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Romashka [77]
3 years ago
11

Nearest hundred the number 324

Mathematics
2 answers:
makvit [3.9K]3 years ago
6 0
The nearest of 324 is 300
tangare [24]3 years ago
5 0
300 because it's not at 50 or above 50 yet
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Rewrite the expression using Distributive Property, then simplify<br><br> -1(x-9)+4x
pentagon [3]
First ask yourselfhat needs to be dristributed.
That would be -1 (x-9)
To distribute you have to multiply the number out side of the parentheses (-1) by each term inside the parentheses( x and -9)
-1×x=-x
-1×-9=9
Now the expression is
-x+9+4x
To simplify you have to combine like terms (4x and-x)
4x-x=3x
Your answer is 9+3x or 3(3+x).
5 0
3 years ago
A rectangular box is designed to have a square base and an open top. The volume is to be 500in.3 What is the minimum surface are
mr_godi [17]

The minimum surface area that such a box can have is 380 square

<h3>How to determine the minimum surface area such a box can have?</h3>

Represent the base length with x and the bwith h.

So, the volume is

V = x^2h

This gives

x^2h = 500

Make h the subject

h = 500/x^2

The surface area is

S = 2(x^2 + 2xh)

Expand

S = 2x^2 + 4xh

Substitute h = 500/x^2

S = 2x^2 + 4x * 500/x^2

Evaluate

S = 2x^2 + 2000/x

Differentiate

S' = 4x - 2000/x^2

Set the equation to 0

4x - 2000/x^2 = 0

Multiply through by x^2

4x^3 - 2000 = 0

This gives

4x^3= 2000

Divide by 4

x^3 = 500

Take the cube root

x = 7.94

Substitute x = 7.94 in S = 2x^2 + 2000/x

S = 2 * 7.94^2 + 2000/7.94

Evaluate

S = 380

Hence, the minimum surface area that such a box can have is 380 square

Read more about surface area at

brainly.com/question/76387

#SPJ1

5 0
2 years ago
If Darlene reads 5/8 of a 56 page book how many pages did she read?
Marina86 [1]
Darlene read 35 pages of the book.
5 0
3 years ago
Read 2 more answers
Please can I have an explanation also, I am terrible at these kinds of questions!
wlad13 [49]

Answer:

<em>The fraction of the beads that are red is</em>

Step-by-step explanation:

<u>Algebraic Expressions</u>

A bag contains red (r), yellow (y), and blue (b) beads. We are given the following ratios:

r:y = 2:3

y:b = 5:4

We are required to find r:s, where s is the total of beads in the bag, or

s = r + y + b

Thus, we need to calculate:

\displaystyle \frac{r}{r+y+b}       \qquad\qquad    [1]

Knowing that:

\displaystyle \frac{r}{y}=\frac{2}{3}      \qquad\qquad    [2]

\displaystyle \frac{y}{b}=\frac{5}{4}

Multiplying the equations above:

\displaystyle \frac{r}{y}\frac{y}{b}=\frac{2}{3}\frac{5}{4}

Simplifying:

\displaystyle \frac{r}{b}=\frac{5}{6}       \qquad\qquad    [3]

Dividing [1] by r:

\displaystyle \frac{r}{r+y+b}=\displaystyle \frac{1}{1+y/r+b/r}

Substituting from [2] and [3]:

\displaystyle \frac{r}{r+y+b}=\displaystyle \frac{1}{1+3/2+6/5}

Operating:

\displaystyle \frac{r}{r+y+b}=\displaystyle \frac{1}{\frac{10+3*5+6*2}{10}}

\displaystyle \frac{r}{r+y+b}=\displaystyle \frac{10}{10+15+12}

\displaystyle \frac{r}{r+y+b}=\displaystyle \frac{10}{37}

The fraction of the beads that are red is \mathbf{\frac{10}{37}}

8 0
2 years ago
Find the minimum and maximum value of the function on the given interval by comparing values at the critical points and endpoint
Kitty [74]

Answer:

maximum: y = 1

minimum: y = 0.

Step-by-step explanation:

Here we have the function:

y = f(x) =  √(1 + x^2 - 2x)

we want to find the minimum and maximum in the segment [0, 1]

First, we evaluate in the endpoints, which are 0 and 1.

f(0)  =√(1 + 0^2 - 2*0) = 1

f(1) = √(1 + 1^2 - 2*1) = 0

Now let's look at the critical points (the zeros of the first derivate)

To derivate our function, we can use the chain rule:

f(x) = h(g(x))

then

f'(x) = h'(g(x))*g(x)

Here we can define:

h(x) = √x

g(x) = 1 + x^2 - 2x

Then:

f(x) = h(g(x))

f'(x)  =  1/2*( 1 + x^2 - 2x)*(2x - 2)

f'(x) = (1 + x^2 - 2x)*(x - 1)

f'(x) = x^3 - 3x^2 + x - 1

this function does not have any zero in the segment [0, 1] (you can look it in the image below)

Thus, the function does not have critical points in the segment.

Then the maximum and minimum are given by the endpoints.

The maximum is 1 (when x = 0)

the minimum is 0 (when x = 1)

7 0
3 years ago
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