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3 years ago

The first row are input, x the second row is output, y .

Mathematics

1 answer:

Stolb23 [73]3 years ago

4 0

Answer:

Domain is also called input you got this

Step-by-step explanation:

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Can somebody help me with this pleas.

Yuki888 [10]

Answer:

what do you need help with <33

Step-by-step explanation:

8 0

3 years ago

Read 2 more answers

Answer number 1 and use the lattice board I set up

Aneli [31]

6*3=18
5*3=15
6*4=24
5*4=20
then you would add the columns and you would get 1,904
It would be easier to show you.

7 0

3 years ago

Work out the area of this trapezium

skelet666 [1.2K]

Step-by-step explanation:

Area = ½ × b × h

= ½ × (7.9 + 10.8) × 6.2

= ½ × 18.7 × 6.2

= ½ × 115.94

= 57.97 cm²

So, the area of that trapezium is 57.97 cm²

Hope it helpful and useful :)

6 0

3 years ago

If the sum of the zereos of the quadratic polynomial is 3x^2-(3k-2)x-(k-6) is equal to the product of the zereos, then find k?

lys-0071 [83]

Answer:

Step-by-step explanation:

So I'm going to use vieta's formula.

Let u and v the zeros of the given quadratic in ax^2+bx+c form.

By vieta's formula:

1) u+v=-b/a

2) uv=c/a

We are also given not by the formula but by this problem:

3) u+v=uv

If we plug 1) and 2) into 3) we get:

-b/a=c/a

Multiply both sides by a:

-b=c

Here we have:

a=3

b=-(3k-2)

c=-(k-6)

So we are solving

-b=c for k:

3k-2=-(k-6)

Distribute:

3k-2=-k+6

Add k on both sides:

4k-2=6

Add 2 on both side:

4k=8

Divide both sides by 4:

k=2

Let's check:

$3x^2-(3k-2)x-(k-6) \text{ with }k=2$ :

$3x^2-(3\cdot 2-2)x-(2-6)$

$3x^2-4x+4$

I'm going to solve $3x^2-4x+4=0$ for x using the quadratic formula:

$\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

$\frac{4\pm \sqrt{(-4)^2-4(3)(4)}}{2(3)}$

$\frac{4\pm \sqrt{16-16(3)}}{6}$

$\frac{4\pm \sqrt{16}\sqrt{1-(3)}}{6}$

$\frac{4\pm 4\sqrt{-2}}{6}$

$\frac{2\pm 2\sqrt{-2}}{3}$

$\frac{2\pm 2i\sqrt{2}}{3}$

Let's see if uv=u+v holds.

$uv=\frac{2+2i\sqrt{2}}{3} \cdot \frac{2-2i\sqrt{2}}{3}$

Keep in mind you are multiplying conjugates:

$uv=\frac{1}{9}(4-4i^2(2))$

$uv=\frac{1}{9}(4+4(2))$

$uv=\frac{12}{9}=\frac{4}{3}$

Let's see what u+v is now:

$u+v=\frac{2+2i\sqrt{2}}{3}+\frac{2-2i\sqrt{2}}{3}$

$u+v=\frac{2}{3}+\frac{2}{3}=\frac{4}{3}$

We have confirmed uv=u+v for k=2.

4 0

3 years ago

8 = 4 - 2x - 2x +5 = 17 12= - 3 ( x+ 5)

max2010maxim [7]

$8 = 4 - 2x\\8 + 2x = 4\\2x = -4\\x = -2$

$- 2x +5 = 17\\-2x =12\\x = -6$

$12= - 3 ( x+ 5)\\12 = -3x -15\\-3x =27\\x= -9$

5 0

3 years ago