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3 years ago

Which linear function represents the line given by the point-slope equation y – 2 = 4(x – 3)?

1 answer:

8 0

y = 4x - 10. The linear functions y = 4x - 10 represent the line given by the point-slope equation y - 2 = 4(x - 3).

In order to solve this problem, we have to take the point-slope equation y - 2 = 4(x - 3) and convert it to the linear function form y = mx + b as follow:

y - 2 = 4(x - 3)

y - 2 = 4(x) - (4)(3)

y - 2 = 4x - 12

y - 2 + 2 = 4x -12 +2

y = 4x - 10 (Linear function)

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A circle with area 16pi has a sector with a central angle of 8/5pi radians. What is the area of the sector?

lianna [129]

Answer:

The area of the sector is $12.8\pi\ units^2$

Step-by-step explanation:

we know that

The area of a complete circle (16π units^2) subtends a central angle of 2π radians

using proportion

Find out the area of a sector , if the central angle is equal to 8π/5 radians

$\frac{16\pi}{2\pi}=\frac{x}{(8\pi/5)}\\\\x=8(8\pi/5)\\\\x= 12.8\pi\ units^2$

4 0

3 years ago

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Susan made a deposit of $74 to her bank account she then had $192 in the account how much money was in the account before the de

Aleksandr-060686 [28]

Rewind the "mind movie" of what happened. She added $74 to her account and then had $192. Back it up by subtracting the $74 from $192.

192 - 74 = 118

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3 years ago

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What is the equation in standard form of the line that passes through the point (1, 24) and has a slope of -0.6.

lana [24]

Answer:

0.6x+y=24.6

Step-by-step explanation:

y-y1=m(x-x1)

y-24=-0.6(x-1)

y=-0.6x+0.6+24

y=-0.6x+24.6

y-(-0.6x)=24.6

y+0.6x=24.6

0.6x+y=24.6

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3 years ago

Please help I will give brainlesstQuestion 9 of 10

inessss [21]

Answer: D, x^3 (x-7)

Step-by-step explanation:

plug into G(f(x))

x^3 (x - 7)

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3 years ago

If $8700 is invested at 3% annual simple interest, how much should be invested at 6% annual simple interest so that the total ye

True [87]

First off, see how much 8700 as principal, yields at 3% APR
that is $\bf \qquad \textit{Simple Interest Earned}\\\\
I = Prt\qquad 
\begin{cases}
I=\textit{interest earned}\\
P=\textit{original amount deposited}\to& \$8700\\
r=rate\to 3\%\to \frac{3}{100}\to &0.03\\
t=years\to &1
\end{cases}$

it will yield some amount

subtract that amount from 393
the difference is how much the yield will be on the 6% investment
so

$\bf \qquad \textit{Simple Interest Earned}\\\\
I = Prt\quad 
\begin{cases}
I=\textit{interest earned}\\
P=\textit{original amount deposited}\to& \$8700\\
r=rate\to 3\%\to \frac{3}{100}\to &0.03\\
t=years\to &1
\end{cases}
\\\\\\
\implies \boxed{?}\\\\
-----------------------------\\\\
\textit{how much to invest at 6\%?}
\\\\\\
$

$\bf \qquad \textit{Simple Interest Earned}\\\\
(393-\boxed{?}) = Prt\quad 
\begin{cases}
I=\textit{interest earned}\\
P=\textit{original amount deposited}\to& \$\\
r=rate\to 6\%\to \frac{6}{100}\to &0.06\\
t=years\to &1
\end{cases}
\\\\\\
\textit{solve for "P", to see how much should the Principal be}\\\\
\textit{keep in mind that }P+\boxed{?}=393\leftarrow \textit{both yields added}$

6 0

3 years ago