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4 years ago

5

the graph shows the proportion relationship between the side lengths of a square and it perimeter.Which equation shows the same

relationship between the side lengths and the perimeter of the square

1 answer:

zhannawk [14.2K]4 years ago

5 0

Answer:

second one I believe

Step-by-step explanation:

You might be interested in

Which ordered pair is the solution to the system of equations?

tiny-mole [99]

Answer:

You can use Gaussian Elimination.

Double both sides of the first equation and add the second equation.

6x + 4y = 8

5x - 4y = 3

---------------

11x = 11

x = 1

5 - 4y = 3

-4y = -2

y = 1/2

Step-by-step explanation:

7 0

4 years ago

How do I solve this? I need help

Sedaia [141]

Answer: √51
—————————
a^2 + b^2 = c^2
a^2 + 7^2 = 10^2
a^2 + 49 = 100
a^2 = 51
√(a^2) = √51
a = √51

7 0

3 years ago

What is the percent change of 20 - 60?

Sergeeva-Olga [200]

40% is the answer ...

3 0

3 years ago

Read 2 more answers

3. What three numbers are the Pythagorean triple generated by using 4 for x and 1 for y. Remember to use the formulas:

Doss [256]

Given:

Consider the three number are given by:

$a=x^2-y^2$

$b=2xy$

$c=x^2+y^2$

To find:

The three numbers are the Pythagorean triple generated by using 4 for x and 1 for y.

Solution:

We have,

$a=x^2-y^2$

$b=2xy$

$c=x^2+y^2$

Substituting 4 for x and 1 for y, we get

$a=4^2-1^2$

$a=16-1$

$a=15$

Similarly,

$b=2(4)(1)$

$b=8$

And,

$c=4^2+1^2$

$c=16+1$

$c=17$

The three three numbers are the Pythagorean triple are 8,15 and 17.

Therefore, the correct option is C.

8 0

3 years ago

Suppose that you pick a bit string from the set of all bit strings of length ten. Find the probability that a) the bit string ha

emmasim [6.3K]

Answer:

45/1024
1/4
15/128
193/512
9/512

Step-by-step explanation:

There are 2^10 = 1024 bit strings of length 10.

a) There are 10C2 = 45 ways to have exactly two 1-bits in 10 bits

p(2 1-bits) = 45/1024

__

b) Of the four (4) possibilities for beginning and ending bits (00, 01, 11, 10), exactly one (1) of those is 00.

p(b0=0 & b9=0) = 1/4

__

c) There are 10C7 = 120 ways to have seven 1-bits in the bit string.

p(7 1-bits) = 120/1024 = 15/128

__

d) ∑10Ck {for k=0 to 4} = 386 is the total of the number of ways to have 0, 1, 2, 3, or 4 1-bits in the string. If there are more than that, there won't be more 0-bits than 1-bits

p(more 0 bits) = 386/1024 = 193/512

__

e) The string will have two 1-bits if it starts with 1 and there is a single 1-bit among the other 9 bits. There are 9 ways that can happen, among the 512 ways to have 9 remaining bits.

p(2 1-bits | first is a 1-bit) = 9/512

6 0

3 years ago