How can you tell when a pattern shows counting on by tens?

A ladder 10 m long,leans against a vertical wall at an angle of 70° to the ground.if the ladder slips down the wall 4m,find,corr

lina2011 [118]

Answer:

(a) the new angle the ladder makes with the ground is $32.7^o$

(b) the ladder slipped back about 5 meters

Step-by-step explanation:

Notice that the ladder doesn't change its length in the process.

So let's start from the initial situation , finding the distance from the ground at which the ladder touches the wall when the angle with the ground is 70^o. Notice that this situation is represented by a right angle triangle with the right angle between the wall and the ground (see attached image), and that we can use the sine function to find the side opposite to the 70 degree angle:

$sin(70^o)=\frac{opposite}{hypotenuse} \\sin(70^o)=\frac{h}{10}\\h=10\, sin(70^o) \approx 9.4 \,\,m$

therefore 9.4 meters is approximately the height at which the ladder touches the wall initially.

Now, if the tip of the ladder goes down the wall 4 meters, it is now at 9.4 m - 4 m = 5.4 m from the ground. We can therefore use again the sine function to solve for the new angle:

$sin(x)=\frac{opposite}{hypotenuse} \\sin(x)=\frac{5.4}{10} \\sin(x)=0.54\\x=arcsin(0.54)\\x= 32.7^o$

To answer the second question we need to find the original distance from the wall that the bottom of the ladder was originally, and for that we can use the cosine function:

$cos(70^o)=\frac{adjacent}{hypotenuse} \\cos(70^o)=\frac{x}{10}\\x=10\,cos(70^o)\\x\approx 3.4 \,\,m$

Now fro the new position of the bottom of the ladder relative to the wall:

$cos(32.7^o)=\frac{adjacent}{10} \\adjacent=10\,cos(32.7^o)\\adjacent\approx 8.4\,\,m$

then the difference in between those two distances is what we need:

8.4 m - 3.4 m = 5 m

4 0

3 years ago

Beaker A contains 1 liter which is 15 percent oil and the rest is vinegar, thoroughly mixed up. Beaker B contains 2 liters which

Tanzania [10]

Answer:

Add 1.25 litres of oil.

Step-by-step explanation:

Contents in beaker A implies;

15% of oil in 1 litre = 0.15 litres of oil

So that, there are 0.15 litres of oil and 0.85 litres of vinegar in beaker A.

Contents in beaker B implies:

55% of oil in 2 litres = 1.1 litres of oil

So that, thee are 1.1 litres of oil and 0.9 litres of vinegar in beaker B.

Half of contents of B poured into A implies that beaker A now contains:

0.15 litres + 0.55 litres = 0.7 litres of oil

0.85 litres + 0.45 litres = 1.3 litres of vinegar

The percentage of oil in A now = $\frac{0.7}{2}$ x 100%

= 35%

To increase the percentage of oil to 60%, then:

0.7 litres + 1.25 litres = 1.95 litres of oil

But the new total litres of the content in beaker A = 3.25 litres

So that;

$\frac{1.95}{3.25}$ x 100% = 60%

Therefore, to produce a mixture which is 60% oil, add 1.25 litres of oil.

7 0

3 years ago

What is 17/7 as a whole number?

deff fn [24]

17/7 cannot be put into a whole number, but it can be put into decimal form. The estimated decimal number is 2.43. You just divide 17 by 7 and then round.

3 0

3 years ago

Read 2 more answers

Three ratios that are Equivalent to 11:1

marin [14]

Answer:

1-22:2

2-33:3

3-44:4

Step-by-step explanation:

11+11=22

22+11=33

33+11=44

1+1=2

2+1=3

3+1=4

so, 22:2

33:3

44:4

6 0

3 years ago

Help help this is 10 points

Delicious77 [7]

Answer:

The anwser you are looking for is number 2

Step-by-step explanation:

7 0

3 years ago