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love history [14]
3 years ago
15

Is 8/9 bigger than 1?

Mathematics
1 answer:
Margarita [4]3 years ago
4 0

No. 10/9 would be bigger than 1.


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What’s -0.25 x = -1.25
grin007 [14]

Answer:

x=5

Step-by-step explanation:

divide -0.25 by -1.25

x= 5

plz give me brainliest

7 0
3 years ago
A spinner is divided into four equal sections labeled 1, 2, 3, and 4. Another spinner is divided into three equal sections label
Romashka-Z-Leto [24]
Since we are asked to compute the probability of two event with "or"
statement,  then we will add the probabilities like this:
Denote A the first event and B the second event:
P(A\cup B)=P(A)+P(B)
A="Get an even number":
P(A)= \frac{2}{4}= \frac{1}{2}
B="Get section B":
P(B)= \frac{1}{3}
Conclusion:
P(A\cup B)=P(A)+P(B)= \frac{1}{2} +\frac{1}{3}\\=\frac{5}{6}
The probability is 5/6
8 0
3 years ago
Read 2 more answers
What is 6/9 divided by 1/3
alisha [4.7K]

Answer:

2

Step-by-step explanation:

6/9÷1/3

Keep Change Flip

6/9 x 3/1

Multiply.

6/3

=2

Hope this helps :)

4 0
3 years ago
Read 2 more answers
Find the absolute maximum and minimum values of f on the set D. f(x,y)=2x^3+y^4, D={(x,y) | x^2+y^2<=1}.
castortr0y [4]

Answer:

absolute maximum is f(1, 0) = 2 and the absolute minimum is f(−1, 0) = −2.

Step-by-step explanation:

We compute,

$ f_x = 6x^2, f_y=4y^3 $

Hence, $ f_x = f_y = 0 $  if and only if (x,y) = (0,0)

This is unique critical point of D. The boundary equation is given by

$ x^2+y^2=1$

Hence, the top half of the boundary is,

$ T = \{ x, \sqrt{1-x^2} : -1 \leq x \leq 1\}

On T we have, $ f(x, \sqrt{1-x^2} = 2x^3 +(1-x^2)^2 = x^4 +2x^3-2x^2+1  \text{ for}\ -1 \leq x \leq 1$

We compute

$ \frac{d}{dx}(f(x, \sqrt{1-x^2}))= 4x^3+6x^2-4x = 2x(2x^2+3x-2)=2x(2x-1)(x+2)=0$

0 if and  only if x=0, x= 1/2 or x = -2.

We disregard  $ x = -2 \notin [-1,1]$

Hence, the critical points on T are (0,1) and $(\frac{1}{2}, \sqrt{1-(\frac{1}{2})^2}=\frac{\sqrt3}{2})$

On the bottom half, B, we have

$ f(x, \sqrt{1-x^2})= f(x,-\sqrt{1-x^2})$

Therefore, the critical points on B are (0,-1) and $( 1/2, -\sqrt3/2)  

It remains to  evaluate f(x, y) at the points $ (0,0), (0 \pm1), (1/2, \pm \sqrt3/2) \text{ and}\  (\pm1, 0)$ .

We should consider  latter two points, $(\pm1,0)$, since they are the boundary points for the T and also  B. We compute $ f(0,0)=0, \ \f(0 \pm1)=1, \ \ f(0, \pm \sqrt3/2)=9/16, \ \ f(1,0 )= 2 \text{ and}\ \ f(-1,0)= -2 $

We conclude that the  absolute maximum = f(1, 0) = 2

And the absolute minimum = f(−1, 0) = −2.

6 0
3 years ago
Simplify<br> (y + 1)<br> (y + 1)?
sattari [20]
Y + 1 is the answer that is as most simplified as possible. We just don’t need the brackets
8 0
3 years ago
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