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2 years ago

6

Please help! what is it

1 answer:

Dmitrij [34]2 years ago

3 0

Answer:

105

Step-by-step explanation:

The interior angle for R is supplementary to the 155 degree angle

(they add to 180)

180-155=25

The sum of the interior angles of a triangle is always 180

180-50-25= 105

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Select the better deal in the pair. Then give the unit rate for the better deal. $43 $61.05 15 gal 10 gal

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43 dollars for 10 gallons

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2 years ago

Which is the result of the following equation<br><br> 3x ^2-9=0

pickupchik [31]

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2 years ago

Read 2 more answers

-x=-y+2 and 3y+4=2x solve using matrices

MissTica

$\begin{cases}-x=-y+2\\3y+4=2x\end{cases}\implies\begin{cases}x-y=-2\\2x-3y=4\end{cases}$

In matrix form, this is

$\begin{bmatrix}1&-1\\2&-3\end{bmatrix}\begin{bmatrix}x\\y\end{bmatrix}=\begin{bmatrix}-2\\4\end{bmatrix}$

The coefficient matrix has determinant

$\begin{vmatrix}1&-1\\2&-3\end{vmatrix}=-3+2=-1\neq0$

so it has an inverse, which is

$\begin{bmatrix}1&-1\\2&-3\end{bmatrix}^{-1}=\begin{bmatrix}3&-1\\2&-1\end{bmatrix}$

Multiply both sides by the inverse matrix:

$\begin{bmatrix}3&-1\\2&-1\end{bmatrix}\begin{bmatrix}1&-1\\2&-3\end{bmatrix}\begin{bmatrix}x\\y\end{bmatrix}=\begin{bmatrix}3&-1\\2&-1\end{bmatrix}\begin{bmatrix}-2\\4\end{bmatrix}$

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8 0

2 years ago

A fossil was analyzed and determined to have a carbon-14 level that is 80 % that of living organisms. the half-life of c-14 is 5

liq [111]

Answer:

The fossil is 1860 years old.

Step-by-step explanation:

The equation for the amount of fossil has the following format:

$Q(t) = Q(0)e^{rt}$

In which Q(t) is the amount after t years, Q(0) is the initial amount and r is the rate of change.

Half-life of c-14 is 5730 years.

This means that $Q(5730) = 0.5Q(0)$

So

$Q(t) = Q(0)e^{rt}$

$0.5Q(0) = Q(0)e^{5730r}$

$e^{5730r} = 0.5$

$\ln{e^{5730r}} = \ln{0.5}$

$5730r = \ln{0.5}$

$r = \frac{\ln{0.5}}{5730}$

$r = -0.00012$

So

$Q(t) = Q(0)e^{-0.00012t}$

How old is the fossil?

This is t for which

$Q(t) = 0.8Q(0)$

So

$Q(t) = Q(0)e^{-0.00012t}$

$0.8Q(0) = Q(0)e^{-0.00012t}$

$e^{-0.00012t} = 0.8$

$\ln{e^{-0.00012t}} = \ln{0.8}$

$-0.00012t = \ln{0.8}$

$t = -\frac{\ln{0.8}}{-0.00012}$

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The fossil is 1860 years old.

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2 years ago

What is the LCMs of 6,2,22

kati45 [8]

The LCM of 6,2,and 22 is 66.

Happy studying ^-^

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2 years ago