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3 years ago

The square of a number minus the number = 20, Can You form a quadratic equation and factorise it?

Mathematics

1 answer:

frosja888 [35]3 years ago

7 0

Yes you can ... here is an example x^2-x=20, x^2-X-20 factored is (x-5)(x+4)

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HELP ASAP!!! DUE SOON!!! 7 7/8 - 3 9/16

jeka57 [31]

Answer:

4 5/16

Step-by-step explanation:

7 7/8 - 3 9/16

7 14/16 - 3 9/16

4 5/16

4 0

3 years ago

Read 2 more answers

What is the length of side BC of the triangle? Enter your answer in the box. units Triangle A B C with horizontal side B C. Vert

harina [27]

Answer:

The length of BC = 28 units

Step-by-step explanation:

* In triangle ABC

∵ Angles B and C are congruent

∴ m∠B = m∠C

* In any triangle if two angles are equal in measure, then the triangle is isosceles means the two sides which opposite to the congruent angles are equal in length

∴ AB = AC

∵ AB = 4x - 7

∵ AC = 2x + 7

∴ 4x - 7 = 2x + 7 ⇒ collect like terms

∴ 4x - 2x = 7 + 7

∴ 2x = 14 ⇒ divide both sides by 2

∴ x = 14 ÷ 2 = 7

* Now we can find the length of BC

∵ The length of BC = 4x ⇒ substitute the value of x

∴ The length of BC = 4 × 7 = 28 units

6 0

3 years ago

HELP PLEASE

Gala2k [10]

Answer:

3-11n>21-5n

Step-by-step explanation:

It seems right, so hopefully it is.

7 0

2 years ago

Find gradient xe^y + 4 ln y = x² at (1, 1)

cricket20 [7]

$xe^y+4\ln y=x^2$

Differentiate both sides with respect to x, assuming y = y(x).

$\dfrac{\mathrm d(xe^y+4\ln y)}{\mathrm dx}=\dfrac{\mathrm d(x^2)}{\mathrm dx}$

$\dfrac{\mathrm d(xe^y)}{\mathrm dx}+\dfrac{\mathrm d(4\ln y)}{\mathrm dx}=2x$

$\dfrac{\mathrm d(x)}{\mathrm dx}e^y+x\dfrac{\mathrm d(e^y)}{\mathrm dx}+\dfrac4y\dfrac{\mathrm dy}{\mathrm dx}=2x$

$e^y+xe^y\dfrac{\mathrm dy}{\mathrm dx}+\dfrac4y\dfrac{\mathrm dy}{\mathrm dx}=2x$

Solve for dy/dx :

$e^y+\left(xe^y+\dfrac4y\right)\dfrac{\mathrm dy}{\mathrm dx}=2x$

$\left(xe^y+\dfrac4y\right)\dfrac{\mathrm dy}{\mathrm dx}=2x-e^y$

$\dfrac{\mathrm dy}{\mathrm dx}=\dfrac{2x-e^y}{xe^y+\frac4y}$

If y ≠ 0, we can write

$\dfrac{\mathrm dy}{\mathrm dx}=\dfrac{2xy-ye^y}{xye^y+4}$

At the point (1, 1), the derivative is

$\dfrac{\mathrm dy}{\mathrm dx}\bigg|_{x=1,y=1}=\boxed{\dfrac{2-e}{e+4}}$

4 0

3 years ago

Gm everyone how are yall lol

Anni [7]

Answer:

I'm doing good thanks for asking :)

Step-by-step explanation:

I'm also a bit tired and overworked since I have to do online school

7 0

2 years ago