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3 years ago

5

which equation represents a proportional relationship that has a constant of proportionality equal to 1/5

2 answers:

Serjik [45]3 years ago

6 0

Answer: Hello, a proportional relationship is a relationship where a quantity is proportional to other.

This means that, for example, if the number Y is proportional to the number X, means that the number Y is k times the number X, where k is the constant of proportionality.

This is represented in the next way: Y= k*X

then, if k = 1/5, our relationship is : Y = (1/5)*X.

liraira [26]3 years ago

4 0

Answer:

y = 1/5 x

Step-by-step explanation:

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Find x if segment A B is perpendicular to segment C D in the figure below.

ki77a [65]

If segment AB is perpendicular to segment CD, then a right angle (90°) is formed.

7x + 27 = 90

7x = 63

x = 9

3 0

3 years ago

Read 2 more answers

PLEASE HELP!!!!!!!

andre [41]

Answer:

$\frac{4^{21}}{5^6}$

Step-by-step explanation:

$\left(\frac{4^7}{5^2}\right)^3$

$=\frac{\left(4^7\right)^3}{\left(5^2\right)^3}$

$\left(4^7\right)^3$

$=4^{21}$

$=\frac{4^{21}}{\left(5^2\right)^3}$

$\left(5^2\right)^3$

$5^6$

$=\frac{4^{21}}{5^6}$

6 0

2 years ago

Please awnser thase questions for my prarice test

Vinvika [58]

Answer:

Add 5 for the first one

The 2nd one is 25 and k or 25 thousand.

Step-by-step explanation:

add 5 for the first one because 5 + 5 = 10

9 + 5 = 14 . and 13 + 5 = 18.

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For the 2nd one: 25k means 25 thousand or the Product of 25 and k.

5 0

2 years ago

Use the symbol > to compare -3.5 and -1.5

dybincka [34]

-1.5 is the answer because it is 75

3 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