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Grace [21]
3 years ago
6

Can u help me plz thank u

Mathematics
1 answer:
Nitella [24]3 years ago
5 0

Answer:

A. Neither Sneha nor the credit card company owes money

Step-by-step explanation:

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What is 5 1/4 divided by 7?
NARA [144]
ANSWER: I think it’s 3/4
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Identify the area of segment MNO to the nearest hundredth. HELP PLEASE!! I don't understand it!
katrin2010 [14]

Answer:

  13.98 in²

Step-by-step explanation:

I don't understand it, either.

Point N is part of a "segment" that above and to the right of chord MO. It is the sum of the areas of 3/4 of the circle and a right triangle with 7-inch sides. The larger segment MO to the upper right of chord MO has an area of about 139.95 in², which <u>is not</u> an answer choice.

__

The remaining segment, to the lower left of chord MO does not seem to have anything to do with point N. However, its area is 13.98 in², which <u>is</u> an answer choice. Therefore, we think the question is about this segment, and we wonder why it is called MNO.

The area of a segment is given by the formula ...

  A = (1/2)(θ -sin(θ))r² . . . . . . where θ is the central angle in radians.

Here, we have θ = π/2, r = 7 in, so we can compute the area of the smaller segment MO as ...

  A = (1/2)(π/2 -sin(π/2))(7 in)² = 24.5(π/2 -1) in² ≈ 13.9845 in²

Rounded to hundredths, this is ...

  ≈ 13.98 in²

3 0
3 years ago
610 divided by 20 without a decimal
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Answer:31 or 30.5 orrr 61/2

Step-by-step explanation:

quick mathz

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1. the prices for 6 different jerseys are
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Step-by-step explanation:

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5 0
2 years ago
Find dy/dx if y =x^3+5x+2/x²-1
stiks02 [169]

<u>Differentiate using the Quotient Rule</u> –

\qquad\pink{\twoheadrightarrow \sf \dfrac{d}{dx} \bigg[\dfrac{f(x)}{g(x)} \bigg]= \dfrac{ g(x)\:\dfrac{d}{dx}\bigg[f(x)\bigg] -f(x)\dfrac{d}{dx}\:\bigg[g(x)\bigg]}{g(x)^2}}\\

According to the given question, we have –

  • f(x) = x^3+5x+2
  • g(x) = x^2-1

Let's solve it!

\qquad\green{\twoheadrightarrow \bf \dfrac{d}{dx}\bigg[ \dfrac{x^3+5x+2 }{x^2-1}\bigg]} \\

\qquad\twoheadrightarrow \sf \dfrac{(x^2-1) \dfrac{d}{dx}(x^3+5x+2) - ( x^3+5x+2)  \dfrac{d}{dx}(x^2-1)}{(x^2-1)^2 }\\

\qquad\twoheadrightarrow \sf \dfrac{(x^2-1)(3x^2+5)  -  ( x^3+5x+2) 2x}{(x^2-1)^2 }\\

\qquad\pink{\sf \because \dfrac{d}{dx} x^n = nx^{n-1} }\\

\qquad\twoheadrightarrow \sf \dfrac{3x^4+5x^2-3x^2-5-(2x^4+10x^2+4x)}{(x^2-1)^2 }\\

\qquad\twoheadrightarrow \sf \dfrac{3x^4+5x^2-3x^2-5-2x^4-10x^2-4x}{(x^2-1)^2 }\\

\qquad\green{\twoheadrightarrow \bf \dfrac{x^4-8x^2-4x-5}{(x^2-1)^2 }}\\

\qquad\pink{\therefore  \bf{\green{\underline{\underline{\dfrac{d}{dx} \dfrac{x^3+5x+2 }{x^2-1}}  =  \dfrac{x^4-8x^2-4x-5}{(x^2-1)^2 }}}}}\\\\

7 0
2 years ago
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