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Korolek [52]
3 years ago
13

Use SOHCAHTOA for this. Work out 'm' in 3sf, I need the working out.

Mathematics
1 answer:
Lady bird [3.3K]3 years ago
7 0

Anwer:3.537m

STEP BY STEP EXPLANATIOND:using SOH CAH TOA

First find the opposite

Represent the opposite with x

Tan 33° =x\10

x=10Tan 33°

x=6.494

To find m

Sin 33°=m\6.494 Sin 33°

m=3.5368

m=3.537meteres

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Can someone please help me??
Nesterboy [21]

Answer:

Answers are below

Step-by-step explanation:

x     y

0    $25,001.00

1     $25,000.95

2    $25,000.90

3    $25,000.86

I graphed the function on the graph below and found the values of y for each value of x.

*My graph showed that the initial value of the printer was $25,001.00, instead of $25,000.00*

7 0
3 years ago
PLS HELP ASAP FEJKRKJGR My bad dont mind the blue bubble
kondor19780726 [428]

Answer:

10

Step-by-step explanation:

yellow

6+6+3=15

so

for green

4+4+2=10

8 0
3 years ago
Read 2 more answers
What is the antiderivative of 3x/((x-1)^2)
Maslowich

Answer:

\int \:3\cdot \frac{x}{\left(x-1\right)^2}dx=3\left(\ln \left|x-1\right|-\frac{1}{x-1}\right)+C

Step-by-step explanation:

Given

\int \:\:3\cdot \frac{x}{\left(x-1\right)^2}dx

\mathrm{Take\:the\:constant\:out}:\quad \int a\cdot f\left(x\right)dx=a\cdot \int f\left(x\right)dx

=3\cdot \int \frac{x}{\left(x-1\right)^2}dx

\mathrm{Apply\:u-substitution:}\:u=x-1

=3\cdot \int \frac{u+1}{u^2}du

\mathrm{Expand}\:\frac{u+1}{u^2}:\quad \frac{1}{u}+\frac{1}{u^2}

=3\cdot \int \frac{1}{u}+\frac{1}{u^2}du

\mathrm{Apply\:the\:Sum\:Rule}:\quad \int f\left(x\right)\pm g\left(x\right)dx=\int f\left(x\right)dx\pm \int g\left(x\right)dx

=3\left(\int \frac{1}{u}du+\int \frac{1}{u^2}du\right)

as

\int \frac{1}{u}du=\ln \left|u\right|     ∵ \mathrm{Use\:the\:common\:integral}:\quad \int \frac{1}{u}du=\ln \left(\left|u\right|\right)

\int \frac{1}{u^2}du=-\frac{1}{u}        ∵     \mathrm{Apply\:the\:Power\:Rule}:\quad \int x^adx=\frac{x^{a+1}}{a+1},\:\quad \:a\ne -1

so

=3\left(\ln \left|u\right|-\frac{1}{u}\right)

\mathrm{Substitute\:back}\:u=x-1

=3\left(\ln \left|x-1\right|-\frac{1}{x-1}\right)

\mathrm{Add\:a\:constant\:to\:the\:solution}

=3\left(\ln \left|x-1\right|-\frac{1}{x-1}\right)+C

Therefore,

\int \:3\cdot \frac{x}{\left(x-1\right)^2}dx=3\left(\ln \left|x-1\right|-\frac{1}{x-1}\right)+C

4 0
3 years ago
Solve using substitution<br><br> y=1/2x<br> x+4y=18
Aleks [24]

Answer:

The solution set is (6,3)

Step-by-step explanation:

Begin by multiplying equation 1 by 2. You'll see why in a moment

2y = 2*(1/2) x

2y = x

Substitute this value into equation 2.

2y + 4y = 18                  Combine like terms on the left

6y = 18                          Divide both sides by 6

6y/6 = 18/6                    Do the division

y = 3

To solve for x just use the top equation

y = 1/2x

3 = 1/2x                          Multiply by 2

3*2 = x

x = 6


7 0
3 years ago
The graph below shows the average daily temperature over the period of a year. Explain how each labeled section of the graph rel
klemol [59]
Remark
I'm going to assume you are in the Northern Hemisphere. 

Answer
A looks like you are beginning in Winter. The temperature is down quite a bit. That would be about October where I live. B is spring because the temperature is rising but not at it's peak. That would be about April May here. 

C is summer. Glorious warm summer. June July and August.

D is autumn. The temperature is falling and it is just about at it's low peak again. 
7 0
3 years ago
Read 2 more answers
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