do you mind rephrasing it in english
will
Answer:
50.24 units^2
Step-by-step explanation:
Area of circle formula: 
r^2
r = 4, = 3.14
3.14 * (4^2) = 3.14 * 16 = 50.24
Answer:
The value of x and y that satisfy the equations is x = 2 and y = 1
Step-by-step explanation:
Given
2.5(x−3y)−3=−3x+0.5
3(x+6y)+4=9y+19
Required.
Find x and y
We start by opening all brackets
2.5(x−3y)−3=−3x+0.5 becomes
2.5x - 7.5y - 3 = -3x + 0.5
Collect like terms
2.5x + 3x - 7.5y = 3 + 0.5
5.5x - 7.5y = 3.5 ---- Equation 1
In similar vein, 3(x+6y)+4=9y+19 becomes
3x + 18y + 4 = 9y + 19
Collect like terms
3x + 18y - 9y = 19 - 4
3x + 9y = 15
Multiply through by ⅓
⅓ * 3x + ⅓ * 9y = ⅓ * 15
x + 3y = 5
Make x the subject of formula
x = 5 - 3y
Substitute 5 - 3y for x in equation 1
5.5(5 - 3y) - 7.5y = 3.5
27.5 - 16.5y - 7.5y = 3.5
27.5 - 24y = 3.5
Collect like terms
-24y = 3.5 - 27.5
-24y = -24
Divide through by - 24
y = 1
Recall that x = 5 - 3y.
Substitute 1 for y in this equation
x = 5 - 3(1)
x = 5 - 3
x = 2
Hence, x = 2 and y = 1
Answer:
16 books
Step-by-step explanation:
if Sandra read 5 books and Deacon read 6 and Breanna read 7 and they all one same book, then that means sandra read 4 different books then Deacon and Breanna. Deacon read 5 different books then Sandra and Breanna. Breanna read 6 different books then Deacon and Sandra.
So Sandras 4 books plus Deacons 5 books plus Breannas 6 books equals up to 15 but, now you need to add the one book they all read which makes the answer 16.
(Hopes this helps)
Answer:
See below
Step-by-step explanation:
When we talk about the function 
, the domain and codomain are generally defaulted to be subsets of the Real set. Once 
and 
such that 
for 
. Therefore,
![\[\sqrt{\cdot}: \mathbb R_{\geq 0} \to \mathbb R_{\geq 0} \]](https://tex.z-dn.net/?f=%5C%5B%5Csqrt%7B%5Ccdot%7D%3A%20%5Cmathbb%20R_%7B%5Cgeq%200%7D%20%5Cto%20%5Cmathbb%20R_%7B%5Cgeq%200%7D%20%5C%5D)
![\[x \mapsto \sqrt{x}\]](https://tex.z-dn.net/?f=%5C%5Bx%20%5Cmapsto%20%5Csqrt%7Bx%7D%5C%5D)
But this table just shows the perfect square solutions.
