No, it is not.
4/7 ≈ 0.5714
3/5 = 0.6
Actually, 3/5 is greater than 4/7.
Another way you could do this is by multiplying 4/7 with 5/5 and by multiplying 3/5 with 7/7.
4/7 * 5/5 = 20/35
3/5 * 7/7 = 21/35
Have an awesome day! :)
Answer:
21 C
22 B
23 169
Step-by-step explanation:
21 3m mean 3 times a certain number
22 the negative make the 4 go negative and the -5 positive
23 i did 5^2-4^2= 9
then 6^2 divided by 9+9= 13 i took the 9 from the 5^2 problem then divided it by 6^2
then 13 x 13 = 169
If we evaluate the function at infinity, we can immediately see that:

Therefore, we must perform an algebraic manipulation in order to get rid of the indeterminacy.
We can solve this limit in two ways.
<h3>Way 1:</h3>
By comparison of infinities:
We first expand the binomial squared, so we get

Note that in the numerator we get x⁴ while in the denominator we get x³ as the highest degree terms. Therefore, the degree of the numerator is greater and the limit will be \infty. Recall that when the degree of the numerator is greater, then the limit is \infty if the terms of greater degree have the same sign.
<h3>Way 2</h3>
Dividing numerator and denominator by the term of highest degree:



Note that, in general, 1/0 is an indeterminate form. However, we are computing a limit when x →∞, and both the numerator and denominator are positive as x grows, so we can conclude that the limit will be ∞.
<h3> Learning task 1</h3>
1. <u> </u><u> </u><u>3</u><u>.</u><u> </u><u> </u> 3. <u> </u><u> </u><u>1</u><u>. </u><u> </u>
4. 2
2. <u> </u><u> </u><u> </u><u>5</u><u>.</u><u> </u> 4. <u> </u><u> </u><u>6</u><u>. </u><u> </u>
9. 13
5. <u> </u><u> </u><u> </u><u>3</u><u>. </u> 6. <u> </u><u> </u><u> </u><u>7</u><u>. </u><u> </u>
5. 9
Step by step explanation:
hopefully that's help
Answer:
V≈100.53
Step-by-step explanation:
V=πr2h=π·22·8≈100.53096