A light bulb consumes 1440 watt hours per day. How long does it take to consume 4680 watt hours?

The charge for a bicylce repair was $9.28 for parts, 1/4 hour of labor at $18 per hour, and a $2 shop fee. What was the total co

Bas_tet [7]

9.28 plus 2 is 11.28. 1/4 of 18 is 4.50. Add 4.50 to 9.28 and your total is $15.78.

The equation would be:
9.28+18(.25)+2=c

7 0

2 years ago

a section of ground measures 1.5 cubic meters and contains 0.78 gram of gold. what is the amount of gold per cubic meter? explai

I am Lyosha [343]

The answer

the amount of gold per cubic meter can be found by the ratio

0.78 g /1 .5 cubic meters = 0.52 g / cubic meter,
proof
it was said the amount of gold per cubic meter (Density)

7 0

3 years ago

How many solutions are there for the system of equations of shown on the graph?

bearhunter [10]

Answer:

one solution

(second option listed)

Step-by-step explanation:

We can that these two lines, each representing one equation/function, only meet at one specific value.

In a system of equations, we are essentially looking for a solution that works for both equations.

So, if both lines share a point/value (meaning they intersect), that point is a solution to the system of equations.

Because these lines only overlap at one point, this system of equations has one solution.

3 0

2 years ago

Read 2 more answers

Hello I need help Finding the probability

Stolb23 [73]

Each petal of the region $R$ is the intersection of two circles, both of diameter 10. Each petal in turn is twice the area of a circular segment bounded by a chord of length $5\sqrt2$ , which implies the segment is subtended by an angle of $\dfrac\pi2$ . This means the area of the segment is

$\text{area}_{\text{segment}}=\text{area}_{\text{sector}}-\text{area}_{\text{triangle}}$
$\text{area}_{\text{segment}}=\dfrac{25\pi}4-\dfrac{25}2$

This means the area of one petal is $\dfrac{25\pi}2-25$ , and the area of $R$ is four times this, or $50\pi-100$ .

Meanwhile, the area of $G$ is simply the area of the square minus the area of $R$ , or $10^2-(50\pi-100)=200-50\pi$ .

So

$\mathbb P(X=R)=\dfrac{50\pi-100}{100}=\dfrac\pi2-1$
$\mathbb P(X=G)=\dfrac{200-50\pi}{100}=2-\dfrac\pi2$
$\mathbb P((X=R)\land(X=G))=0$ (provided these regions are indeed disjoint; it's hard to tell from the picture)
$\mathbb P((X=R)\lor(X=G))=\mathbb P(X=R)+\mathbb P(X=G)=1$

4 0

3 years ago

A post is supported by two wires (one on each side going in oppositedirections) creating an angle of 80° between the wires. The

Vladimir [108]

Using the Sine rule,

$\frac{\sin A}{A}=\frac{\sin B}{B}=\frac{\sin C}{C}$ $\begin{gathered} \text{Let A = 14m,} \\ Substituting the variables into the formula,Where the length of the wires are, AP = xm and BP = ym[tex]\begin{gathered} \frac{\sin80^0}{14}=\frac{\sin40^0}{x} \\ \text{Crossmultiply,} \\ x\times\sin 80^0=14\times\sin 40^0 \\ Divide\text{ both sides by }\sin 80^0 \\ x=\frac{14\sin40^0}{\sin80^0} \\ x=9.14m \end{gathered}$

Hence, the length of wire AP (x) is 9.14m.

For wire BP (y)m,

Sum of angles in a triangle is 180 degrees,

$A^0+P^0+B^0=180^0$ $\begin{gathered} \text{Where A}^0=\text{ unknown,} \\ P^0=80^0\text{ and,} \\ B^0=40^0 \\ A^0+80^0+40^0=180^0 \\ A^0+120^0=180^0 \\ A^0=180^0-120^0 \\ A^0=60^0 \end{gathered}$

Using the side rule to find the length of wire BP,

$\begin{gathered} \frac{\sin 60^0}{y}=\frac{\sin 80^0}{14} \\ \text{Crossmultiply,} \\ y\times\sin 80^0=14\times\sin 60^0 \\ \text{Didive both sides by }\sin 80^0 \\ y=\frac{14\times\sin 60^0}{\sin 80^0} \\ y=12.31m \end{gathered}$

Hence, the length of wire BP (y) is 12.31m

Therefore, the length of the wires are (9.14m and 12.31m).

4 0

11 months ago