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Kay [80]
3 years ago
13

Which system of equations has exactly one solution?

Mathematics
2 answers:
Cerrena [4.2K]3 years ago
5 0

Answer:

2x+y = 3

5x-y= 11

has exactly one solution

Step-by-step explanation:

Hope this Helped :)

Hunter-Best [27]3 years ago
4 0
If you put them in desmos online it works
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Need help on geometry please right answer only I’m giving out 40 points
Juliette [100K]
Answer: first option 392.699 square feet.


Explanation:

1) The shape of the sidewalk is an ring with exterior radius equal to the radious of the fountain + 5 feet and inner radius equal to the radius of the fountain.

2) The area of such ring is equal to the area of the outer circle less the area of the inner circle (the fountain)

Area of a circle = π × r²

Area of the outer circle: π (10ft +  5 ft)² = π (15 ft)² = 225 π ft²

Area of the inner circle = π (10ft)² = 100 π ft²

Area of the ring  (sidewald) = 225π ft² - 100π ft² = 125π ft² = 392.699 ft²
6 0
3 years ago
Read 2 more answers
To the nearest foot, what is the radius of the Ferris wheel if you traveled 157 feet? feet
Fittoniya [83]

Answer:

50 feet

Step-by-step explanation:

on edg

6 0
3 years ago
Read 2 more answers
The output is eleven more then the input
enot [183]
The input is known as x and the output is known as y in math.
So you’re stating that we’re making the equation equal the output, y, and we’re adding based off of the word “more”. Now that we know we’re adding, we have to find what we are adding together. We are adding x and 11 is that it equals y. Note that x and y could be many different variables and values.
Y= x+11
5 0
3 years ago
Inverse laplace of [(1/s^2)-(48/s^5)]
Katen [24]
**Refresh page if you see [ tex ]**

I am not familiar with Laplace transforms, so my explanation probably won't help, but given that for two Laplace transform F(s) and G(s), then \mathcal{L}^{-1}\{aF(s)+bG(s)\} = a\mathcal{L}^{-1}\{F(s)\}+b\mathcal{L}^{-1}\{G(s)\}

Given that \dfrac{1}{s^2} = \dfrac{1!}{s^2} and -\dfrac{48}{s^5} = -2\cdot\dfrac{4!}{s^5}

So you have \mathcal{L}^{-1}\left\{\dfrac{1}{s^2} - 2\cdot\dfrac{4!}{s^5}\right\} = \mathcal{L}^{-1}\left\{\dfrac{1}{s^2}\right\} - 2\mathcal{L}^{-1}\left\{\dfrac{4!}{s^5}\right\}

From Table of Laplace Transform, you have \mathcal{L}\{t^n\} = \dfrac{n!}{s^{n+1}} and hence \mathcal{L}^{-1}\left\{\dfrac{n!}{s^{n+1}}\right\} = t^n

So you have \mathcal{L}^{-1}\left\{\dfrac{1}{s^2}\right\} - 2\mathcal{L}^{-1}\left\{\dfrac{4!}{s^5}\right\} = \boxed{t-2t^4}.

Hope this helps...
7 0
3 years ago
Find the distance between the points (10,10) and (–5,–5).
Eddi Din [679]

Answer:

Distance= 21.21

Step-by-step explanation:

3 0
3 years ago
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