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3 years ago

Last year, Nick went bowling several times and earned an average score

1 answer:

4 0

Answer: 150%

Explanation:

1. Turn it into a fraction. (123/82)

2. Divide the numerator (123) by the denominator (82). (123÷82=1.5)

3. Multiply the decimal (1.5) by 100 and add “%” at the end to make it a percentage. (1.5x100=150)(150=150%)

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PLEASE HELP ME: the ratio of cars to trucks on a highway at any given time is 12 to 8. If there are always more than 105 cars on

andre [41]

Answer:

Step-by-step explanation:

First, set up the ratio of c:t, which is 12:8. Then set up your new ratio of c:t, which is 105:?. The way I find the easiest is to divide 105 by 12, which is 8.75, then multiply that by 8, which is 70. To check if it's right, just divide 12 by 8, and 105 by 70, and if they're the same number, you've got it right.

Ps. I'm guessing you meant to ask the minimum number of trucks, not cars.

5 0

2 years ago

What are the focus and directrix of the parabola that is the graph of the function f(x)=x²?

Hitman42 [59]

Answer:

Step-by-step explanation:

Given

$y=x^2$

comparing it with standard equation $x^2=4ay$

so 4a=1

$a=\frac{1}{4}$

so Focus of parabola is (0,0)

directrix

y=-a

here $a=\frac{1}{4}$

$y=-\frac{1}{4}$

5 0

3 years ago

The angle of elevation from me to the top of a hill is 51 degrees. The angle of elevation from me to the top of a tree is 57 deg

julia-pushkina [17]

Answer:

Approximately $101\; \rm ft$ (assuming that the height of the base of the hill is the same as that of the observer.)

Step-by-step explanation:

Refer to the diagram attached.

Let $\rm O$ denote the observer.
Let $\rm A$ denote the top of the tree.
Let $\rm R$ denote the base of the tree.
Let $\rm B$ denote the point where line $\rm AR$ (a vertical line) and the horizontal line going through $\rm O$ meets. $\angle \rm B\hat{A}R = 90^\circ$ .

Angles:

Angle of elevation of the base of the tree as it appears to the observer: $\angle \rm B\hat{O}R = 51^\circ$ .
Angle of elevation of the top of the tree as it appears to the observer: $\angle \rm B\hat{O}A = 57^\circ$ .

Let the length of segment $\rm BR$ (vertical distance between the base of the tree and the base of the hill) be $x\; \rm ft$ .

The question is asking for the length of segment $\rm AB$ . Notice that the length of this segment is $\mathrm{AB} = (x + 20)\; \rm ft$ .

The length of segment $\rm OB$ could be represented in two ways:

In right triangle $\rm \triangle OBR$ as the side adjacent to $\angle \rm B\hat{O}R = 51^\circ$ .
In right triangle $\rm \triangle OBA$ as the side adjacent to $\angle \rm B\hat{O}A = 57^\circ$ .

For example, in right triangle $\rm \triangle OBR$ , the length of the side opposite to $\angle \rm B\hat{O}R = 51^\circ$ is segment $\rm BR$ . The length of that segment is $x\; \rm ft$ .

$\begin{aligned}\tan{\left(\angle\mathrm{B\hat{O}R}\right)} = \frac{\,\rm {BR}\,}{\,\rm {OB}\,} \; \genfrac{}{}{0em}{}{\leftarrow \text{opposite}}{\leftarrow \text{adjacent}}\end{aligned}$ .

Rearrange to find an expression for the length of $\rm OB$ (in $\rm ft$ ) in terms of $x$ :

$\begin{aligned}\mathrm{OB} &= \frac{\mathrm{BR}}{\tan{\left(\angle\mathrm{B\hat{O}R}\right)}} \\ &= \frac{x}{\tan\left(51^\circ\right)}\approx 0.810\, x\end{aligned}$ .

Similarly, in right triangle $\rm \triangle OBA$ :

$\begin{aligned}\mathrm{OB} &= \frac{\mathrm{AB}}{\tan{\left(\angle\mathrm{B\hat{O}A}\right)}} \\ &= \frac{x + 20}{\tan\left(57^\circ\right)}\approx 0.649\, (x + 20)\end{aligned}$ .