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3 years ago

12

What is .64 repeating as a fraction

1 answer:

Lerok [7]3 years ago

6 0

It is about 2 over 3 which equals .67

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At a special sale, all pens are sold at one price and all pencils at another price.If 3 pens and 2 pencils are sold for 47 cents

mezya [45]

3 pen + 2 pencil = .47
2 pen + 3 pencil = .38

2(3pen + 2 pencil = .47)
-3(2pen + 3 pencil = .38)

6 pen + 4 pencil = .94
-6 pen - 9 pencil = -1.14
- 5 pencil = -.2
pencil = .04

3pen + 2(.04) = .47
3pen + .08 = .47
3 pen + .08 - .08 = .47 -.08
3 pen = .39
pen = .13

Pen = 13 cents and pencil = 4 cents

4 0

3 years ago

Find the area of the convex polygon with vertices (0,5), (-1,2), (4,4), (-3,-4) and (2,0).

Afina-wow [57]

The area of the convex polygon is 43/2 square units

<h3>How to determine the area of the convex polygon?</h3>

The vertices are given as:

(0,5), (-1,2), (4,4), (-3,-4) and (2,0)

The area is then calculated as:

$A = \frac 12(\left[\begin{array}{cc}x_1&x_2\\y_1&y_2\end{array}\right] + \left[\begin{array}{cc}x_2&x_3\\y_2&y_3\end{array}\right] + ....+\left[\begin{array}{cc}x_n&x\\y_n&y\end{array}\right] )$

So, we have:

$A = \frac12 * (|-1 * 5 - 2 * 0 + 0 * 4 - 5 * 4 + 4 * 0 - 4 * 2 + 2 * -4 - 0 * -3 -3 * 2 + 4 * 1|)$

Evaluate

$A = \frac12 * (|-43|)$

Remove the absolute bracket

$A = \frac12 * 43$

This gives

$A = \frac{43}2$

Hence, the area of the convex polygon is 43/2 square units

Read more about convex polygon at:

brainly.com/question/14522707

#SPJ1

7 0

2 years ago

Pls help me answer this

eimsori [14]

Answer:

11/25

Step-by-step explanation:

You would add the black, white and red which equal 50. Next you would get 22(red) over 50 (total) 22/50 = 11/25 when you simplify

7 0

3 years ago

Plz help meeeeeeeeee

Brums [2.3K]

Y = 2.2x

to check let’s plug in numbers

6.6 = 2.2(3)
6.6 = 6.6

since it’s true it’s most likely the right answer :)

6 0

3 years ago

Determine whether the function f(x) -1 / x^2 + x^4 is even odd or neither

kirill [66]

If f(-x) = f(x) then the function is even.

If f(-x) = -f(x) then the function is odd.

We have: $f(x)=\dfrac{-1}{x^2+x^4}$

calculate f(-x):

$f(-x)=\dfrac{-1}{(-x)^2+(-x)^4}=\dfrac{-1}{(-1x)^2+(-1x)^4}\\\\=\dfrac{-1}{(-1)^2x^2+(-1)^4x^4}=\dfrac{-1}{1x^2+1x^4}=\dfrac{-1}{x^2+x^4}=f(x)$

f(-x) = f(x), therefore your answer: The function f(x) is even.

If we have: $f(x)=\dfrac{-1}{x^2}+x^4$

$f(-x)=\dfrac{-1}{(-x)^2}+(-x)^4=\dfrac{-1}{x^2}+x^4$

f(-x) = f(x)

f(x) is even

5 0

3 years ago