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3 years ago

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a man 1.30 m tall is on a top of a building. he observed a car on the road at an angle of 60 degrees if the building is 30 m hig

h,how far is the car from the building?

1 answer:

romanna [79]3 years ago

7 0

Given :

A man 1.30 m tall is on a top of a building. He observed a car on the road at an angle of 60 degrees if the building is 30 m high.

To Find :

How far is the car from the building.

Solution :

Let, distance between car and the building is x m.

So,

$tan \ \theta = \dfrac{height\ of\ building + height\ of\ man}{distance \ between \ car \ and \ building}\\\\tan \ 60^o = \dfrac{30+1.3}{x}\\\\x = \dfrac{31.3}{1.73}\ m\\\\x = 18.09\ m$

Hence, this is the required solution.

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7. A man makes a simple discount note for $6,200, at an ordinary bank discount rate of 8.84%, for 40 days. What is the effective

Elina [12.6K]

Answer:

The effective interest rate, rounded to the nearest tenth, is 0.1%.

Step-by-step explanation:

The banker's rule is the simple interest formula.

The simple interest formula is given by:

$E = P*I*t$

In which E are the earnings, P is the principal(the initial amount of money), I is the interest rate(yearly) and t is the time, in years.

The effective interest rate is given by the following formula:

$E_{IR} = \frac{E}{P}$ .

In this problem, we have that:

A man makes a simple discount note for $6,200, at an ordinary bank discount rate of 8.84%, for 40 days. We consider that the year has 360 days. This means that $P = 6200, I = 0.0884, t = \frac{40}{360} = \frac{1}{9}$ .

So

$E = 6200*0.0884*\frac{1}{9} = 60.9$

The effective interest rate is

$E_{IR} = \frac{E}{P} = \frac{60.9}{6200} = 0.0098 = 0.001$

The effective interest rate, rounded to the nearest tenth, is 0.1%.

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3 years ago

(1 point) consider the function f(t)=⎧⎩⎨⎪⎪⎪⎪0,−5,−6,6,t<00≤t<11≤t<7t≥7;f(t)={0,t<0−5,0≤t<1−6,1≤t<76,t≥7; 1. wr

sergij07 [2.7K]

$f(t)=\begin{cases}0&\text{for }t$

Recall that

$u(t)=\begin{cases}0&\text{for }t$

Take it one piece at a time. For $t\ge0$ , we can scale $u(t)$ by -5:

$-5u(t)=\begin{cases}0&\text{for }t$

If we shift the argument by 1 and scale by -5, we have

$-5u(t-1)=\begin{cases}0&\text{for }t$

so if we subtract this from $-5u(t)$ , we'll end up with

$-5u(t)+5u(t-1)=\begin{cases}0&\text{for }t$

For the next piece, we can add another scaled and shifted step like

$-6u(t-1)+6u(t-7)=\begin{cases}0&\text{for }t$

so that

$-5u(t)+5u(t-1)-6u(t-1)+6u(t-7)=\begin{cases}0&\text{for }t$

For the last piece, we add one more term:

$6u(t-7)=\begin{cases}0&\text{for }t$

and so putting everything together, we get $f(t)$ :

$f(t)\equiv-5u(t)+5u(t-1)-6u(t-1)+6u(t-7)+6u(t-7)$
$f(t)\equiv-5u(t)-u(t-1)+12u(t-7)$

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Its perpendicular, if you need me to explain lmk

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3 years ago

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madreJ [45]

Answer:

x=2√5

Step-by-step explanation:

<u>1st method:</u>

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sin(30)x=√5

x=√5/sin(30)

x=2√5

<u>2nd method:</u>

cos(60)=adjacent/hypotenuse=√5/x

cos(60)x=√5

x=√5/cos(60)

x=2√5

<u>3rd method:</u>

This is a 30-60-90 triangle, so we can multiply the short side, √5, by 2, to get x=2√5

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