Step-by-step explanation:
i think it will help you please give me Brainlist bro
Sphere Surface Area = <span> 4 • <span>π <span>• r²
For it to equal 16 PI, then radius must equal 2
4*PI*2*2 = 16 PI
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Sphere Volume = <span> 4/3 • <span>π <span>• r³
</span></span></span>
Sphere Volume = <span> 4/3 • <span>π <span>• 2^3
</span></span></span>
Sphere Volume = <span> 4/3 *PI * 8
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Sphere Volume = <span> 32 / 3 PI
</span>
Sphere Volume = <span> 10.666 PI cubic feet AND I think that is answer B
which SHOULD read 10 (2/3) PI ft^3
</span>
Answer:
a. 
b. 
c. 
d. 
Step-by-step explanation:
The sample space associated with the random experiment of throwing a dice is is the equiprobable space {R1, R2, R3, R4, R5, R6}. Then,
a. The conditional probability that 3 is rolled given that the roll is greater than 1? 
b. What is the conditional probability that 6 is rolled given that the roll is greater than 3? 
c. What is P [GIE], the conditional probability that the roll is greater than 3 given that the roll is even? 
d. Given that the roll is greater than 3, what is the conditional probability that the roll is even? 
Answer:-1
Step-by-step explanation:You do reverse operation so negative 2Y means multiplication so you divide two -2 sides so positive two divided by -2 is -1
First, we are going to find the vertex of our quadratic. Remember that to find the vertex

of a quadratic equation of the form

, we use the vertex formula

, and then, we evaluate our equation at

to find

.
We now from our quadratic that

and

, so lets use our formula:




Now we can evaluate our quadratic at 8 to find

:




So the vertex of our function is (8,-72)
Next, we are going to use the vertex to rewrite our quadratic equation:



The x-coordinate of the minimum will be the x-coordinate of the vertex; in other words: 8.
We can conclude that:
The rewritten equation is

The x-coordinate of the minimum is 8