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Ksenya-84 [330]
3 years ago
5

What is the constant variation in the equation 3y=6x? Explain. PLEASE HELP IM DESPERATE

Mathematics
1 answer:
lbvjy [14]3 years ago
4 0
Hi Desperate!

anyways the equation for constant variation is y=kx
3y=6x is it, but we don't want the 3.
so we divide on both sides to cancel out the 3.
y=2x
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Camila Cabello wants to order cheesecakes over the internet. Each cheesecake costs $15.99 and shipping for the entire order is
ladessa [460]

Answer:

she can purchase 5 cheesecakes

Step-by-step explanation:

15.99*5=79.95

if we add 9.99 for the shipping of thw whole order

79.95+9.99=89.94

6 0
3 years ago
Adena used a qualateral in her ary design. The quadrlateral has no sides ,and only one pair of opposite sides parallel what shap
Radda [10]

Answer: The quadrilateral used here is a trapezium.




3 0
4 years ago
What are the solutions of the equation x4 – 5x2 - 36 = 0? Use factoring to solve.
balandron [24]

Answer:

x = -2i and x = 2i

x = 3 and x = - 3

Step-by-step explanation:

x^4 - 5x^2 - 36 = 0

let u = x^2 and substitute

we have:

u^2 - 5u - 36 = 0

(u - 9)(u + 4) = 0

u = 9 and u = - 4

substitute back for u

x^2 = 9

x = 3 and x = - 3

x^2 = -4 -- i am going to assume you've learned imaginary numbers; otherwise, you cannot take the square root of a negative number.

x = -2i and x = 2i

5 0
3 years ago
A 4 metre ladder is placed against a vertical wall.
Black_prince [1.1K]

Answer:

Original position: base is 1.5 meters away from the wall and the vertical distance from the top end to the ground let it be y and length of the ladder be L.

Step-by-step explanation:

By pythagorean theorem, L^2=y^2+(1.5)^2=y^2+2.25 Eq1.

Final position: base is 2 meters away, and the vertical distance from top end to the ground is y - 0.25 because it falls down the wall 0.25 meters and length of the ladder is also L.

By pythagorean theorem, L^2=(y -0.25)^2+(2)^2=y^2–0.5y+ 0.0625+4=y^2–0.5y+4.0625 Eq 2.

Equating both Eq 1 and Eq 2: y^2+2.25=y^2–0.5y+4.0625

y^2-y^2+0.5y+2.25–4.0625=0

0.5y- 1.8125=0

0.5y=1.8125

y=1.8125/0.5= 3.625

Using Eq 1: L^2=(3.625)^2+2.25=15.390625, L=(15.390625)^1/2= 3.92 meters length of ladder

Using Eq 2: L^2=(3.625)^2–0.5(3.625)+4.0625

L^2=13.140625–0.90625+4.0615=15.390625

L= (15.390625)^1/2= 3.92 meters length of ladder

<em>hope it helps...</em>

<em>correct me if I'm wrong...</em>

4 0
3 years ago
Given part of the equation of a parabola in root form, y=a(x-3)(x+4). What is the value of ‘a’ if the parabola passes through th
Elis [28]

<em><u>Answer:</u></em>

The answer is: a = -2.

<em><u>Step-by-step explanation:</u></em>

A = -2, because after graphing the equation in Desmos and plugging in various values for a, I found that when a = -2, the equation passed through the point (-3,12), so that's the answer. I'll attach a photo of the graph.

Hope this helps! Feel free to give me Brainliest if you feel this helped. Have a good day, and good luck on your assignment. :)

5 0
3 years ago
Read 2 more answers
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