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sladkih [1.3K]
3 years ago
5

Which is smaller? A. 0.1456 B. 0.0321

Mathematics
2 answers:
Lerok [7]3 years ago
7 0
It should be answer B. With 0.0321 there is a zero both before and after the decimal, while answer A only has one before the decimal.
andrey2020 [161]3 years ago
5 0
B is smaller.
Look in the tenths place, A has a 1 and B has a 0. 
Which number is greater? 1 is greater than 0.
Also, A has one tenth while B has a fraction of a tenth.
So, B. 0.0321 is smaller.
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You and two friends visit a pizza shop. There are 15 toppings. 5 toppings are meat and the rest are vegetable. Each of you order
andrew-mc [135]
We know that
total toppings=15
<span>meat toppings=5
</span>
<span>The first person has a
5/15 chance of getting a meat topping.

The second person has a
4/14 chance because there are no repeats,

The third person would have a
3/13 chance.
Since there are three multiply
5/15 * 4/14 * 3/13.=60/2730
</span>60/2730-------> divide by 30 both members------> 2/91

the answer is
2/91
8 0
3 years ago
$3.00 is ____ % of $1.80
natta225 [31]
1.666666666666667%

3.00/1.80


1.666666666666667 as a fraction is 5/3
8 0
3 years ago
I need help with part b. I feel like there’s a catch, I want to do the first derivative test, however, I feel like there is a be
Sladkaya [172]

Answer:

The fifth degree Taylor polynomial of g(x) is increasing around x=-1

Step-by-step explanation:

Yes, you can do the derivative of the fifth degree Taylor polynomial, but notice that its derivative evaluated at x =-1 will give zero for all its terms except for the one of first order, so the calculation becomes simple:

P_5(x)=g(-1)+g'(-1)\,(x+1)+g"(-1)\, \frac{(x+1)^2}{2!} +g^{(3)}(-1)\, \frac{(x+1)^3}{3!} + g^{(4)}(-1)\, \frac{(x+1)^4}{4!} +g^{(5)}(-1)\, \frac{(x+1)^5}{5!}

and when you do its derivative:

1) the constant term renders zero,

2) the following term (term of order 1, the linear term) renders: g'(-1)\,(1) since the derivative of (x+1) is one,

3) all other terms will keep at least one factor (x+1) in their derivative, and this evaluated at x = -1 will render zero

Therefore, the only term that would give you something different from zero once evaluated at x = -1 is the derivative of that linear term. and that only non-zero term is: g'(-1)= 7 as per the information given. Therefore, the function has derivative larger than zero, then it is increasing in the vicinity of x = -1

6 0
3 years ago
Solve –16t² +144= 0 to find the number of seconds, t, it takes for an object dropped from 144 ft above the ground to hit the gro
Vlad1618 [11]

Answer:

h(t) = -16t2 + 144

h(1) = -16(12) + 144 = 128 ft

h(2) = -16(22) + 144 = 80 ft

h(2) - h(1) = 80 - 128 = -48 ft

It fell 48 ft between t = 1 and t = 2 seconds.

It reaches the ground when h(t) = 0

0 = -16t2 + 144

t = √(144/16) s = 3s

It reaches the ground 3s after being dropped.

Step-by-step explanation:

6 0
2 years ago
Rewrite the expression in the form 2n​
SashulF [63]

Answer:

two times a number

Step-by-step explanation:

5 0
3 years ago
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