What is the quantity proportional to 18/6

You roll a standard number cube once. Find P(0)

Sphinxa [80]

??? Im lost in this question, could you be more specific?

7 0

3 years ago

Geometry B: fill in the blank

olga_2 [115]

Answer:

The circumference of a circle is the distance around the entire circle, while arc length, s, is the distance between two points on the circle.

C = 2πr = 2π( 5 ) = 10π ≈ 31.4 units

3.14 is an approximation for π.

s = θ/360° (C)

mAB = 60°/360° ( 31.4)

= 1/6(31.4) ≈5.2 units

3 0

2 years ago

Solve the system of equations y = 2x + 3 and 4x - 2y = -6 using a graphical method.

emmasim [6.3K]

Answer:

-1

Step-by-step explanation:

You are asked to solve this set by substitution. Note that the first equation defines

the value of y in terms of x. Therefore, you can use the right side of this equation

(2x + 3) as a substitution for y in the second equation. When you substitute 2x +3 for y

in the second equation, the result is:

.

2x + 3 = 4x + 7

.

Let's set the goal of getting all the terms that contain x isolated on the left side of

the equal sign and all the constants by themselves on the right side. Begin by getting rid

of the +3 on the left side by subtracting 3 from the left side. But if you subtract

3 from the left side you must also subtract 3 from the right side. These subtractions

result in the following sequence:

.

2x + 3 - 3 = 4x + 7 - 3

.

Combining numbers on the left side and on the right side simplifies the equation to:

.

2x = 4x + 4

.

Now, in a similar fashion, let's get rid of the 4x on the right side by subtracting

4x from the right side. When we do this subtraction, to keep the equation in balance

we must also subtract 4x from the left side. This results in:

.

2x - 4x = 4x - 4x + 4

.

On both sides of the equation combine terms containing x to get:

.

-2x = 4

.

Finally, solve for x by dividing both sides of the equation by -2 because -2 is the multiplier

of x. This results in:

.

x = 4/-2 = -2

.

So now that we know x = -2, we can return to either one of the original equations

in the equation set and substitute -2 for x. Then we can solve for y. Let's return

to the first equation and substitute -2 for x. This begins with:

.

y = 2x + 3

.

Then substituting -2 for x gives:

.

y = 2*(-2) + 3 = -4 + 3 = -1

7 0

3 years ago

Multiply (x-4)(x^2-5x-6) ...?

frez [133]

See attached photo for answer.

6 0

3 years ago

Read 2 more answers

Use the quadratic formula to solve the following equation -3x^2-x-3=0

tigry1 [53]

Answer:

$x=-\frac{1}{6}-\frac{\sqrt{35}}{6} i \text { and } x=-\frac{1}{6}+\frac{\sqrt{35}}{6} i$ are two roots of equation $-3 x^{2}-x-3=0$

Solution:

Need to solve given equation using quadratic formula.

$-3 x^{2}-x-3=0$

General form of quadratic equation is $a x^{2}+b x+c=0$

And quadratic formula for getting roots of quadratic equation is

$x=\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}$

In our case b = -1 , a = -3 and c = -3

Calculating roots of the equation we get

$\begin{array}{l}{x=\frac{-(-1) \pm \sqrt{(-1)^{2}-4(-3)(-3)}}{2 \times-3}} \\\\ {x=-\frac{1}{6} \pm\left(-\frac{\sqrt{-35}}{6}\right)}\end{array}$

Since $b^{2}-4 a c$ is equal to -35, which is less than zero, so given equation will not have real roots and have complex roots.

$\begin{array}{l}{\text { Hence } x=-\frac{1}{6}-\frac{\sqrt{35}}{6} i \text { and } x=-\frac{1}{6}+\frac{\sqrt{35}}{6} i \text { are two roots of equation - }} \\ {3 x^{2}-x-3=0}\end{array}$

8 0

3 years ago