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4 years ago

Which equation represents a line that passes through (2,-2) and has a slope of 3?

Mathematics

1 answer:

notka56 [123]4 years ago

4 0

Y=3x-8 is the answer ， maybe u can find it in this equations

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Rachna borrowed a certain sum at the rate of 15% per annum. if she paid at the end of the two years Rs.1290 as interest Compound

Natalija [7]

Given info:

Compound Interest = Rs.1290

Rate of Interest = 15% p.a

Time = 2 years

<h3>Formula we have to know:-</h3>

$\dag{\underline{\boxed{\sf{C.I = P \bigg(1 + \dfrac{R}{100} \bigg)^{n} - 1}}}}$

<u>Where</u>

C.I = Compound Interest

P = Principle

R = Rate of Interest

N = Time

$\textsf{ \underline{Solution-}}\\$

Here

C.I = Rs.1290

R = 15%

N = 2 years

Principle = ?

Now,Calculating the sum (Principle) borrowed by Rachna

$\quad{: \implies{\sf{C.I = \Bigg[P \bigg(1 + \dfrac{R}{100} \bigg)^{n} - 1 \Bigg]}}}$

Substituting the given values

$\quad{: \implies{\sf{1290 = \Bigg[P \bigg(1 + \dfrac{15}{100} \bigg)^{2} - 1 \Bigg]}}}$

$\quad{: \implies{\sf{1290 = \Bigg[P \bigg(\dfrac{(1 \times 100) + 15}{100} \bigg)^{2} - 1 \Bigg]}}}$

$\quad{: \implies{\sf{1290 = \Bigg[P \bigg(\dfrac{115}{100} \bigg)^{2} - 1 \Bigg]}}}$

$\quad{: \implies{\sf{1290 = \Bigg[P \bigg(\dfrac{115}{100} \times \dfrac{115}{100} \bigg) - 1 \Bigg]}}}$

$\quad{: \implies{\sf{1290 = \Bigg[P \bigg(\dfrac{13225}{10000} \bigg) - 1 \Bigg]}}}$

$\quad{: \implies{\sf{1290 = \Bigg[P \bigg( \cancel{\dfrac{13225}{10000}} \bigg) - 1 \Bigg]}}}$

$\quad{: \implies{\sf{1290 = \Bigg[P \bigg({1.3225 - 1} \bigg) \Bigg]}}}$

$\quad{: \implies{\sf{1290 = P \times {0.3225}}}}$

$\quad{: \implies{\sf{\dfrac{1290}{0.3225} = P}}}$

$\quad{: \implies{\sf{\dfrac{1290 \times 1000}{0.3225 \times 1000} = P}}}$

$\quad{: \implies{\sf{\dfrac{12900000}{3225} = P}}}$

$\quad{: \implies{\sf{\cancel{\dfrac{12900000}{3225}} = P}}}$

$\quad{: \implies{\sf{Rs.4000 = P}}}$

$\quad{\dag{\underline{\boxed{\tt{\blue{Principle} = \purple{Rs.4000 }}}}}}$

$\begin{gathered}\end{gathered}$

<u>Hence,</u>

The sum (Principle) is Rs.4000.

3 0

3 years ago

If m is the midpoint of segment JK, JM equals 7x minus 32, and MK equals 89 minus 4x, what is MK?

KonstantinChe [14]

Step-by-step explanation:

Imagine this segment, given that m is the midpoint of segment JK...

J________M________K

JM = 7x-32 MK = 89 - 4x (given)

If m is the midpoint, then JM and MK are the same lengths (equal to each other). We can use that to first find what <em>x</em> is.

$7x -32 = 89 - 4x\\$

Steps to find x (we have to get x by itself):

$7x - 32 + 4x = 89 - 4x +4x$

$11x -32 = 89$

$11x -32 +32 = 89 +32$

$11x = 121$

$\frac{11x}{11} = \frac{121}{11}$

$x=11$ (we haven't found the answer yet! we only found x)

Now that we found x, we can find segment MK. Now we know that:

$MK = 89 - 4x$

$x= 11$

Let's plug x into the equation:

$MK = 89 - 4(11)\\MK = 89- (44)\\MK = 45$

Answer: $MK= 45$

4 0

3 years ago

AP Calculus redo! I know the answer but can't figure out exactly how to get there. Thank you! I want to work through the steps.

Monica [59]

You're approximating

$\displaystyle\int_1^5 x^2\,\mathrm dx$

with a Riemann sum, which comes in the form

$\displaystyle\int_a^b f(x)\,\mathrm dx=\lim_{n\to\infty}\sum_{i=1}^nf(x_i)\Delta x_i$

where $x_i$ are sample points chosen according to some decided-upon rule, and $\Delta x_i$ is the distance between adjacent sample points in the interval.

The simplest way of approximating the definite integral is by partitioning the interval into equally-spaced subintervals, in which case $\Delta x=\dfrac{b-a}n$ , and since $[a,b]=[1,5]$ , we have

$\Delta x=\dfrac{5-1}n=\dfrac4n$

Using the right-endpoint method, we approximate the area under $f(x)$ with rectangles whose heights are determined by their right endpoints. These endpoints are chosen by successively adding the subinterval length to the starting point of the interval of integration.

So if we had $n=4$ subintervals, we'd split up the interval of integration as

$[1,5]=[1,2]\cup[2,3]\cup[3,4]\cup[4,5]$

Note that the right endpoints follow a precise pattern of

$2=1+\dfrac44$
$3=1+\dfrac84$
$4=1+\dfrac{12}4$
$5=1+\dfrac{16}4$

The height of each rectangle is then given by the values above getting squared (since $f(x)=x^2$ ). So continuing with the example of $n=4$ , the Riemann sum would be

$\displaystyle\sum_{i=1}^4\left(1+\dfrac{4i}4\right)^2\dfrac44$

For $n=5$ ,

$\displaystyle\sum_{i=1}^5\left(1+\dfrac{4i}5\right)^2\dfrac45$

and so on, so that the definite integral is given exactly by the infinite sum

$\displaystyle\lim_{n\to\infty}\sum_{i=1}^n\left(1+\dfrac{4i}n\right)^2\dfrac4n$

4 0

3 years ago

The College Board originally scaled SAT scores so that the scores for each section were approximately normally distributed with

likoan [24]

Answer:

Percentage of students who scored greater than 700 = 97.72%

Step-by-step explanation:

We are given that the College Board originally scaled SAT scores so that the scores for each section were approximately normally distributed with a mean of 500 and a standard deviation of 100.

Let X = percentage of students who scored greater than 700.

Since, X ~ N( $\mu, \sigma^{2}$ )

The z probability is given by;

Z = $\frac{X-\mu}{\sigma}$ ~ N(0,1) where, $\mu$ = 500 and $\sigma$ = 100

So, P(percentage of students who scored greater than 700) = P(X > 700)

P(X > 700) = P( $\frac{X-\mu}{\sigma}$ < $\frac{700-500}{100}$ ) = P(Z < 2) = 0.97725 or 97.72% Therefore, percentage of students who scored greater than 700 is 97.72%.

8 0

3 years ago

20 points and will mark brainliest

mixer [17]

Answer:

$a \leqslant x \leqslant e$

Step-by-step explanation:

The domain is simply the farthest left ppint to the farthest right point on the graph.

4 0

4 years ago