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3 years ago

What is the y - intercept of a line that has a slope of 1/4 and passes through point (8, 3)

1 answer:

7 0

Substitute the point coordinates into the point slope formula: $y-y_1=m(x-x_1)$ . Substitute the slope 1/4 for m, 3 for y_1, and 8 for x_1.

$y-3=1/4(x-8)$ , distribute 1/4 inside the parentheses.

$y-3=1/4x-2$ , add 3 to both sides.

$y=1/4x+1$ ; the slope-intercept form of a line is y = mx + b, where b is the y-intercept, so in this line the y-intercept is A. 1.

You might be interested in

Write down the place value of 7 in each of these numbers

gizmo_the_mogwai [7]

Answer:

You haven't given the numbers for us to find the numbers place in. Here are my examples of the place values that "7" could be in:

17- ones

71- tens

711-hundreds

7,111-thousands

71,111-ten thousands

711,111-hundred thousands

7,111,111-one millions

Step-by-step explanation:

Hope this helps! Pls give brainliest!

7 0

3 years ago

You wish to buy a new skateboard. To earn money, you sell your baseball cards to

zzz [600]

So first you sell your baseball cards for $2.50 and you need $98 THEN

The minimum cards u need to sell is
98/2.50 which is 39.2 but you can’t sell cards in fraction so the final answer is 40

4 0

3 years ago

Read 2 more answers

At an interest rate of 8% compounded annually, how long will it take to double the following investments?

Paladinen [302]

Let's see, if the first one has a Principal of $50, when it doubles the accumulated amount will then be $100,

recall your logarithm rules for an exponential,

$\bf \textit{Logarithm of exponentials}\\\\
log_{{ a}}\left( x^{{ b}} \right)\implies {{ b}}\cdot log_{{ a}}(x)\\\\
-------------------------------\\\\
\qquad \textit{Compound Interest Earned Amount}
\\\\
$

$\bf A=P\left(1+\frac{r}{n}\right)^{nt}
\quad 
\begin{cases}
A=\textit{accumulated amount}\to &\$100\\
P=\textit{original amount deposited}\to &\$50\\
r=rate\to 8\%\to \frac{8}{100}\to &0.08\\
n=
\begin{array}{llll}
\textit{times it compounds per year}\\
\textit{annnually, thus once}
\end{array}\to &1\\
t=years
\end{cases}
\\\\\\
100=50\left(1+\frac{0.08}{1}\right)^{1\cdot t}\implies 100=50(1.08)^t
\\\\\\
\cfrac{100}{50}=1.08^t\implies 2=1.08^t\implies log(2)=log(1.08^t)
\\\\\\
$

$\bf log(2)=t\cdot log(1.08)\implies \cfrac{log(2)}{log(1.08)}=t\implies 9.0065\approx t\\\\
-------------------------------\\\\
$

now, for the second amount, if the Principal is 500, the accumulated amount is 1000 when doubled,

$\bf \qquad \textit{Compound Interest Earned Amount}
\\\\
A=P\left(1+\frac{r}{n}\right)^{nt}
\quad 
\begin{cases}
A=\textit{accumulated amount}\to &\$1000\\
P=\textit{original amount deposited}\to &\$500\\
r=rate\to 8\%\to \frac{8}{100}\to &0.08\\
n=
\begin{array}{llll}
\textit{times it compounds per year}\\
\textit{annnually, thus once}
\end{array}\to &1\\
t=years
\end{cases}
\\\\\\
1000=500\left(1+\frac{0.08}{1}\right)^{1\cdot t}\implies 1000=500(1.08)^t
\\\\\\
$

$\bf \cfrac{1000}{500}=1.08^t\implies 2=1.08^t\implies log(2)=log(1.08^t)
\\\\\\
log(2)=t\cdot log(1.08)\implies \cfrac{log(2)}{log(1.08)}=t\implies 9.0065\approx t\\\\
-------------------------------$

now, for the last, Principal is 1700, amount is then 3400,

$\bf \qquad \textit{Compound Interest Earned Amount}
\\\\
A=P\left(1+\frac{r}{n}\right)^{nt}
\quad 
\begin{cases}
A=\textit{accumulated amount}\to &\$3400\\
P=\textit{original amount deposited}\to &\$1700\\
r=rate\to 8\%\to \frac{8}{100}\to &0.08\\
n=
\begin{array}{llll}
\textit{times it compounds per year}\\
\textit{annnually, thus once}
\end{array}\to &1\\
t=years
\end{cases}$

$\bf 3400=1700\left(1+\frac{0.08}{1}\right)^{1\cdot t}\implies 3400=1700(1.08)^t
\\\\\\
\cfrac{3400}{1700}=1.08^t\implies 2=1.08^t\implies log(2)=log(1.08^t)
\\\\\\
log(2)=t\cdot log(1.08)\implies \cfrac{log(2)}{log(1.08)}=t\implies 9.0065\approx t$

8 0

4 years ago

In triangle opq right angled at p op=7cm,oq-pq=1 determine the values of sinq and cosq

soldi70 [24.7K]

Answer:

see explanation

Step-by-step explanation:

let pq = x

given oq - pq = 1 then oq = 1 + x

Using Pythagoras' identity, then

(oq)² = 7² + x²

(1 + x)² = 49 + x² ( expand left side )

1 + 2x + x² = 49 + x² ( subtract 1 from both sides )

2x + x² = 48 + x² ( subtract x² from both sides )

2x = 48 ( divide both sides by 2 )

x = 24 ⇒ pq = 24

and oq = 1 + x = 1 + 24 = 25 ← hypotenuse

sinq = $\frac{opposite}{hypotenuse}$ = $\frac{7}{25}$

cosq = $\frac{adjacent}{hypotenuse}$ = $\frac{24}{25}$

6 0

4 years ago

Solve for z N (17+ x ) = 34x - r

BigorU [14]

Answer:

Step-by-step explanation:

N(17+x)=34x−r

17n+xn=34x-r

xn= 34x - r - 17n

xn-34x= -r - 17n

x(n-34)= -r - 17n

x= (-r - 17n)/(n-34)

7 0

3 years ago