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3 years ago

5

Find the area of the circle with the center C (1, -2) and passes through the point (4, 2). Round your answer to the nearest tent

h.

1 answer:

Firlakuza [10]3 years ago

7 0

Answer:

Step-by-step explanation:

r=√((4-1)²+(2+2)²)=√(9+16)=5

area of circle=π r²=3.14×5²=78.50=78.5

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One side of triangle twice the length of the shortest side and the 3rd side 14 more feet than length of shortest side, find dime

kifflom [539]

Let's say the sides are "a" first one, "b" second one and "c" the third one

ok, let's say the shortest is "a"
and let's say "b" is -> "<span> twice the length of the shortest side"
so b = 2a

and let's say "c" is -> "</span><span>3rd side 14 more feet than length of shortest side"
so c = a + 14

we know the perimeter ( sum of all sides ) is 154

</span> $\bf \begin{array}{llll}
a&b&c\\
&\downarrow&\downarrow \\
&2a&a+14\\
\downarrow &\downarrow &\downarrow \\
\end{array}\\
\ a+b+\ c=154\\\\
a+(2a)+(a+14)=154$
<span>
solve for "a"</span>

5 0

3 years ago

If m+n=0, what is the value of n?

LUCKY_DIMON [66]

Answer: n= 0 - m

Step-by-step explanation:

To find value of n, isolate it by subtracting m from both sides.

6 0

3 years ago

Write in logarithmic form: <img src="https://tex.z-dn.net/?f=5%5E%5Cfrac%7B-1%7D%7B2%7D%20%3D%20%5Cfrac%7B%5Csqrt%7B5%7D%20%7D%7

Ket [755]

Answer:

Left-hand side: $\displaystyle -\frac{1}{2}\, \ln(5)$ .

Right-hand side: $\displaystyle \frac{1}{2}\, \ln(5) - \ln(5)$ .

Step-by-step explanation:

Apply the logarithm power rule: $\ln \left(x^{a}\right) = a\, \ln (x)$ for all $x >0$ .

This property is not only true for logarithm to the base $e$ , but for other bases, as well.

Take the logarithm (to the base $e$ ) of the left-hand side of this equation:

$\displaystyle \ln \left(5^{-1/2}\right) = (-1/2)\, \ln(5)$ .

For the right-hand side of this equation, consider the logarithm quotient rule:

$\displaystyle \ln \left(\frac{a}{b}\right) = \ln(a) - \ln (b)$ for all $a> 0$ and $b > 0$ .

Indeed, on the right-hand side of this equation, $\sqrt{5} > 0$ and $5 > 0$ . Therefore:

$\displaystyle \ln\left(\frac{\sqrt{5}}{5}\right) = \ln\left(\sqrt{5}\right) - \ln(5)$ .

This expression could be further simplified. Notice that $\sqrt{x}$ is equivalent to $x^{1/2}$ for all $x \ge 0$ . (Think about how $\sqrt{x} \cdot \sqrt{x} =x$ whereas $x^{1/2} \cdot x^{1/2} = x^{(1/2) + (1/2)} = x$ .)

Therefore, $\ln \left(\sqrt{5}\right)$ would be equivalent to $\ln\left(5^{1/2}\right)$ . Apply the logarithm power rule to show that $\displaystyle \ln\left(5^{1/2}\right) = \frac{1}{2}\, \ln(5)$ .

$\begin{aligned} \text{R.H.S.} &= \ln\left(\frac{\sqrt{5}}{5}\right) \\ &= \ln\left(\sqrt{5}\right) - \ln(5) \\ &= \frac{1}{2}\, \ln(5) - \ln (5) = -\frac{1}{2}\, \ln(5)\end{aligned}$ .

Indeed, the left-hand side of this equation matches the right-hand side.

7 0

3 years ago

The missing term in the following polynomial has a degree of 5 and a coefficient of 16. Which statement best describes the polyn

dexar [7]

5+6=y+m so it will be 3

5 0

3 years ago

Read 2 more answers

An architect presents a 3 inch wide by 4 inch deep by 3 inch tall model of a new central campus dorm . If the final building fou

IRISSAK [1]

Answer:

Building depth = 168 ft

Step-by-step explanation:

we must find the relationship (scale) of the model with respect to the building .

wide-m (model) = 3 in

deep-m(model) = 4 in

wide-b (building) = 126 feet

deep-b(building) = x

the measures must be standardized

let's convert feet to inches

12 in → 1 feet

3 in → ?, then

3 in → 0.25 feet and 4 in → 1/3 feet

now deep-m = x

$x=\frac{(1/3)ft*126ft}{0.25ft}=168 ft$

8 0

3 years ago