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3 years ago

Tom's age is 5 more than twice Dic's age.

1 answer:

4 0

If we break the answer down, we can solve this pretty easily.
We can represent Dic's age with x.
Tom's age is 5 MORE (+5) than TWICE (2x) Dic's age (x)
Therefore, our answer is 5+2x (Five more than twice Dic's age)

The correct answer is C: 5+2x

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From a point A that is 8.20 m above level ground, the angle of elevation of the top of a building s 31 deg 20 min and the angle

Mariulka [41]

Answer:

30.11 meters ( approx )

Step-by-step explanation:

Let x be the distance of a point P ( lies on the building ) from the top of the building such that AP is perpendicular to the building and y be the distance of the building from point A, ( shown in the below diagram )

Given,

Point A is 8.20 m above level ground,

So, the height of the building = ( x + 8.20 ) meters,

Now, 1 degree = 60 minutes,

⇒ $1\text{ minute } =\frac{1}{60}\text{ degree }$

$20\text{ minutes }=\frac{20}{60}=\frac{1}{3}\text{ degree}$

$50\text{ minutes }=\frac{50}{60}=\frac{5}{6}\text{ degree}$

By the below diagram,

$tan ( 12^{\circ} 50') = \frac{8.20}{y}$

$tan(12+\frac{5}{6})^{\circ}=\frac{8.20}{y}$

$tan (\frac{77}{6})^{\circ}=\frac{8.20}{y}$

$\implies y=\frac{8.20}{tan (\frac{77}{6})^{\circ}}$

Now, again by the below diagram,

$tan (31^{\circ}20')=\frac{x}{y}$

$tan(31+\frac{1}{3})=\frac{x}{y}$

$\implies x=y\times tan(\frac{94}{3})=\frac{8.20}{tan (\frac{77}{6})^{\circ}}\times tan(\frac{94}{3})^{\circ}=21.9142943216\approx 21.91$

Hence, the height of the building = x + 8.20 = 30.11 meters (approx)

8 0

2 years ago

Which equation matches the graph?

ipn [44]

The equation that matches the graph is the second one !

4 0

3 years ago

Plz help i'm having troubled understanding how to do this

Lunna [17]

Answer:

1/2 - 2/6 = 1/6

Step-by-step explanation:

to answer two that are different simply take the lesser up or higher down example: 1/2-2/4 this would be zero because 2 x 1/2 = 2/4

8 0

2 years ago

Read 2 more answers

Mason walked on Monday, Wensday, and Friday. Use these clues to find how far he walked each day. These distances were six-eights

krok68 [10]

Answer:

Monday he walked $\frac{1}{4}miles$

Wednesday he walked $\frac{6}{8}miles$

Friday he walked $\frac{1}{6}miles$

Step-by-step explanation:

Given Mason walked on Monday, Wednesday and Friday. These distances were six-eights mile, one-fourth mile, and one-sixth mile. He did not walk the farthest on Monday. He walked less on Friday than Monday. we have to find how far he walked each day.

distances are $\frac{6}{8}, \frac{1}{4}, \frac{1}{6}$ that are 0.75, 0.25 and 0.17 respectively.