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3 years ago

Anybody know the answer?

Mathematics

1 answer:

dsp733 years ago

8 0

So since its a right triangle, we can use the pythagorean theorem, which is $leg^2+leg^2=hypotenuse^2$ . Since we know that 2 and 3 are our legs and x is our hypotenuse, we can solve the equation as such:

$2^2+3^2=x^2\\ 4+9=x^2\\ 13=x^2\\ \sqrt{13}=x$

Since √13 can't be further simplified, that is your final answer.

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What does -8(6x - 2) equal? How do I get my answer?

Free_Kalibri [48]

Answer:

-48x+16

Step-by-step explanation:

-8(6x - 2) first multiply each term in the parentheses by 8 to get -8x6x-8x(-2) and calculate the product to get your answer which is -48x+16

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3 years ago

Read 2 more answers

Please help!!! Answer needed fast<3

zhannawk [14.2K]

Answer:

-5 + 6 = 1

Step-by-step explanation:

The model number may be used to simplify -5 + 6 = 1

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2 years ago

What is 63 over 75 as a decimal?

marissa [1.9K]

Answer:

0.84

Step-by-step explanation:

Use a calculator. That would be 0.84.

3*21

Note that 63/75 = ---------- = 21/25. Multiplying top and bottom by 4 results

3/25 in 0.84 (so this answer checks out ok)

3 0

3 years ago

Tres amigos: Andrés, Braulio y Ciro, juegan 3 apuestas

skelet666 [1.2K]

Answer:

La persona que ganó más dinero fue Braulio por S/.16

Step-by-step explanation:

Deje que la cantidad inicial sea Andrés X, Braulio Y y Ciro Z

Luego, dado que cada uno pierde una apuesta en el orden presentado, tenemos;

Después de la primera apuesta, tenemos;

Andrés X-Y-Z, Braulio (Y + Y), Ciro (Z + Z)

En la segunda apuesta, tenemos;

2 (X-Y-Z), 2Y - (X-Y-Z) - 2Z, 2Z + 2Z

2 (X-Y-Z), 3Y - X -Z, 4Z

En el tercero

4 (X-Y-Z), 6Y - 2X - 2Z, 4Z-2 (X-Y-Z) - (3Y - X -Z)

4 (X-Y-Z), 6Y - 2X - 2Z, 7Z - X - Y

4 (X-Y-Z) = 0.48 .............................. (1)

6Y - 2X - 2Z = 0.56 ...................... (2)

7Z - X - Y = 0.28 ............................ (3)

Multiplicamos las ecuaciones (2) por 2 y restamos la ecuación (1), tenemos;

12Y - 4X - 4Z - 4X + 4Y + 4Z = 0.56 * 2 - 0.48 = 0.64

16Y - 8X = 0.64 ...................... (4)

Multiplicamos las ecuaciones (2) por 7/2 y sumamos a la ecuación (3), tenemos;

21Y - 7X - 7Z + 7Z - X - Y = 7/2 * 0.56 + 0.28

20Y - 8X = 2.24 ...................... (5)

Restando la ecuación (4) de la ecuación (5) tenemos;

20Y - 8X - 16Y - 8X = 2.24 - 0.64 =

4Y = 1.6

Y = 1.6 / 4 = 0.4

De la ecuación (5), tenemos;

20Y - 8X = 20 * 0.4 - 8X = 2.24

8X = 8 - 2.24 = 5.76

X = 5,76 / 8 = 0,72

De la ecuación (3), tenemos;

7Z - X - Y = 7Z - 0.72 - 0.4 = 0.28

7Z = 1.4

Z = 1.4 / 7 = 0.2

Por lo tanto tienen inicialmente

X₁ = 0.72

Y₁ = 0.4

Z₁ = 0.2

Después de jugar tienen;

X₂ = 0.48

Y₂ = 0.56

Z₂ = 0.28

Los cambios son

Andrés 0.48 - 0.72 = -0.24

Braulio; 0.56 - 0.4 = 0.16

Ciro 0.28 - 0.2 = 0.08

La persona que ganó más dinero fue Braulio por S/.16

7 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