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3 years ago

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I NEED THIS ANSWER ASAP: A 10 foot ladder is leaning against a house. The base of the ladder is 4 feet from the house. How far d

oes the ladder reach up the house?

1 answer:

ad-work [718]3 years ago

4 0

Answer:The ladder and the ground form a right triangle. Use the Pythagorean Theorem to find the other leg of the triangle.

Explanation Steps: a^2 + b^2 = c^2

a^2 + 4^2 = 15^2

a^2 + 16 = 225

a^2 = 209

a = 14.5 feet

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I don't understand this question

kotegsom [21]

Answer: The correct answer is the third choice.

Step-by-step explanation: The question is asking you to look at pieces of data from the graph. The best way to do this question is to look at each question to determine whether the statement is true or false.

In looking at the each choice:

First one - The statement is true. Looking at the graph, a 40 and 50 year old person has the same average rate.

Second - The statement is true. Looking at the graph, you can see a steady decline from a 10 year old to a 20 year old person.

Third - The statement is false. A 10 year old and 20 year old do not have the same rate.

Fourth - The statement is true. A 20 and 30 year old have the same heart rate.

6 0

3 years ago

What is the answer to<br> 56 is 28% of

klio [65]

In mathematics, 'is' would be the equal sign. And 'of' would be the multiplcation sign. The missing number would be any variable.

56 is 28% of a number would be translated algebraically to:

$56 = \frac{28}{100} \times x$

Solving for x, we get:

$56 = \frac{28}{100} \times x$

Divide 28 on both sides

$2 = \frac{x}{100}$

Multiply 100 on both sides

$x = 200$

56 is 28% of 200

5 0

3 years ago

Read 2 more answers

A batting machine uses an automatic baseball feeder. During baseball

Inessa05 [86]

The feeders in battling machine are represented in proportions and fractions.

The equation that represents the problem is: $\mathbf{\frac 16x + 15 = \frac 23x}$
The feeder can hold <em>30 baseballs</em>, when full

The given parameters are:

<em /> $\mathbf{Initial = \frac 16x}$ <em> ------ 1/6 full</em>

<em /> $\mathbf{Additional = 15}$ <em> --- baseballs added</em>

<em /> $\mathbf{Final = \frac 23x}$ <em> ---- 2/3 full</em>

<em />

So, the equation that represents the problem is:

$\mathbf{Initial + Additional = Final}$

So, we have:

$\mathbf{\frac 16x + 15 = \frac 23x}$

The number of baseballs it can hold is calculated as follows:

$\mathbf{\frac 16x + 15 = \frac 23x}$

Multiply through by 6

$\mathbf{x + 90 = 4x}$

Collect like terms

$\mathbf{4x - x = 90 }$

$\mathbf{3x = 90 }$

Divide through by 3

$\mathbf{x = 30 }$

Hence, the feeder can hold 30 baseballs, when full

Read more about proportions and fractions at:

brainly.com/question/20337104

5 0

2 years ago

How do I evaluate the expression -(p+q)2 /(-6) for p=2 andq=4

Dimas [21]

Plug in the values of p and q where you see them in the equation:

-(2+4)2 / (-6) - Distribute the -1
(-2-4)2 / (-6) - Distribute the 2
(-4-8) / (-6) - Subtract what's inside the parenthesis
(-12) / (-6) - Divide
The answer is 2

3 0

4 years ago

Learning Thoery In a learning theory project, the proportion P of correct responses after n trials can be modeled by p = 0.83/(1

elena-s [515]

Answer:

a) $P(n=3) = \frac{0.83}{1+e^{-0.2(3)}}= \frac{0.83}{1+ e^{-0.6}} = 0.536$

b) $P(n=7) = \frac{0.83}{1+e^{-0.2(7)}}= \frac{0.83}{1+ e^{-1.4}} = 0.666$

c) $0.75 =\frac{0.83}{1+e^{-0.2n}}$

$1+ e^{-0.2n} = \frac{0.83}{0.75}= \frac{83}{75}$

$e^{-0.2n} = \frac{83}{75}-1= \frac{8}{75}$

$ln e^{-0.2n} = ln (\frac{8}{75})$

$-0.2 n = ln(\frac{8}{75})$

And then if we solve for t we got:

$n = \frac{ln(\frac{8}{75})}{-0.2} = 11.19 trials$

d) If we find the limit when n tend to infinity for the function we have this:

$lim_{n \to \infty} \frac{0.83}{1+e^{-0.2t}} = 0.83$

So then the number of correct responses have a limit and is 0.83 as n increases without bound.

Step-by-step explanation:

For this case we have the following expression for the proportion of correct responses after n trials:

$P(n) = \frac{0.83}{1+e^{-0.2t}}$

Part a

For this case we just need to replace the value of n=3 in order to see what we got:

$P(n=3) = \frac{0.83}{1+e^{-0.2(3)}}= \frac{0.83}{1+ e^{-0.6}} = 0.536$

So the number of correct reponses after 3 trials is approximately 0.536.

Part b

For this case we just need to replace the value of n=7 in order to see what we got:

$P(n=7) = \frac{0.83}{1+e^{-0.2(7)}}= \frac{0.83}{1+ e^{-1.4}} = 0.666$

So the number of correct responses after 7 weeks is approximately 0.666.

Part c

For this case we want to solve the following equation:

$0.75 =\frac{0.83}{1+e^{-0.2n}}$

And we can rewrite this expression like this:

$1+ e^{-0.2n} = \frac{0.83}{0.75}= \frac{83}{75}$

$e^{-0.2n} = \frac{83}{75}-1= \frac{8}{75}$

Now we can apply natural log on both sides and we got:

$ln e^{-0.2n} = ln (\frac{8}{75})$

$-0.2 n = ln(\frac{8}{75})$

And then if we solve for t we got:

$n = \frac{ln(\frac{8}{75})}{-0.2} = 11.19 trials$

And we can see this on the plot attached.

Part d

If we find the limit when n tend to infinity for the function we have this:

$lim_{n \to \infty} \frac{0.83}{1+e^{-0.2t}} = 0.83$

So then the number of correct responses have a limit and is 0.83 as n increases without bound.

5 0

3 years ago