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Jlenok [28]
3 years ago
6

Which table of ordered pairs represents a proportional relationship?

Mathematics
2 answers:
mixer [17]3 years ago
6 0

for all my smoth head fam he said D

:)

dedylja [7]3 years ago
5 0

Answer:

A 2-column table with 3 rows. Column 1 is labeled x with entries 14, 28, 42. Column 2 is labeled y with entries 2, 4, 6.

Step-by-step explanation:

Examining the response options:

A 2- compartment chart with 3 rows. Column 1 is designated x with records 1, 5, 6. compartment 2 is designated y with records 5, 20, 30.

No, as y = 25 not 20 and x = 5 .

A 2- compartment chart with 3 rows. Column 1 is identified x with approaches 48, 32, 16. compartment 2 is designated y with listings 4, 2, 1.

No, y =64 not 48 and x= 4.

A 2- compartment chart with 3 lines. compartment 1 is designated x with records 0, 1, 3. compartment 2 is designated y with records 90, 70, 25.

No, y= 25+3*45= 160 not 90 and x = 3.

A 2- compartment chart with 3 lines. compartment 1 is designated x with records 42, 28, 14. compartment 2 is designated y with records 2, 4, 6.

Yes, as 2*7 = 14, 4*7= 28, 6*7= 42.

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The design for the palladium window shown includes a semicircular shape at the top. The bottom is formed by squares of equal siz
9966 [12]

Answer:

(a)\ Area = 3765.32

(b)\ Area = 4773

Step-by-step explanation:

Given

A_1 = 169in^2 --- area of each square

Shade = 4in

See attachment for window

Solving (a): Area of the window

First, we calculate the dimension of each square

Let the length be L;

So:

L^2 = A_1

L^2 = 169

L = \sqrt{169

L=13

The length of two squares make up the radius of the semicircle.

So:

r = 2 * L

r = 2*13

r = 26

The window is made up of a larger square and a semi-circle

Next, calculate the area of the larger square.

16 small squares made up the larger square.

So, the area is:

A_2 = 16 * 169

A_2 = 2704

The area of the semicircle is:

A_3 = \frac{\pi r^2}{2}

A_3 = \frac{3.14 * 26^2}{2}

A_3 = 1061.32

So, the area of the window is:

Area = A_2 + A_3

Area = 2704 + 1061.32

Area = 3765.32

Solving (b): Area of the shade

The shade extends 4 inches beyond the window.

This means that;

The bottom length is now; Initial length + 8

And the height is: Initial height + 4

In (a), the length of each square is calculated as: 13in

4 squares make up the length and the height.

So, the new dimension is:

Length = 4 * 13 + 8

Length = 60

Height = 4*13 + 4

Height = 56

The area is:

A_1 = 60 * 56 = 3360

The radius of the semicircle becomes initial radius + 4

r = 26 + 4 = 30

The area is:

A_2 = \frac{3.14 * 30^2}{2} = 1413

The area of the shade is:

Area = A_1 + A_2

Area = 3360 + 1413

Area = 4773

7 0
2 years ago
Which number property is illustrated in the Identy ?
zhuklara [117]
The property of this equation is associative property
4 0
1 year ago
(18 months). hi I'm gothy and I just needed help with this problem. you don't have to include the answer but I'd appreciate it​
natulia [17]

Answer:

  • $93.68

Step-by-step explanation:

<u>Price of computer:</u>

  • $1486.25

<u>Warranty </u>

  • $199.99

<u>Total</u>

  • $1486.25 + $199.99 = $1686.24

<u>Payment per month:</u>

  • $1686.24/ 18 = $93.68

8 0
3 years ago
What is the coefficient of x2y3 in the expansion of (2x + y)5?
Zigmanuir [339]

Option C:

The coefficient of x^{2} y^{3} is 40.

Solution:

Given expression:

(2 x+y)^{5}

Using binomial theorem:

(a+b)^{n}=\sum_{i=0}^{n}\left(\begin{array}{l}n \\i\end{array}\right) a^{(n-i)} b^{i}

Here a=2 x, b=y

Substitute in the binomial formula, we get

(2x+y)^5=\sum_{i=0}^{5}\left(\begin{array}{l}5 \\i\end{array}\right)(2 x)^{(5-i)} y^{i}

Now to expand the summation, substitute i = 0, 1, 2, 3, 4 and 5.

$=\frac{5 !}{0 !(5-0) !}(2 x)^{5} y^{0}+\frac{5 !}{1 !(5-1) !}(2 x)^{4} y^{1}+\frac{5 !}{2 !(5-2) !}(2 x)^{3} y^{2}+\frac{5 !}{3 !(5-3) !}(2 x)^{2} y^{3}

                                                            $+\frac{5 !}{4 !(5-4) !}(2 x)^{1} y^{4}+\frac{5 !}{5 !(5-5) !}(2 x)^{0} y^{5}

Let us solve the term one by one.

$\frac{5 !}{0 !(5-0) !}(2 x)^{5} y^{0}=32 x^{5}

$\frac{5 !}{1 !(5-1) !}(2 x)^{4} y^{1} = 80 x^{4} y

$\frac{5 !}{2 !(5-2) !}(2 x)^{3} y^{2}= 80 x^{3} y^{2}

$\frac{5 !}{3 !(5-3) !}(2 x)^{2} y^{3}= 40 x^{2} y^{3}

$\frac{5 !}{4 !(5-4) !}(2 x)^{1} y^{4}= 10 x y^{4}

$\frac{5 !}{5 !(5-5) !}(2 x)^{0} y^{5}=y^{5}

Substitute these into the above expansion.

(2x+y)^5=32 x^{5}+80 x^{4} y+80 x^{3} y^{2}+40 x^{2} y^{3}+10 x y^{4}+y^{5}

The coefficient of x^{2} y^{3} is 40.

Option C is the correct answer.

5 0
2 years ago
Sonji bought a combination lock that opens with a four-digit number created using the digits 0 through 9. The same digit cannot
Alex777 [14]

Answer:

504 combos

Step-by-step explanation:

Cannot use digits twice and last is '7'

 9 digits for first place choice  

     8 digits for second place

         7 digits for third place

               only  '7' for fourth place       9 x 8 x7 = 504 combos

8 0
1 year ago
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