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Anarel [89]
3 years ago
13

The resultant of two forces acting on a body has a magnitude of 80 pounds. The angles between the resultant and the forces are 2

0Á and 52Á. Find the magnitude of the larger force.
Mathematics
1 answer:
kifflom [539]3 years ago
5 0
<span>The basic idea is that you form a parallelogram with those two vectors as the two different side lengths another way to see it: start at the tip of one vector and move in the same direction as the other vector (and the same length as the other vector)

</span><span>With any parallelogram, the adjacent angles are supplementary</span>

180-52-20= 108 degrees
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Elodia [21]

Answer:

hrs =4

Step-by-step explanation:

550-11

200-x

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550x=200×11

550÷550=2200÷560

x=4

3 0
3 years ago
Write the equation of the circle graphed below
Mnenie [13.5K]

Answer:

Step-by-step explanation:

Comment

The basic equation of a circle is

(x - a)^2 + (y - b)^2 = r^2

a and b are the coordinates of the center of the circle

r is the radius.

Development of the equation

r = 1 - 1/2 = 1/2. r is the distance between the red dot and the blue dot.

So far what you have is (x - a)^2 + (y - b)^2 = (1/2)^2

The center is the red dot.

x = 1

y = - 1

Answer: (x - 1)^2 + (y + 1/2)^2 = 1/4

7 0
2 years ago
_________________<br>√(-√2-(-√2))²+(√3-2√3)²
xz_007 [3.2K]
_________________
√(-√2-(-√2))²+(√3-2√3)²= 3
7 0
2 years ago
What is the domain of the function represented by the graph?
OLga [1]

Answer:

I believe its ALL REAL NUMBERS

Step-by-step explanation:

im not sure because they didnt put in a picture of the graph

7 0
3 years ago
If a factory continuously dumps pollutants into a river at the rate of the quotient of the square root of t and 45 tons per day,
julsineya [31]
<h2>Hello!</h2>

The answer is:

The first option, the amount dumped after 5 days is 0.166 tons.

<h2>Why?</h2>

To solve the problem, we need to integrate the given expression and evaluate using the given time.

So, integrating we have:

\int\limits^5_0 {\frac{\sqrt{t} }{45} } \, dt=\int\limits^5_0 {\frac{1}{45} (t)^{\frac{1}{2} } } \, dt\\\\\int\limits^5_0 {\frac{1}{45} (t)^{\frac{1}{2} } } \ dt=\frac{1}{45}\int\limits^5_0 {t^{\frac{1}{2} } } } \ dt\\\\\frac{1}{45}\int\limits^5_0 {t^{\frac{1}{2} } } } \ dt=(\frac{1}{45}*\frac{t^{\frac{1}{2}+1} }{\frac{1}{2} +1})/t(5)-t(0)\\\\(\frac{1}{45}*\frac{t^{\frac{1}{2}+1} }{\frac{1}{2} +1})/t(5)-t(0)=(\frac{1}{45}*\frac{t^{\frac{3}{2}} }{\frac{3}{2}})/t(5)-t(0)

(\frac{1}{45}*\frac{t^{\frac{3}{2}} }{\frac{3}{2}})/t(5)-t(0)=(\frac{1}{45}*\frac{2}{3}*t^{\frac{3}{2} })/t(5)-t(0)\\\\(\frac{1}{45}*\frac{2}{3}*t^{\frac{3}{2} })/t(5)-t(0)=(\frac{2}{135}*t^{\frac{3}{2}})/t(5)-t(0)\\\\(\frac{2}{135}*t^{\frac{3}{2}})/t(5)-t(0)=(\frac{2}{135}*5^{\frac{3}{2}})-(\frac{2}{135}*0^{\frac{3}{2}})\\\\(\frac{2}{135}*5^{\frac{3}{2}})-(\frac{2}{135}*0^{\frac{3}{2}})=\frac{2}{135}*11.18-0=0.1656=0.166

Hence, we have that the amount dumped after 5 days is 0.166 tons.

Have a nice day!

5 0
3 years ago
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