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3 years ago

Find the missing length marked with a question mark. UVW-USR

Mathematics

1 answer:

mariarad [96]3 years ago

3 0

Given:

$\Delta UVW\sim \Delta USR$

To find:

The missing length marked with a question mark.

Solution:

Let x be the missing value.

We have,

$\Delta UVW\sim \Delta USR$

Corresponding sides of similar triangles are proportional, so

$\dfrac{UV}{US}=\dfrac{VW}{SR}=\dfrac{UW}{UR}$

Using this, we get

$\dfrac{UV}{US}=\dfrac{UW}{UR}$

On substituting the values, we get

$\dfrac{70}{x}=\dfrac{55}{44}$

$\dfrac{70}{x}=\dfrac{5}{4}$

On cross multiplication, we get

$70\times 4=5\times x$

$280=5x$

$\dfrac{280}{5}=x$

$56=x$

Therefore, the missing length is 56 units.

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A consumer is considering two different purchasing options for the car of their choice. The first option, which is leasing, is d

AVprozaik [17]

Answer:

y=250x+4000
y=400x+400

Step-by-step explanation:

Given that:

x represents the number of months of ownership; and
y represents the total paid for the car after ‘x' months.

First Option (Leasing)

250x - y + 4000 = 0

Expressing the equation in the Slope-Intercept Form y=mx+b, we have:

y=250x+4000

Second Option (Financing)

$400 for 0 months of ownership, (0,400), and $4400 for 10 months of ownership, (10, 4400).

First, we determine the slope of the line joining (0,400) and (10,4400)

$Slope, m= \dfrac{4400-400}{10-0}= \dfrac{4000}{10}=400$

We have:

y=400x+b

When y=400, x=0

400=400(0)+b

b=400

Therefore, the Slope-Intercept Form of the second option is:

y=400x+400

Significance

In the first option, there is a down payment of $4000 and a monthly payment of $250.
In the second option, there is a down payment of $400 and a monthly payment of $400.

Part B

We notice from the graph that after 24 months, the cost for leasing and financing becomes the same ($10,000). Therefore, a consumer will be better off financing since the downpayment for leasing is higher.

i.e

When x=0, y=$4000 for leasing
When x=0, y=$400 for financing

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4 years ago

A lighthouse is located on an island 33 miles from the closest point on a straight shoreline. If the lighthouse light rotates cl

Zinaida [17]

Question:

A lighthouse is located on an island 3 miles from the closest point on a straight shoreline. If the lighthouse light rotates clockwise at a constant rate of 9 revolutions per minute, how fast does the beam of light move towards the point on the shore closest to the island when it is 52 miles from that point

Answer:

The beam of light moves at $16278\pi$ miles/min

Step-by-step explanation:

This question is illustrated with the attached image

Taking the instructions in the question, one at a time.

A revolution of 9 per minute implies that:

$\frac{d\theta}{dt} = \frac{9 * 2\pi\ rad}{1\ min}$

$\frac{d\theta}{dt} = 18\pi \frac{rad}{min}$

Take tan of the angle in the attachment:

$tan(\theta) =\frac{52}{3}$

Differentiate both sides with respect to time

$\frac{d\ tan(\theta)}{dt} =\frac{52}{3} * \frac{dx}{dt}$

Rewrite as:

$\frac{d\ tan(\theta)}{d\theta} * \frac{d\theta}{dt} =\frac{52}{3} * \frac{dx}{dt}$

In calculus:

$sec^2(\theta) =\frac{d\ tan(\theta)}{d\theta}$ -- Chain rule

So:

$sec^2(\theta) *\frac{d\theta}{dt} =\frac{52}{3} * \frac{dx}{dt}$

In trigonometry:

$sec^2(\theta) = tan^2(\theta) + 1$

So:

$(tan^2(\theta) + 1)\frac{d\theta}{dt} =\frac{52}{3} * \frac{dx}{dt}$

Recall that:

$tan(\theta) =\frac{x}{3}$

$((\frac{52}{3})^2 + 1)\frac{d\theta}{dt} =\frac{1}{3} * \frac{dx}{dt}$

$(\frac{52^2}{9} + 1)\frac{d\theta}{dt} =\frac{1}{3} * \frac{dx}{dt}$

$(\frac{2704}{9} + 1)\frac{d\theta}{dt} =\frac{1}{3} * \frac{dx}{dt}$

$(\frac{2704+9}{9})\frac{d\theta}{dt} =\frac{1}{3} * \frac{dx}{dt}$

$(\frac{2713}{9})\frac{d\theta}{dt} =\frac{1}{3} * \frac{dx}{dt}$

Recall that: $\frac{d\theta}{dt} = 18\pi \frac{rad}{min}$

$(\frac{2713}{9}) * 18\pi =\frac{1}{3} * \frac{dx}{dt}$

$2713 * 2\pi =\frac{1}{3} * \frac{dx}{dt}$

Multiply both sides by 3

$3 * 2713 * 2\pi =\frac{1}{3} * \frac{dx}{dt} * 3$

$3 * 2713 * 2\pi =\frac{dx}{dt}$

$16278\pi =\frac{dx}{dt}$

$\frac{dx}{dt} = 16278\pi$ miles/min

Hence:

The beam of light moves at $16278\pi$ miles/min

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Cross multiply:
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Final answer: x=10.

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