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4 years ago

Which of the following pair(s) of circles have l as a common internal tangent? Select ALL that apply.

Mathematics

2 answers:

koban [17]4 years ago

8 0

Answer:

A and B

A and C

Step-by-step explanation:

Common internal tangents intersect the segment joining the centers of the two circles.

Tangent I is common internal tangent to circles A and B as the line joining the centers of circles will intersect the tangent I.

Tangent I is also a common internal tangent to circle A and C.The line joining the centers of circles A and C intersects tangent I .

Tangent I is not a common internal tangent to circles B and C,

kkurt [141]4 years ago

7 0

It looks like only A and B, but could be trick question?

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We will use compound interest formula to solve our given problem.

$A=P(1+\frac{r}{n})^{nt}$ , where,

A = Final amount after t years,

P = Principal amount,

r = Annual interest rate in decimal form,

n = Number of times interest is compounded per year,

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$8\%=\frac{8}{100}=0.08$

$300,000=100,000(1+\frac{0.08}{1})^{1\cdot t}$

$300,000=100,000(1.08)^{t}$

$\frac{300,000}{100,000}=\frac{100,000(1.08)^{t}}{100,000}$

$3=(1.08)^{t}$

$(1.08)^{t}=3$

Let us take natural log on both sides of equation.

$\text{ln}((1.08)^{t})=\text{ln}(3)$

Using natural log property $\text{ln}(a^b)=b\cdot \text{ln}(a)$ , we will get:

$t\cdot \text{ln}(1.08)=\text{ln}(3)$

$\frac{t\cdot \text{ln}(1.08)}{\text{ln}(1.08)}=\frac{\text{ln}(3)}{\text{ln}(1.08)}$

$t=\frac{1.0986122886681097}{0.0769610411361283}$

$t=14.274914586$

Upon rounding to nearest tenth of year, we will get:

$t\approx 14.3$

Therefore, it will take approximately 14.3 years until the account holds $300,000.

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