Find the slope of the line perpendicular to the line y=6x+11

In a math class, 12 people reported their grades on the last test. Here are their responses:

Murrr4er [49]

Answer:

(a) 23, 35, 28, 33, 5, 12, 40, 25, 20, 18, 1, 16

1, 5, 12, 16, 18, 20 23, 25, 28, 33, 35, 40

n = 12 M = (20 + 23)/ 2 = 21.5

Q1 = (12 + 16) / 2 = 14 Q3 = (28 + 33) /2 = 30.5

(b) 22, 33, 25, 28, 5, 12, 35, 23, 20, 18, 1, 40, 16

1, 5, 12, 16, 18, 20 | 22 | 23, 25, 28, 33, 35, 40

n = 13 M = 22 Q1 = 14 Q3 = 30.5

(c) 20, 28, 23, 25, 3, 5, 33, 22, 18, 16, 40, 1, 35, 12

1, 3, 5, 12, 16, 18, 20, 22, 23, 25, 28, 33, 35, 40

n = 14 M = (20 + 22)/2 = 21 Q1 = 12 Q3 = 28

(d) 20, 28, 23, 25, 3, 5, 30, 22, 18, 40, 16, 35, 1, 33, 12

1, 3, 5, 12, 16, 18, 20 | 22 | 23, 25, 28, 30, 33, 35, 40

n = 15 M = 22 Q1 = 12 Q3 = 30

hope this helps :)

5 0

3 years ago

Y=(x+5)(2x-3) in standard form

mestny [16]

the step by step explanation is to expand the binomial

3 0

3 years ago

Solve for x radical forms

vova2212 [387]

Answer:

in right angled triangle ABC

by using Pythagoras law

h²=p²+b²

x²=7²+10²

x=√149=12.21cm

8 0

3 years ago

A student placement center has request from six students for interviews regarding employment with a particular consulting firm,

Nikolay [14]

Use the image below for your answer please

3 0

3 years ago

Which of the following is a like radical to RootIndex 3 StartRoot 7x EndRoot?

lana [24]

Answer:

A. 4 (RootIndex 3 StartRoot 7 x EndRoot) or $4(\sqrt[3]{7x})$

Step-by-step explanation:

Given:

A radical whose value is, $r_1=\sqrt[3]{7x}$

Now, we need to find the like radical for $r_1$ .

Let the like radical be $r_2$ .

As per the definition of like radicals, like radicals are those that can be expressed as multiples of each other.

So, if two radicals $r_1\ and\ r_2$ are like radicals, then

$r_1 = n \times r_2
$

Where, 'n' is a real number.

Here, $r_1=\sqrt[3]{7x}$

Now, let us check all the options .

Option A:

4 (RootIndex 3 StartRoot 7 x EndRoot) or $r_2=4\sqrt[3]{7x}$

Now, we observe that $r_2$ is a multiple of $r_1$ because

$r_2=4\times \sqrt[3]{7x}\\\\ r_2=4\times r_1..............(r_1=\sqrt[3]{7x})$

Therefore, option A is correct.

Option B:

StartRoot 7 x EndRoot or $r_2=\sqrt{7x}$

As the above radical is square root and not a cubic root, this option is incorrect.

Option C:

x (RootIndex 3 StartRoot 7 EndRoot) or $r_2=x\sqrt[3]{7}$

As the term inside the cubic root is not same as that of $r_1$ , this option is also incorrect.

Option D:

7 StartRoot x EndRoot or $r_2=7\sqrt{x}$

As the above radical is square root and not a cubic root, this option is incorrect.

Therefore, the like radical is option (A) only.

4 0

3 years ago

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