I need help, can anyone please help me?? I'll mark you BRAINLIEST if you answer correctly. :). Fill in the blanks.

Mathematics

1 answer:

pav-90 [236]3 years ago

5 0

(x+4)²=-63

....brainliest would be appreciated

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A circle has a diameter with endpoints at 3 – 5i and –8 + 2i. What is the center of the circle?

bagirrra123 [75]

So we got the real axis and the imaginary axis
we just need to find the average of the 2 points

remember
midpoint of (x1,y1) and (x2,y2) is

((x1+x2)/2,(y1+y2)/2)
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average of 3 and -8 is -5/2
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Two forces with magnitudes of 150 and 100 pounds act on an object at angles of 30° and 120°, respectively. Find the direction an

lyudmila [28]

<h2>Answer:</h2>

$Magnitude=180.27 \ lbf \\ \\ Direction=63.69 \ degrees$

<h2>Step-by-step explanation:</h2>

We have two forces as follows:

<u>First force:</u>

Magnitude: 150 pounds

Angle: 30°

<u>First force:</u>

Magnitude: 100 pounds

Angle: 120°

So the components can be found as follows:

$F_{1x}=150cos(30)=129.90 \ lbf\\F_{1y}=150sin(30) = 75 \ lbf \\ \\ F_{2x}=100cos(120)=-50 \ lbf\\F_{2y}=100sin(120) = 86.60 \ lbf$

So the components of the resultant force can be found by adding each component of the individual forces as follows:

$R_{x}=\Sigma F_{x} \\ R_{y}=\Sigma F_{y} \\ \\ R_{x}=129.90-50=79.90 \ lbf \\ R_{y}=75+86.60=161.6 \ lbf$

Finally, the magnitude and direction of the resultant force is:

$Magnitude \rightarrow R=\sqrt{R_{x}^2+R_{y}^2}=\sqrt{79.90^2+161.6^2}=180.27 \ lbf \\ \\ Direction \rightarrow \theta=tan^{-1}(\frac{R_{y}}{R_{x}})=tan^{-1}(\frac{161.6}{79.90})=63.69 \ degrees$