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3 years ago

What is the length of the hypotenuse?

Mathematics

2 answers:

Akimi4 [234]3 years ago

8 0

=>(hypotenuse)^2=(base)^2+(perpendicular)^2

=>(h)^2=(15)^2+(8)^2

=>(h)^2=289

=>h=√289

=>h=17 yd

Hope it helps you

deff fn [24]3 years ago

4 0

225+64=289 is your answer

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MissTica

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3 years ago

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Can I get some help, it is due in 15 minutes and I need help. Thanks, any help appreciated

koban [17]

Well, I'm way past the 15 min mark, but here's how to do the question.

With this, you will need to use the distance formula, $\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$ , on XY, YZ, and ZX.

XY: $\sqrt{(3-1)^2+(1-6)^2}$

Firstly, solve inside the parentheses: $\sqrt{(2)^2+(-5)^2}$

Next, solve the exponents: $\sqrt{4+25}$

Next, solve the addition, and XY's distance will be √29

(The process is the same with the other 2 sides, so I'll go through them real quickly)

YZ:

$\sqrt{(6-3)^2+(3-1)^2}\\ \sqrt{(3)^2+(2)^2}\\ \sqrt{9+4}\\ \sqrt{13}$

ZX:

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Now that we got the 3 sides, we can add them up: $\sqrt{29}+\sqrt{13} +\sqrt{34} =14.8$

In short, your answer is 14.8, or the second option.

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3 years ago

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Vilka [71]

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3 years ago

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2 years ago

Marcos had 15 coins in nickels and quarters. He had 3 more quarters than nickels. He wrote a system of equations to represent th

hodyreva [135]

Answer:

There are 6 coins in nickels and 9 coins in quarters.

Step-by-step explanation:

Now, to find the solution.

According to question:

The number of nickels = $x$ .

The number of quarters = $y.$

So, the total number of coins:

$x+y=15.$

He had 3 more quarters than nickels.

Thus,

$y=x+3$ .....(1)

Now, to get the solution:

$x+y=15$

Substituting the equation (1):

$x+(x+3)=15$

$x+x+3=15$

$2x+3=15$

Subtracting both sides by 3 we get:

$2x=12$

Dividing both sides by 2 we get:

$x=6.$

The number of nickels = 6.

Now, substituting the value of $x$ in equation (1):

$y=x+3\\\\y=6+3\\\\y=9.$

The number of quarters = 9.

Thus, there are 6 coins in nickels and 9 coins in quarters.

6 0

3 years ago