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3 years ago

What is 119.63 rounded to the nearest whole number ?

Mathematics

2 answers:

Evgen [1.6K]3 years ago

4 0

120

bc its grater than four in this case forty

Ghella [55]3 years ago

3 0

119.63 rounded to the nearest whole number is 120.

A whole number is NOT a decimal so to answer this, look to the number in the tenths place because it is next to the number you are rounding off to in the ones place and determine whether or not it is less than five. If it is less than five, the number in the ones place remains the same; if it is greater than or equal to five, it goes up. In this case, 6>5, so you round up.

Since the number in the ones place is 9, it becomes a zero and the number in the tens place becomes a 2 so you are one number higher than 119. The final answer is 120.

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The function f(x) = 2x2 + 6x - 12 has a domain consisting of the integers from -2 to 1, inclusive. Which set represents the corr

Dimas [21]

Answer:

[-16,-4]

Step-by-step explanation:

We are given that

$f(x)=2x^2+6x-12$

Domain=[-2,1]

We have to find the set of range values of f(x).

Substitute x=-2

$f(-2)=2(-2)^2+6(-2)-12$

$f(-2)=-16$

Substitute x=1

$f(1)=2(1)+6(1)-12=-4$

Range of f(x)=[-16,-4]

Hence, the set of range values of f(x)

[-16,-4]

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3 years ago

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DerKrebs [107]

Answer:

The answer to the question provided is letter D.

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3 years ago

If f(x) = 2x−5, then what is the value of f(2) + f(5)

DochEvi [55]

First of all f(2)=2(2)-5=-1 and f(5)=2(5)-5=5
so
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7 0

3 years ago

Find the average value of f over the region D. f(x, y) = 3xy, D is the triangle with vertices (0, 0), (1, 0), and (1, 9)

MakcuM [25]

The average value of f over the region D is 243/4

To answer the question, we need to know what the average value of a function is

<h3>What is the average value of a function?</h3>

The average value of a function f(x) over an interval [a,b] is given by

$\frac{1}{b - a} \int\limits^b_a {f(x)} \, dx$

Now, given that we require the average value of f(x,y) = 3xy over the region D where D is the triangle with vertices (0, 0), (1, 0), and (1, 9).

x is intergrated from x = 0 to 1 and the interval is [0,1] and y is integrated from y = 0 to y = 9

So, $\frac{1}{b - a} \int\limits^b_a {f(x,y)} \, dA = \frac{1}{1 - 0} \int\limits^1_0 \int\limits^9_0 {3xy} \, dxdy \\= \frac{3}{1} \int\limits^1_0 {x} \,dx\int\limits^9_0 {y} \,dy\\ = \frac{3}{1} [\frac{x^{2} }{2} ]^{1}_{0}[\frac{y^{2} }{2} ]^{9}_{0} \\= 3[\frac{1^{2} }{2} - \frac{0^{2}}{2} ] [\frac{9^{2} }{2} - \frac{0^{2}}{2} ] \\= 3[\frac{1}{2} - 0 ][\frac{81}{2} - 0 ]\\= \frac{81}{2} X3 X \frac{1}{2} \\= \frac{243}{4}$