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3 years ago

What is the difference between perimeter and area of polygons?

1 answer:

3 0

Answer: The perimeter is when you add all of the sides together, and the area is when you multiply the sides together

Step-by-step explanation:

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Find the value of each trigonometric ratio

Orlov [11]

The value of each trigonometric ratio is

$$\sin A=\frac{15}{17}, \ \cos A=\frac{8}{17}, \ \tan A =\frac{15}{8}$

$$\csc A=\frac{17}{15}, \ \sec A=\frac{17}{8}, \ \cot A =\frac{8}{15}$

Solution:

The given triangle is right triangle.

AC (hypotenuse) = 34, AB (adjacent) = 16, BC (opposite) = 30

<u>To find the trigonometric ratios:</u>

Using trigonometric formulas for right triangle,

$$\sin \theta=\frac{\text { opposite }}{\text { hypotenuse }}$

$$\sin A=\frac{BC}{AC}$

$$\sin A=\frac{30}{34}=\frac{15}{17}$

$$\cos \theta=\frac{\text { adjacent }}{\text { hypotenuse }}$

$$\cos A=\frac{AB}{AC}$

$$\cos A=\frac{16}{34}=\frac{8}{17}$

$$\tan \theta=\frac{\text { opposite }}{\text { adjacent }}$

$$\tan A=\frac{BC}{AB}$

$$\tan A=\frac{30}{16}=\frac{15}{8}$

$$\csc A =\frac{1}{\sin A}$

$$\csc A =\frac{17}{15}$

$$\sec A =\frac{1}{\cos A}$

$$\sec A =\frac{17}{8}$

$$\cot A =\frac{1}{\tan A}$

$$\cot A =\frac{8}{15}$

Hence the value of each trigonometric ratio is

$$\sin A=\frac{15}{17}, \ \cos A=\frac{8}{17}, \ \tan A =\frac{15}{8}$

$$\csc A=\frac{17}{15}, \ \sec A=\frac{17}{8}, \ \cot A =\frac{8}{15}$

4 0

3 years ago

Two point charges q and 4q are at x = 0 and x = L, respectively, and free to move. A third charge isplaced so that the entire th

Ivanshal [37]

Answer:

The answers to the question are

Magnitude = 4/9·q

Sign = Opposite in sign to those of q and 4·q, that is, -ve where q and 4·q are +ve

x coordinate L/3

or $+\frac{4}{9} q$ at x coordinate = $\frac{L}{3}$

Step-by-step explanation:

To solve the question, we note that

Force between charges is given by

$F =k\frac{q_1q_2}{r^2}$ , therefore the force between the two charges q and 4q is

$F = k\frac{q(4q)}{(L^2)} = k\frac{4q^2}{L^2}$

For equilibrium, the charge on the third charge p, will be opposite to those of q and 4·q and the location will be between 0 and L

Therefore the force between the p and q placed at a distance d from q = $F(pq) = k\frac{pq}{d^2}$ and the force between p and 4q = $F(4qp) = k\frac{4pq}{(L-d)^2}$

For equilibrium, these two forces should be equal, therefore

$k\frac{qp}{d^2}$ = $k\frac{4pq}{(L-d)^2}$ which gives $\frac{qp}{d^2} = \frac{4pq}{(L-d)^2}$ and by cross multiplying, we have

(L-d)²× p×q = d²× 4×p×q → (L-d)² = d²× 4 = (L-d)² - d²× 4 = 0 or

L² - 2·d·L -3·d² = 0, which could be factored as

(L+d)×(L-3·d) = 0 Which gives either L = -d or L = 3·d

Since L > d as d is in between 0 and L, then the correct solution is L = 3·d

Since the system is in equilibrium then $\frac{4q^2}{L^2} = \frac{pq}{d^2}$ or $\frac{4q^2}{(3d)^2} = \frac{pq}{d^2}$

Cancelling like terms gives $\frac{4}9 q= p$

Therefore the magnitude of p = 4/9·q

The location of p is L/3 from the charge q

7 0

3 years ago

What is the result of adding these two equations?

andrew-mc [135]

Answer:

When adding these two equations we get the value of x is $\frac{14}{3}$ and y is $\frac{52}{3}$

Therefore $x=\frac{14}{3}$ and and $y=\frac{52}{3}$

Step-by-step explanation:

Given two equations are $5x-y=6\hfill (1)$ $-2x+y=8\hfill (2)$ To find the addition of these two given equations :

Adding the equations (1) and (2) we get

5x-y=6

-2x+y=8

_______

3x=14

$x=\frac{14}{3}$

Substitute the value $x=\frac{14}{3}$ in equation (1) we get

$5(\frac{14}{3})-y=6$

$-y=6-\frac{70}{3}$

$-y=\frac{18-70}{3}$

$-y=-\frac{52}{3}$

Therefore $y=\frac{52}{3}$

When adding these two equations we get the value of x is $\frac{14}{3}$

Therefore $x=\frac{14}{3}$ and $y=\frac{52}{3}$

5 0

3 years ago

I don’t know how to graph two equations that 2w+2l=228 and l=w+42 and calculate the w and l.

Reika [66]

Solve for w and l by using the system of equations.

2l=-2w+228

l=w+42

Substitute w+42 for l in the first equation.

2(w+42)=-2w+228

2w+84=-2w+228

4w=144

w=36

Now substitute 36 for w in any equation to get l=78.

To graph the equations, first make them equal l.

l=-w+114 and l=w+42 are your equations. You ca plug in values for w to plot points on the graph.

5 0

3 years ago

Right on. Naya's graph has a slope of -3, so the

zloy xaker [14]

B I hope this helps a lot for your test

3 0

3 years ago