All categories

3 years ago

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.

You might be interested in

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:

Price of computer:

$1486.25

Warranty

$199.99

Total

$1486.25 + $199.99 = $1686.24

Payment per month:

$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