1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
maksim [4K]
3 years ago
11

a 10 foot ladder is leaning against a wall if the ladder reaches 8 feet up the wall how far is the base of the ladder from the w

all.

Mathematics
1 answer:
kolezko [41]3 years ago
4 0

Answer:

6ft

Step-by-step explanation:

This requires a little use of the Pythagorean Theorem.

The length of the leaning ladder is the hypothenuse of the right triangle formed by the wall and the ladder.

Hyp. = 10 ft Say this is C

The ladder height up the wall = 8 ft. Say this A.

We need to find the other leg, say B ,of the right triangle.

C^2 = A^2 + B^2

10^2 = 8^2 + B^2

100 = 64 + B^2 => B^2 = 100 - 64 = 36

B^2 = 36 so, B = 6 The positive square root since length is positive.

B = 6 ft This is how far the ladder bottom is from the wall.

Double Check: 10^2 = 8^2 + 6^2 or 100 = 64 + 36 Ok. As, it should. This check could be done mentally.

In addition, this is a Pythagorean Triple. (6, 8, 10).

Answer: 6 ft.

COPIED:)

You might be interested in
How do we convert kilogram to gram​
Sloan [31]

<u>1kg=1000g</u>

7kg*1000=7000g

10kg*1000=10000g

15g*1000=15000g

5 0
3 years ago
How many letters are in this sentence that you are reading right now?
Mila [183]

Answer:

56 letters

Step-by-step explanation:

6 0
2 years ago
Read 2 more answers
Which of the following could be the interpretation of the distance vs time graph shown?
Brums [2.3K]

The slope of the curve is positive between the times t=0\text{ s}t=0 st, equals, 0, start a text, space, s, end text, and t=2 \text{ s}t=2 st, equals, 2, start a text, space, s, end text since the slope is directed upward. This means the acceleration is positive.

The slope of the curve is negative between t=2 \text{ s}t=2 st, equals, 2, start a text, space, s, end text, and t=8 \text{ s}t=8 st, equals, 8, start a text, space, s, end text since the slope is directed downward. This means the acceleration is negative.

At t=2\text{ s}t=2 st, equals, 2, start a text, space, s, end text, the slope is zero since the tangent line is horizontal. This means the acceleration is zero at that moment.

Concept check: Is the object whose motion is described by the graph above speeding up or slowing down at time t=4\text{ s}t=4 st, equals, 4, start a text, space, s, end text?

[Show me the answer.]

What does the area under a velocity graph represent?

The area under a velocity graph represents the displacement of the object. To see why to consider the following graph of motion that shows an object maintaining a constant velocity of 6 meters per second for a time of 5 seconds.

To find the displacement during this time interval, we could use this formula

\Delta x=v\Delta t=(6\text{ m/s})(5\text{ s})=30\text{ m}Δx=vΔt=(6 m/s)(5 s)=30 mdelta, x, equals, v, delta, t, equals, left parenthesis, 6, start text, space, m, slash, s, end text, right parenthesis, left parenthesis, 5, start text, space, s, end text, right parenthesis, equals, 30, start text, space, m, end text

which gives a displacement of 30\text{ m}30 m30, start text, space, m, end text.

Now we're going to show that this was equivalent to finding the area under the curve. Consider the rectangle of the area made by the graph as seen below.

The area of this rectangle can be found by multiplying the height of the rectangle, 6 m/s, times its width, 5 s, which would give

\text{ area}=\text{height} \times \text{width} = 6\text{ m/s} \times 5\text{ s}=30\text{ m} area=height×width=6 m/s×5 s=30 mstart text, space, a, r, e, a, end text, equals, start text, h, e, i, g, h, t, end text, times, start text, w, i, d, t, h, end text, equals, 6, start text, space, m, slash, s, end text, times, 5, start text, space, s, end text, equals, 30, start text, space, m, end text

This is the same answer we got before for the displacement. The area under a velocity curve, regardless of the shape, will equal the displacement during that time interval. [Why is this still true when velocity isn't constant?]

\text{area under curve}=\text{displacement}area under curve=displacementstart text, a, r, e, a, space, u, n, d, e, r, space, c, u, r, v, e, end text, equals, start text, d, i, s, p, l, a, c, e, m, e, n, t, end text

[Wait, aren't areas always positive? What if the curve lies below the time axis?]

What do solved examples involving velocity vs. time graphs look like?

Example 1: Windsurfing speed change

A windsurfer is traveling along a straight line, and her motion is given by the velocity graph below.

Select all of the following statements that are true about the speed and acceleration of the windsurfer.

(A) Speed is increasing.

(B) Acceleration is increasing.

(C) Speed is decreasing.

(D) Acceleration is decreasing.

Options A, speed increasing, and D, acceleration decreasing, are both true.

The slope of a velocity graph is the acceleration. Since the slope of the curve is decreasing and becoming less steep this means that the acceleration is also decreasing.

It might seem counterintuitive, but the windsurfer is speeding up for this entire graph. The value of the graph, which represents the velocity, is increasing for the entire motion shown, but the amount of increase per second is getting smaller. For the first 4.5 seconds, the speed increased from 0 m/s to about 5 m/s, but for the second 4.5 seconds, the speed increased from 5 m/s to only about 7 m/s.

Example 2: Go-kart acceleration

The motion of a go-kart is shown by the velocity vs. time graph below.

A. What was the acceleration of the go-kart at time t=4\text{ s}t=4 st, equals, 4, start a text, space, s, end text?

B. What was the displacement of the go-kart between t=0\text{ s}t=0 st, equals, 0, start text, space, s, end text and t=7\text{ s}t=7 st, equals, 7, start text, space, s, end text?

A. Finding the acceleration of the go-kart at t=4\text{ s}t=4 st, equals, 4, start a text, space, s, end text

We can find the acceleration at t=4\text{ s}t=4 st, equals, 4, start a text, space, s, end text by finding the slope of the velocity graph at t=4\text{ s}t=4 st, equals, 4, start a text, space, s, end text.

\text{slope}=\dfrac{\text{rise}}{\text{run}}slope=  

run

7 0
3 years ago
Please help! Will give all my points! Due in 30 mins!
Blababa [14]

Answer:

WELP 30 mins passed

Step-by-step explanation:

7 0
2 years ago
Carl scored 35% of his basketball teams 40 points during a game. How many points did Carl score
ExtremeBDS [4]

Answer:

14

Step-by-step explanation:

If 5 percent = 2

35/7 (7 being number that goes into 40)

2x7 = 14

Hope this helps

4 0
3 years ago
Other questions:
  • Answers to 6,7,8 pleaseee?
    5·1 answer
  • Need help with questions................​
    10·1 answer
  • Write the following as a fraction in simple form <br><br>1.6​
    7·2 answers
  • It took malcom 6 minutes to walk 500m from the bus stop to the opera house. How fastwas he travelling
    6·2 answers
  • Evaluate the expression.<br> (6 − 3)!
    5·2 answers
  • What is the slope of the line through (1,0) and (3,8)​
    8·1 answer
  • Jordan deposited $500 in a savings account that earns 8% interest compounded annually. At the end of 2 years, she transferred th
    11·1 answer
  • Need help on this question loves
    6·2 answers
  • Jayden's family watched 8 hours of TV last week. This week, they watchec
    11·1 answer
  • 4) Cody invests $2,733 in a retirement
    12·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!