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3 years ago

Find the missing angle in the triangle (type in numbers only)

Mathematics

1 answer:

Brums [2.3K]3 years ago

4 0

Answer:42 degrees

Step-by-step explanation:

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4. If TV = 14x - 8, find TU. 9x+2 5

Musya8 [376]

Answer:

x = 3

Explanation:

The complete question is shown in the attached diagram

Collinear points are defined as points that lie on the same straight line

Since points U, T and V are given to be collinear, this means that these three points lie on the same straight line

Now, we are given that point U is between points T and V

This would mean that the length of segment TV can be written as the sum of the segments TU and UV

Therefore:

TV = TU + UV

We are given that:

TV = 14x - 8

TU = 9x + 2

UV = 5

Substitute with the givens in the above mentioned equation and solve for x as follows:

TV = TU + UV

14x - 8 = 9x + 2 + 5

14x - 8 = 9x + 7

14x - 9x = 7 + 8

5x = 15

x = 3

Hope this helps :)

5 0

4 years ago

Plisss help on my assesment

Anon25 [30]

Answer:

The first one is equivalent [the 6x+48 = 2(3x+24)] and the second one is NOT equivalent [the 7x+21 ≠ 2(5x+3)]

Step-by-step explanation:

Just follow distributive property to solve these. You can ignore the first expression in both until you have to compare the answers.

1. 6x+48 and 2(3x+24)

2(3x+24) ---> 2(3x) + 2(24) ---> 6x + 48

Bring in the first expression ~ 6x+48 and 6x+48

They are the same, so they are equivalent

2. 7x+21 and 2(5x+3)

2(5x+3) ---> 2(5x) + 2(3) ----> 10x + 6

Bring in the first expression ~ 7x+21 and 10x + 6

They are NOT the same, so they are NOT equivalent

6 0

3 years ago

the histogram shows the result of survey asking students how many windows are in their homes how mnay studengs have less than 10

serious [3.7K]

Answer:

4 students

Step-by-step explanation:

Although I am not sure if I am completely correct. I’ll try my best.

We know:

The numbers 0-12 on the frequency table is the number of windows.

The numbers on the bottom is the amount of students.

What we can see:

As we can see..

the 0-4 section in the table shows that only 0-4 students have 9 windows.
The 5-9 sections shows that only 5-9 students have 10 windows.
The 10-14 sections shows that 10-14 students have 0 windows.
The 15-19 sections shows that 15-19 students have 4 windows.
The 20-24 sections shows that 20-24 students have 2 windows.

Calculations:

Since the number of students that have the amount of windows less than 10 is 4 (0-4 section) and 0 (10-14 section)

Add.

4+0

= 4

Therefore, 4 students have less than 10 windows.

4 0

3 years ago

I need help figuring out someone's annual pay.

insens350 [35]

Answer:

He makes 24,000 a year

Step-by-step explanation:

First figure out how much he makes a month without the extra $50 every day he works. which is $600, then add the 4 weeks (28 days) worth of the $50 every day, which is $1400, then add the $600 to get $2000 a month, then times that by twelve, which is $24,000.

4 0

4 years ago

Combine into a single logarithm. 3log(x+y)+2log(x-y)-log(x^2 +y^2)

seropon [69]

Answer:

$3\log _{10}\left(x+y\right)+2\log _{10}\left(x-y\right)-\log _{10}\left(x^2+y^2\right)=\log _{10}\left(\frac{\left(x+y\right)^3\left(x-y\right)^2}{x^2+y^2}\right)$

Step-by-step explanation:

Given the expression

$3log\left(x+y\right)+2log\left(x-y\right)-log\left(x^2\:+y^2\right)$

solving to write into a single logarithm

$3log\left(x+y\right)+2log\left(x-y\right)-log\left(x^2\:+y^2\right)$

$\mathrm{Apply\:log\:rule}:\quad \:a\log _c\left(b\right)=\log _c\left(b^a\right)$

$3\log _{10}\left(x+y\right)=\log _{10}\left(\left(x+y\right)^3\right)$

$=\log _{10}\left(\left(x+y\right)^3\right)+2\log _{10}\left(x-y\right)-\log _{10}\left(x^2+y^2\right)$

$\mathrm{Apply\:log\:rule}:\quad \:a\log _c\left(b\right)=\log _c\left(b^a\right)$

$2\log _{10}\left(x-y\right)=\log _{10}\left(\left(x-y\right)^2\right)$

$=\log _{10}\left(\left(x+y\right)^3\right)+\log _{10}\left(\left(x-y\right)^2\right)-\log _{10}\left(x^2+y^2\right)$

$\mathrm{Apply\:log\:rule}:\quad \log _c\left(a\right)+\log _c\left(b\right)=\log _c\left(ab\right)$

$\log _{10}\left(\left(x+y\right)^3\right)+\log _{10}\left(\left(x-y\right)^2\right)=\log _{10}\left(\left(x+y\right)^3\left(x-y\right)^2\right)$

$=\log _{10}\left(\left(x+y\right)^3\left(x-y\right)^2\right)-\log _{10}\left(x^2+y^2\right)$

$\mathrm{Apply\:log\:rule}:\quad \log _c\left(a\right)-\log _c\left(b\right)=\log _c\left(\frac{a}{b}\right)$

$\log _{10}\left(\left(x+y\right)^3\left(x-y\right)^2\right)-\log _{10}\left(x^2+y^2\right)=\log _{10}\left(\frac{\left(x+y\right)^3\left(x-y\right)^2}{x^2+y^2}\right)$

$=\log _{10}\left(\frac{\left(x+y\right)^3\left(x-y\right)^2}{x^2+y^2}\right)$

Thus,

$3\log _{10}\left(x+y\right)+2\log _{10}\left(x-y\right)-\log _{10}\left(x^2+y^2\right)=\log _{10}\left(\frac{\left(x+y\right)^3\left(x-y\right)^2}{x^2+y^2}\right)$

6 0

3 years ago