Janson has quarters and dimes in his pocket. he counted 42 coins that added up to $7.50

Mathematics

2 answers:

vlada-n [284]3 years ago

7 0

Answer:

28 quarters and 5 dimes

7(Q)+5(D)

7(4)+5(10)

28Q 50D Q=quarters D=dimes

it would be more efficient to do the substitution method bc its faster and easier

scoray [572]3 years ago

4 0

I think Jason has 57

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From a window 33.0 ft above the street, the angle of elevation to the top of the building across the street is 45.0° and

puteri [66]

Answer:

Approximately $105\; \rm ft$ (assuming that the street is level.)

Step-by-step explanation:

Refer to the diagram attached. Consider the height of this building as the sum of two parts: the part below the window of the observer, and the part above the window of the observer.

The question states that the window of the observer is $33\; \rm ft$ above the street. Assume that the street is level. The height of the part of that building below the window of this observer would also be $33\; \rm ft\!$ .

Make use of the angle $18^{\circ}$ to calculate the horizontal distance between that building and the window of the observer.

Refer to the diagram attached. There are two right triangles in this diagram. Consider the one on the bottom (the one with the angle that measures $18^{\circ}$ .) The two sides of this right triangle adjacent to the right angle correspond to:

the height of the opposite building below the window of the observer (the green line segment on the right-hand side): $33\; \rm ft$ , and
the horizontal distance between the two buildings (the horizontal dashed line in the middle.

The angle that measures $18^{\circ}$ is adjacent to the side that corresponds to the horizontal distance between the two buildings. The tangent of this angle would relate the lengths of the two sides:

$\displaystyle \tan(18^{\circ}) = \frac{33\; \rm ft}{\text{horizontal distance}} \quad \genfrac{}{}{0}{}{\leftarrow \text{opposite}}{\leftarrow \text{adjacent}}$ .

Hence, the horizontal distance between the two buildings would be:

$\begin{aligned}\text{horizontal distance} = \frac{33\; \rm ft}{\tan(18^{\circ})}\end{aligned}$ .

In the other right triangle in this diagram (the one with the angle that measures $45^{\circ}$ , the two sides adjacent to the right angle correspond to:

the height of the opposite building above the window of the observer, and
the horizontal distance between the two buildings.

Again, the tangent of the $45^{\circ}$ angle would correspond to:

$\displaystyle \tan(45^{\circ}) = \frac{\text{height of opposite building above the window}}{\text{horizontal distance}} \quad \genfrac{}{}{0}{}{\leftarrow \text{opposite}}{\leftarrow \text{adjacent}}$ .

Rearrange to find the height of the building above the window of the observer:

$\begin{aligned}& \text{height of opposite building above the window} \\ =\; & (\tan(45^{\circ})) \, (\text{horizontal distance}) \\ =\; & \tan(45^{\circ}) \cdot \frac{33\; \rm ft}{\tan(18^{\circ})} \\ \approx\; &102\; \rm ft\end{aligned}$ .