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3 years ago

Helpppppppppppp!!!!!!

Mathematics

1 answer:

Ipatiy [6.2K]3 years ago

8 0

Answer:

1- (-1,4)

2- (3,8)

Step-by-step explanation:

PRETTY SURE

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Helppp needed ( answer quickly)

ale4655 [162]

Answer:

C. 57

Step-by-step explanation:

all triangles add up to 180. the tick marks show that ∠D and ∠ E are congruent so you do this and solve for x:

(4x + 1) + (4x + 1) + (5x - 4) = 180

13x - 2 = 180

13x = 182

x = 182/13

x = 14

4(14) + 1 = 57

3 0

3 years ago

24 + (-9) - 44 + 33 - 17<br> word problem form

pashok25 [27]

Twenty four plus (negative nine) minus forty-four plus thirty three minus seventeen.

7 0

2 years ago

What does he mean on question one

Katarina [22]

You plug -2 in for the function k(p) and add it to the function g(w), getting
(-2+3)*(-2-7)+(-2-5)^2=1*-9+49=40 for a - I challenge you to do B on your own!

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

2 years ago

Choose ALL the right triangles.

11Alexandr11 [23.1K]

Answer:

Bottom left corner and bottom right corner

Step-by-step explanation:

6 0

2 years ago

Read 2 more answers