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18 days ago

A 12-centimeter rod is held between a flashlight and a wall as shown. Find the length of the shadow on

Mathematics

1 answer:

lawyer [9.2K]18 days ago

7 0

Answer:

48 cm

Step-by-step explanation:

Parameters given:

Distance of the rod from the wall = 45 cm

Distance of the rod from the light source = 15 cm

Rod length = 12 cm

It can be observed that <DAM and <BAF are congruent.

Additionally, <DMA and <BFM are congruent as they represent corresponding angles.

To determine the shadow length, we will consider the equation

$\frac{DM}{BF} = \frac{AM}{A.F}$

Where:

DM = half the length of the rod = ½*12 = 6

A.F = 15 + 45 = 60 cm

AM = 15 cm

Thus,

$\frac{DM}{BF} = \frac{AM}{A.F}$

$= \frac{6}{BF} = \frac{15}{60}$

After cross-multiplying, we get:

15 * B.F = 60 * 6

15 * B.F = 360

$BF = \frac{360}{15}$

BF = 24 cm

The shadow on the wall is thus:

2 * BF

= 2 * 24

= 48 cm

The shadow's length on the wall is 48 cm.

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Wyatt is going to a carnival that has games and rides. Each game costs $1.25 and each ride costs $2.75. Wyatt spent $20.25 altog

Inessa [9000]

Answer:

Wyatt participated in 3 games and enjoyed 6 rides

Step-by-step explanation:

Let x represent the number of games

And let y represent the number of rides

Each game costs $1.25

Each ride costs $2.75

The total expenses incurred by Wyatt amounted to $20.25

Thus, the equation becomes: x1.25 + y2.75 = 20.25..... Equation 1

The number of rides he enjoyed is double the games he competed in

Y = 2x... Equation 2

By substituting the value of y into equation 1

x1.25 + y2.75 = 20.25

x1.25 + 2(x)2.75 = 20.25

x1.25 + x5.5 = 20.25

x6.75= 20.25

x= 20.25/6.75

X= 3

Y= 2x

Y= 2(3)

Y= 6

Thus, Wyatt participated in 3 games and enjoyed 6 rides

5 0

11 days ago

Harper knows he is 50 yards from school. The map on his phone shows that the school is 34 inch from his current location. How fa

PIT_PIT [9117]

To solve this question, follow the steps outlined below:

1. Since 1 yard equals 36 inches, to convert 50 yards to inches involves:

$(50)(36)=1800in$

2. If he is standing 50 yards away from the school, and the map indicates that the school is 34 inches away from him, when it shows 3 inches the actual distance is:

$\frac{(1800in)(3in)}{34in}=158.82in$

3. If you express this back in yards:

$\frac{158.82}{36} =4.41 yd$

Consequently, the answer is: 158.82 inches or 4.41 yards.

8 0

1 month ago

Read 2 more answers

893 hundreds x 85 tens =

tester [8842]

The result is 75,905.

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4 days ago

Read 2 more answers

Point T is the midpoint of RS, W is the midpoint of RT, and Z is the midpoint of WS. If the length of TZ is x, find the lengths

PIT_PIT [9117]

Let RS be denoted as y.

Given that T is the midpoint of RS, this results in RT = TS = $\frac{y}{2}$

And W acting as the midpoint of RT, leads to

thus RW = WT = $\frac{\frac{y}{2} }{2} = \frac{y}{4}$

Given that Z is the midpoint of WS.

Meaning WZ = ZS = $\frac{WS}{2} = \frac{WT+TS}{2} = \frac{(\frac{y}{4} +\frac{y}{2} )}{2} =\frac{3y}{8}$

Consequently, TZ = TS - ZS = $\frac{y}{2} -\frac{3y}{8} = \frac{4y-3y}{8} = \frac{y}{8}$

However, TZ is defined as x.

Thus, $\frac{y}{8} = x$

y = 8x = RS.

A) Length of RW = $\frac{y}{4} = \frac{8x}{4} = 2x$

B) Length of WZ = $\frac{3y}{8} =\frac{3*8x}{8} = 3x$

C) Length of RS = y = 8x

D) Length of ZS = WZ = 3x.

An image is provided for clarification.

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4 days ago

Two planes A and B can each fly through still air at 550 mph. The distance from Paris to Los Angeles is 5000 miles. Wind blows f

PIT_PIT [9117]

Answer:2688.63 miles

Step-by-step explanation:

Given

The distance from Paris to Los Angeles $d=5000 miles$

Wind traveling at 100 mph from Los Angeles to Paris

The velocity of the plane through still air is $v=550 mph$

Relative velocity of Plane A is $v_a=550-100 =450 mph$ since Plane A is heading against the wind

Relative velocity of Plane B is $v_b=550+100=650 mph$

Plane A departs at noon

Distance covered by Plane A in 1 hour $=450\times 1=450 miles$

The total separation distance between the planes is $5000-450=4550 miles$

Assuming they meet after t hours

Distance covered by Plane B in that time frame $=v_b\times t=650\times t$

Distance traveled by Plane A in t hours is $=v_a\times t=450 t$

The total distance equals 4550 miles

$4550=650t+450t$

$t=\frac{4550}{1100}$

$t=4.13 hr$

Distance traveled by B is $=650\times 4.136=2688.63 miles$

thus, they meet at a point 2688.63 miles away from Los Angeles.

5 0

1 month ago