we get,

S = (100 * 5) - ( $\frac{1}{2}$ * 12 * 5²)

On solving we get,

S = 500 - 150 = 350 m,

So,

Plane will be at 350m distance.

Now,

For t₃ = 10s,

Putting values in above equation we get,

S = (100 * 10) - ( $\frac{1}{2}$ * 12 * 10²)

On solving we get,

S = 1000 - 600 = 400 m,

So,

Plane will be at 400m distance.

Hence, we can say that the distance covered by plane when t = 2s will be 176m and when t = 5s will be 350m and when t = 10s will be 400m found using equation of motion.

To learn more about equation of motion click here

brainly.com/question/12114762

#SPJ4

4 0

1 year ago

Write an equation in slope-intercept form of the line that passes through the given points.

Andre45 [30]

Answer:

$y=\frac{1}{3}x-3$