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2 years ago

In the figure shown, FG is tangent to circle D.

Mathematics

1 answer:

romanna [79]2 years ago

8 0

Answer:

Option B

Step-by-step explanation:

From the figure attached,

Circle D is drawn with the radius = DG or DE

A tangent FG has been drawn at a point G on the circle from an external point F.

By theorem,

Radius of a circle is always perpendicular to the tangent, drawn to the circle from an external point.

Therefore, DG ⊥ FG.

Option B will be the correct option.

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The ratio of money in Obi's wallet to Rudy's wallet one day was 5:2. Obi spent ?20 that day. Obi now had ?8 less than Rudy. How

BlackZzzverrR [31]

Answer:

Altogether, they had ?28

Step-by-step explanation:

Had Obi spent only ?12, they would have been even. Thus ?12 is the same as 5-2 = 3 "ratio units". So, each "ratio unit" is worth ?12/3 = ?4.

Then Obi started the day with 5·4 = 20, and Obi started the day with 2·4 = 8. Obi ended with 0, which is 8 less than Rudy.

Altogether, they started with 20 + 8 = 28 of whatever a ? is.

8 0

2 years ago

Read 2 more answers

The harmonic motion of a particle is given by f(t) = 2 cos(3t) + 3 sin(2t), 0 ≤ t ≤ 8. (a) When is the position function decreas

iren [92.7K]

For the last part, you have to find where $f'(t)$ attains its maximum over $0\le t\le8$ . We have

$f'(t)=-6\sin3t+6\cos2t$

so that

$f''(t)=-18\cos3t-12\sin2t$

with critical points at $t$ such that

$-18\cos3t-12\sin2t=0$

$3\cos3t+2\sin2t=0$

$3(\cos^3t-3\cos t\sin^2t)+4\sin t\cos t=0$

$\cos t(3\cos^2t-9\sin^2t+4\sin t)=0$

$\cos t(12\sin^2t-4\sin t-3)=0$

So either

$\cos t=0\implies t=\dfrac{(2n+1)\pi}2$

$12\sin^2t-4\sin t-3=0\implies\sin t=\dfrac{1\pm\sqrt{10}}6\implies t=\sin^{-1}\dfrac{1\pm\sqrt{10}}6+2n\pi$

where $n$ is any integer. We get 8 solutions over the given interval with $n=0,1,2$ from the first set of solutions, $n=0,1$ from the set of solutions where $\sin t=\dfrac{1+\sqrt{10}}6$ , and $n=1$ from the set of solutions where $\sin t=\dfrac{1-\sqrt{10}}6$ . They are approximately

$\dfrac\pi2\approx2$

$\dfrac{3\pi}2\approx5$

$\dfrac{5\pi}2\approx8$

$\sin^{-1}\dfrac{1+\sqrt{10}}6\approx1$

$2\pi+\sin^{-1}\dfrac{1+\sqrt{10}}6\approx7$

$2\pi+\sin^{-1}\dfrac{1-\sqrt{10}}6\approx6$

4 0

2 years ago

How many weeks of flea treatment would Jim’s dog get from one package if each treatment only lasted 3 weeks In a normal year whe

4vir4ik [10]

Solution

Number of weeks in a year = 52 weeks

If in a normal year when each package of flea treatment lasts for 4 weeks, then in a year there Jim's dog will have to be treated for

$\frac{52}{4} =13 times$

Where as, when the fleas are bad in a year, the treatment lasts for only 3 weeks.

Then in a year Jim's dog would get

$\frac{52}{3} =17 times$

So Jim's dog will get 17- 13 =4 treatments more.

4 treatments that are made in 3 weeks each will be 4×3 =12 weeks more treatment

4 0

2 years ago

Question 1 (1 point) The yearly interest payable on a deposit of $250 at 5.5% p.a. simple interest is: A $137.50 B $13.75 C $12.

dlinn [17]

Answer:

B: $13.75

Step-by-step explanation:

5.5% of $250 = $13.75

Therefore, the simple interest is $13.75.

6 0

1 year ago

What are these fractions in simplest form?

Anton [14]

$\frac{16y^{3}}{20y^{4}} = \frac{4}{5y}$
$\frac{6xy}{16y} = \frac{3x}{8}$
$\frac{abc}{10abc} = \frac{1}{10}$
$\frac{mn^{2}}{pm^{5}n} = \frac{n}{pm^{4}}$
$\frac{12h^{3}k}{16h^{2}k^{2}} = \frac{3h}{4k}$
$\frac{8x}{10y} = \frac{4x}{5y}$
$\frac{24n^{2}}{28n} = \frac{6n}{7}$
$\frac{30hxy}{54kxy} = \frac{5h}{9k}$
$\frac{5jh}{15jh^{3}} = \frac{1}{3h^{2}}$
$\frac{20s^{2}t^{3}}{16st^{5}} = \frac{5s}{4t^{2}}$

8 0

2 years ago