Ask question

Ask question

All categories

dimulka [17.4K]

3 years ago

13

The Carolina Castaways and Raleigh Rockets scored a total of 96 points. The Rockets scored one third as many points as the Casta

ways.

2 answers:

zysi [14]3 years ago

8 0

The castaways scored 32 pts

Hope it helps

11Alexandr11 [23.1K]3 years ago

4 0

I believe it's 32
96/1(1/3)
96/3
32

You might be interested in

[asy] size(7cm); draw((1.8,0)--(3.2,0),Arrows); draw((2,0.03)--(2,-0.03)); draw((3,0.03)--(3,-0.03)); for(real i=15/7; i<2.9;

shutvik [7]

Answer:

2 2/7

Step-by-step explanation:

3 0

2 years ago

if the equation y=3/2 x+5 were to be graphed in the xy-plane above, at what point would the graph intersect the line shown

adoni [48]

Answer:

if you go on https://www.geogebra.org/ it will graph it for you

Step-by-step explanation:

3 0

2 years ago

Please helppp it’s urgent , I’ll give a crown !!

mars1129 [50]

Answer:

Slide to the left then rotate 90° to the right then slide down then rotate 180° to the left/ right.

8 0

2 years ago

Solve for x. 3x−82=x−6 Enter your answer in the box. x =

My name is Ann [436]

Get like terms together. you might have to move across the equals sign

3x-x=-6+82

2x=76

x=38

6 0

2 years ago

Solve the following equation:

Rama09 [41]

Complete the square.

$z^4 + z^2 - i\sqrt 3 = \left(z^2 + \dfrac12\right)^2 - \dfrac14 - i\sqrt3 = 0$

$\left(z^2 + \dfrac12\right)^2 = \dfrac{1 + 4\sqrt3\,i}4$

Use de Moivre's theorem to compute the square roots of the right side.

$w = \dfrac{1 + 4\sqrt3\,i}4 = \dfrac74 \exp\left(i \tan^{-1}(4\sqrt3)\right)$

$\implies w^{1/2} = \pm \dfrac{\sqrt7}2 \exp\left(\dfrac i2 \tan^{-1}(4\sqrt3)\right) = \pm \dfrac{2+\sqrt3\,i}2$

Now, taking square roots on both sides, we have

$z^2 + \dfrac12 = \pm w^{1/2}$

$z^2 = \dfrac{1+\sqrt3\,i}2 \text{ or } z^2 = -\dfrac{3+\sqrt3\,i}2$

Use de Moivre's theorem again to take square roots on both sides.

$w_1 = \dfrac{1+\sqrt3\,i}2 = \exp\left(i\dfrac\pi3\right)$

$\implies z = {w_1}^{1/2} = \pm \exp\left(i\dfrac\pi6\right) = \boxed{\pm \dfrac{\sqrt3 + i}2}$

$w_2 = -\dfrac{3+\sqrt3\,i}2 = \sqrt3 \, \exp\left(-i \dfrac{5\pi}6\right)$

$\implies z = {w_2}^{1/2} = \boxed{\pm \sqrt[4]{3} \, \exp\left(-i\dfrac{5\pi}{12}\right)}$

3 0

1 year ago