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sergey [27]
3 years ago
8

Write and solve an equation to find x. Then find the measure of angle YXZ.

Mathematics
1 answer:
Verdich [7]3 years ago
5 0

This is so simple so here...

A right angle is 90 degrees.

We learn that WXZ is 35 degrees.

90-35=55

YXZ is 55 degrees. I hope I could help!

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Find the 14th term of the geometric sequence 5 -10 20
Anna [14]

Answer:

  a14 = -40,960

Step-by-step explanation:

The sequence has a first term of 5 and a common ratio of -10/5 = -2, so the n-th term is given by ...

  an = a1·r^(n-1)

  an = 5·(-2)^(n-1)

The 14th term is then ...

  a14 = 5·(-2)^(14-1) = -5·2^13 = -5·8192

  a14 = -40,960

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What is the least common multiple of 2, 6, 8, and 12?
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24 because it’s the lowest multiple of 12 that’s also in 2,6 and 8
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Two different cars each depreciate to 60% of their respective original values. The first car depreciates at an annual rate of 10
zvonat [6]

The approximate difference in the ages of the two cars, which  depreciate to 60% of their respective original values, is 1.7 years.

<h3>What is depreciation?</h3>

Depreciation is to decrease in the value of a product in a period of time. This can be given as,

FV=P\left(1-\dfrac{r}{100}\right)^n

Here, (<em>P</em>) is the price of the product, (<em>r</em>) is the rate of annual depreciation and (<em>n</em>) is the number of years.

Two different cars each depreciate to 60% of their respective original values. The first car depreciates at an annual rate of 10%.

Suppose the original price of the first car is x dollars. Thus, the depreciation price of the car is 0.6x. Let the number of year is n_1. Thus, by the above formula for the first car,

0.6x=x\left(1-\dfrac{10}{100}\right)^{n_1}\\0.6=(1-0.1)^{n_1}\\0.6=(0.9)^{n_1}

Take log both the sides as,

\log 0.6=\log (0.9)^{n_1}\\\log 0.6={n_1}\log (0.9)\\n_1=\dfrac{\log 0.6}{\log 0.9}\\n_1\approx4.85

Now, the second car depreciates at an annual rate of 15%. Suppose the original price of the second car is y dollars.

Thus, the depreciation price of the car is 0.6y. Let the number of year is n_2. Thus, by the above formula for the second car,

0.6y=y\left(1-\dfrac{15}{100}\right)^{n_2}\\0.6=(1-0.15)^{n_2}\\0.6=(0.85)^{n_2}

Take log both the sides as,

\log 0.6=\log (0.85)^{n_2}\\\log 0.6={n_2}\log (0.85)\\n_2=\dfrac{\log 0.6}{\log 0.85}\\n_2\approx3.14

The difference in the ages of the two cars is,

d=4.85-3.14\\d=1.71\rm years

Thus, the approximate difference in the ages of the two cars, which  depreciate to 60% of their respective original values, is 1.7 years.

Learn more about the depreciation here;

brainly.com/question/25297296

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2 years ago
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Step-by-step explanation:

Hope this helps!

8 0
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tino4ka555 [31]

If we let p be the number of pages read in 1 hour, then in 7 hours Doug reads 7p pages. We know that this number of pages is 231, so we have

7p=231 \iff p = \dfrac{231}{7} = 33

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