Y=x2+8x+10. complete the square

Write an<br> equation in point slope form that passes through (25,32) with a slope of -4

Margaret [11]

Answer:

y-32=-4(x-25) point slope form

Step-by-step explanation:

y-y1=m(x-x1)

y-32=-4(x-25)

3 0

3 years ago

The rectangle below has an area of 8x^5+12x^3+20x^2.

Y_Kistochka [10]

Answer:

The width of the rectangle is 4x^24x

2

4, x, squared

The length of the rectangle is 2x^3+3x+52x

3

+3x+5

Step-by-step explanation:

5 0

4 years ago

Read 2 more answers

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

Please help me! I’ll mark as brilliant

IrinaVladis [17]

Answer: 30 i believe

Step-by-step explanation:

3 0

3 years ago

1) The graph of g(x) is shown below. Which statement given below is FALSE?

Mnenie [13.5K]

<h3>Answer: Choice D </h3>

=======================================================

Explanation:

Let's go through the answer choices one by one to see which are true, and which are false.

Choice A) This is true because as we approach x = 2 from the left hand side, the y values get closer to y = 1 from the top
Choice B) This is true. As we get closer to x = 4 on the left side, the blue curve is heading downward forever toward negative infinity. So this is what y is approaching when x approaches 4 from the left side.
Choice C) This is true also. The function is continuous at x = -3 due to no gaps or holes at this location, so that means its limit here is equal to the function value.
Choice D) This is false. The limit does exist and we find it by approaching x = -4 from either side, and we'll find that the y values are approaching y = -2. In contrast, the limit at x = 2 does not exist because we approach two different y values when we approach x = 2 from the left and right sides (approach x = 2 from the left and you get closer to y = 1; approach x = 2 from the right and you get closer to y = -2). So again, the limit does exist at x = -4; however, the function is not continuous here because its limiting value differs from its function value.
Choice E) This is true because the function curve approaches the same y value from either side of x = 6. Therefore, the limit at x = 6 exists.

7 0

3 years ago