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3 years ago

7

Name the three similar triangles correspondingly. Start with the original right triangle LMK, then the remaining two right trian

gles created by the altitude MN.

2 answers:

nasty-shy [4]3 years ago

6 0

Answer:△LMK △MNK △LNM

im pretty sure thats it

aleksandr82 [10.1K]3 years ago

5 0

Answer:

Step-by-step explanation:

You might be interested in

A rocking horse has a weight limit of 60pounds. What percentage of the weight limit is 33 pounds? What percentage of the weight

bixtya [17]

Answer:

The percentage is <u>55%</u> when weight limit is 33 pounds.

The percentage is <u>190%</u> when weight limit is 114 pounds.

The 95% of the limit weighs <u>57</u> pounds.

Step-by-step explanation:

Given:

Weight limit of rocking horse = 60 pounds.

We need to find:

a) What percentage of the weight limit is 33 pounds.

b)What percentage of the weight limit is 114 pounds.

c) What weight is 95% of the limit.

Now Solving for a we get;

Weight limit = 33 pounds

Now percentage of the weight limit can be calculated by dividing given weight limit with total weight limit and then multiplying by 100 we get;

framing in equation form we get;

percentage of the weight limit = $\frac{33}{60}\times100 = 55\%$

Hence The percentage is <u>55%</u> when weight limit is 33 pounds.

Now Solving for b we get;

Weight limit = 114 pounds

Now percentage of the weight limit can be calculated by dividing given weight limit with total weight limit and then multiplying by 100 we get;

framing in equation form we get;

percentage of the weight limit = $\frac{114}{60}\times100 = 190\%$

Hence The percentage is <u>190%</u> when weight limit is 114 pounds.

Now Solving for c we get;

Percentage Weight = 95%

Now Amount of weight limit can be calculated by multiplying Percentage weight limit with total weight limit and then Dividing by 100 we get;

framing in equation form we get;

percentage of the weight limit = $\frac{95}{100}\times60 = 57 \pounds$

Hence The 95% of the limit weighs <u>57</u> pounds.

4 0

4 years ago

Felix has completed 6 laps of the race in 6 minutes. If he continues to drive at the same speed, how many more minutes will it t

Blababa [14]

Answer:

36 minutes

Step-by-step explanation:

6 0

3 years ago

Suppose an unknown radioactive substance produces 8000 counts per minute on a Geiger counter at a certain time, and only 500 cou

mariarad [96]

Answer:

The half-life of the radioactive substance is of 3.25 days.

Step-by-step explanation:

The amount of radioactive substance is proportional to the number of counts per minute:

This means that the amount is given by the following differential equation:

$\frac{dQ}{dt} = -kQ$

In which k is the decay rate.

The solution is:

$Q(t) = Q(0)e^{-kt}$

In which Q(0) is the initial amount:

8000 counts per minute on a Geiger counter at a certain time

This means that $Q(0) = 8000$

500 counts per minute 13 days later.

This means that $Q(13) = 500$ . We use this to find k.

$Q(t) = Q(0)e^{-kt}$

$500 = 8000e^{-13k}$

$e^{-13k} = \frac{500}{8000}$

$\ln{e^{-13k}} = \ln{\frac{500}{8000}}$

$-13k = \ln{\frac{500}{8000}}$

$k = -\frac{\ln{\frac{500}{8000}}}{13}$

$k = 0.2133$

So

$Q(t) = Q(0)e^{-0.2133t}$

Determine the half-life of the radioactive substance.

This is t for which Q(t) = 0.5Q(0). So

$Q(t) = Q(0)e^{-0.2133t}$

$0.5Q(0) = Q(0)e^{-0.2133t}$

$e^{-0.2133t} = 0.5$

$\ln{e^{-0.2133t}} = \ln{0.5}$

$-0.2133t = \ln{0.5}$

$t = -\frac{\ln{0.5}}{0.2133}$

$t = 3.25$

The half-life of the radioactive substance is of 3.25 days.

7 0

3 years ago

I don’t understand this problem

Galina-37 [17]

Answer: 20

Step-by-step explanation:

a^2=b^2 - c^2. i think i could be wrong

6 0

3 years ago

2/3p-14=p+13<br> How do I solve this

Aleks04 [339]

Add 14 to both sides
2/3p=p+27

Subtract p from both sides
-1/3p= 27

Multiply both sides by -3
p= -81

Final answer: p=-81

7 0

4 years ago

Read 2 more answers