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3 years ago

7

I have an old alarm clock, on which the hands move at only 95% of the speed that they should do.

1 answer:

erica [24]3 years ago

3 0

9514 1404 393

Answer:

a) 9:45 pm

b) 10:57 pm

Step-by-step explanation:

In the 15 hours between 7:30 am and 10:30 pm, the clock hands will have moved 15×0.95 = 14.25 hours = 14 hours and 15 minutes.

At 10:30 pm, the clock indicates 9:45 pm.

__

b) In the 9 hours between 10:30 pm and 7:30 am, the clock will have moved 9×0.95 = 8.55 hours, or 8 hours 33 minutes. That is, the clock needs to be set 27 minutes ahead of the actual time.

The clock need to be set to 10:57 pm.

_____

<em>Alternate solution</em>

At 95% speed, the clock loses 5%×60 minutes = 3 minutes each hour. In the 15 hours from morning to night, the clock will be 45 minutes slow. It will indicate 9:45 at 10:30 pm.

In the 9 hours from night to morning, the clock will lose 27 minutes so needs to be set 27 minutes fast. It should be set to 10:57 pm at 10:30 pm.

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Triangle JKL has vertices J(2,5), K(1,1), and L(5,2). Triangle QNP has vertices Q(-4,4), N(-3,0), and P(-7,1). Is (triangle)JKL

Tems11 [23]

Answer:

Yes they are

Step-by-step explanation:

In the triangle JKL, the sides can be calculated as following:

J(2;5); K(1;1)

=> JK = $\sqrt{(1-2)^{2} + (1-5)^{2} } = \sqrt{(-1)^{2}+(-4)^{2} } = \sqrt{1+16}=\sqrt{17}$

J(2;5); L(5;2)

=> JL = $\sqrt{(5-2)^{2} + (2-5)^{2} } = \sqrt{3^{2}+(-3)^{2} } = \sqrt{9+9}=\sqrt{18} = 3\sqrt{2}$

K(1;1); L(5;2)

=> KL = $\sqrt{(5-1)^{2} + (2-1)^{2} } = \sqrt{4^{2}+1^{2} } = \sqrt{1+16}=\sqrt{17}$

In the triangle QNP, the sides can be calculate as following:

Q(-4;4); N(-3;0)

=> QN = $\sqrt{[-3-(-4)]^{2} + (0-4)^{2} } = \sqrt{1^{2}+(-4)^{2} } = \sqrt{1+16}=\sqrt{17}$

Q (-4;4); P(-7;1)

=> QP = $\sqrt{[-7-(-4)]^{2} + (1-4)^{2} } = \sqrt{(-3)^{2}+(-3)^{2} } = \sqrt{9+9}=\sqrt{18} = 3\sqrt{2}$

N(-3;0); P(-7;1)

=> NP = $\sqrt{[-7-(-3)]^{2} + (1-0)^{2} } = \sqrt{(-4)^{2}+1^{2} } = \sqrt{16+1}=\sqrt{17}$

It can be seen that QPN and JKL have: JK = QN; JL = QP; KL = NP

=> They are congruent triangles

7 0

3 years ago

Read 2 more answers

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:

$\sqrt{(1-6)^2+(6-3)^2}\\ \sqrt{(-5)^2+(3)^2}\\ \sqrt{25+9}\\ \sqrt{34}$

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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