What is the length of the dotted line in the diagram below? Leave your answer in simplest radical form.

Mathematics

2 answers:

Alex_Xolod [135]3 years ago

8 0

Answer:

The length of the dotted line is √93.

Step-by-step explanation:

<em>Use Pythagorean Theorem to solve the hypotenuse of the top triangle.</em>

8² + 5² = √89

<em>√89 is now known as the measurement of the bottom line as well. We can visualize the dotted line as the hypotenuse of a triangle and solve for it using Pythagorean Theorem again.</em>

√89² + 2² = √93

Pachacha [2.7K]3 years ago

6 0

Answer:

$\sqrt{93}$

Step-by-step explanation:

8^2+5^2=c^2

64+25=c^2

89=c^2

c= $\sqrt{89}$ <-- this is the length of the rectangle at the bottom

2^2+( $\sqrt{89}$ )^2=c^2

4+89=c^2

c= $\sqrt{93}$ <-- length of dotted line (diagonal)

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Guys, please help me with this problem.

SashulF [63]

<h2>Answer:</h2>

The number is -2.

<h2>Step-by-step explanation:</h2><h3>Written out, our equation looks like this:</h3><h3>(-5·x)+2=12</h3>

We'll start of by taking our final result of 12 and subtracting 2, as the beginning of the statement says "two more."

<h3>12 minus 2 gives us 10.</h3>

Now, since the statement says "the product of -5," we will take this 10 and divide by -5.

<h3>10 divided by -5 gives us -2.</h3>

This means that our number, x equals -2. Of course, we must input this solution to double check our work.

<h2>Our final answer is x=-2!</h2>

3 0

3 years ago

Read 2 more answers

A scientist inoculates mice, one at a time, with a disease germ until he finds 3 that have contracted the disease. If the prob

UkoKoshka [18]

Answer:

The probability that 8 mice are required is 0.2428.

Step-by-step explanation:

Given : A scientist inoculates mice, one at a time, with a disease germ until he finds 3 that have contracted the disease. If the probability of contracting the disease is two sevenths.

To find : What is the probability that 8 mice are required? The probability that that 8 mice are required is nothing ?

Solution :

Applying binomial distribution,

$P(X=r)=^nC_r p^rq^{n-r}$