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4 years ago

C is a point of tangency. Segments PC and PB have a common point at P. Find the length PC.

Mathematics

2 answers:

Dmitry_Shevchenko [17]4 years ago

8 0

Answer:

Step-by-step explanation:

Tcecarenko [31]4 years ago

6 0

The correct answer is B hope it helps :)

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Evaluate rge following logarithmic expression without the use of a calculator.Enter your answer as a fraction reduced to lowest

UNO [17]

$\bf log_8(\sqrt[7]{64})\qquad 64=8^2\qquad log_8(\sqrt[7]{8^2})\\\\
-----------------------------\\\\
log_{{ a}}{{ a}}^x\implies x\impliedby \textit{using this cancellation rule}\\\\
-----------------------------\\\\
log_8\left( 8^{\frac{2}{7}} \right)\implies \cfrac{2}{7}$

7 0

3 years ago

An answer from the smarter a+7(a+b)-8(a+b)+b???

netineya [11]

Answer:

Step-by-step explanation:

Given

a + 7(a + b) - 8(a + b) + b ← distribute both parenthesis

= a + 7a + 7b - 8a - 8b + b ← collect like terms

= a + 7a - 8a + 7b - 8b + b

= 0

4 0

3 years ago

Read 2 more answers

I'd really appreciate it if anyone could help! :)

PSYCHO15rus [73]

C (third graph, the parabola); To find if it is a function or not draw a vertical line on the graph. If the vertical line could never intercept the figure more than once anywhere on the graph then it is a function.

3 0

3 years ago

Read 2 more answers

If 10000 is invested at an interest rate of 10 per year ,compound semiannually,find the value of the investment after the given

slega [8]

Answer:

Part a) $\$17,958.56$

Part b) $\$32,251.00$

Part c) $\$57,918.16$

Step-by-step explanation:

The complete question is

If $10,000 is invested at an interest rate of 10% per year, compounded semiannually, find the value of the investment after the given number of years. (Round your answers to the nearest cent.)

a)6 years

b)12 years

c)18 years

we know that

The compound interest formula is equal to

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

where

A is the Final Investment Value

P is the Principal amount of money to be invested

r is the rate of interest in decimal

t is Number of Time Periods

n is the number of times interest is compounded per year

Part a) 6 years

we have

$t=6\ years\\ P=\$10,000\\ r=10\%=10/100=0.10\\n=2$

substitute in the formula above

$A=10,000(1+\frac{0.10}{2})^{2*6}$

$A=10,000(1.05)^{12}$

$A=\$17,958.56$

Part b) 12 years

we have

$t=12\ years\\ P=\$10,000\\ r=10\%=10/100=0.10\\n=2$

substitute in the formula above

$A=10,000(1+\frac{0.10}{2})^{2*12}$

$A=10,000(1.05)^{24}$

$A=\$32,251.00$

Part c) 18 years

we have

$t=18\ years\\ P=\$10,000\\ r=10\%=10/100=0.10\\n=2$

substitute in the formula above

$A=10,000(1+\frac{0.10}{2})^{2*18}$

$A=10,000(1.05)^{36}$

$A=\$57,918.16$

7 0

4 years ago

For the following system, use the second equation to make a substitution for y in the first equation. 3x + y = 2 y + 4 = 3x

expeople1 [14]

First, set the second equation with y as the isolated variable. To do this, subtract 4 from both sides to get
$y = 3x - 4$
Now that you know the value of y, put that expression in for y in the first equation to get
$3x + (3x - 4) = 2$
Then, simplify by adding the x terms together to get
$6x - 4 = 2$
add 4 on both sides to isolate the x terms to get
$6x = 6$

then divide 6 from both sides to get
$x = 1$
Using that x value, plug it into one of the equations as x to find y ( I will use y=3x-4)
$y = 3(1) - 4 = 3 - 4 = - 1$
x=1 and y=-1 is the answer

5 0

3 years ago