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3 years ago

How did you figure out the answers ? how did you work it out ?

Mathematics

1 answer:

Maksim231197 [3]3 years ago

7 0

How did who figure out what answer?

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Find an irrational number between -10 and -10.1.

Pavlova-9 [17]

Between two rational numbers there are an infinite number of irrational numbers.

Example:

$- 5 \sqrt{3} -\sqrt{2}= 5 \times 1.7320508 - 1.41421356 \approx -10.0744676\\\\
-10.1 \ \textless \ -10,0744676 \ \textless \ -10$

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2 years ago

Ming is painting a picture. He has 2/3 pint of red paint, 2/8 pint of yellow paint and 2/6 pint of green paint. Which list order

vichka [17]

Answer:

2/3,2/6,2/8

Step-by-step explanation:

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3 years ago

PLEASE HELP!!!!

Paraphin [41]

What is your favorite flavor of ice cream

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3 years ago

You and your friends have just measured the heights of your dogs. The heights are 600mm, 470mm, 170mm, 430mm and 300mm. Find out

Rufina [12.5K]

Answer:

To find the mean, you have to add all numbers together and divide them by the number of data points.

let me show what I mean:

The heights of the dogs are 600,470,170, 430,300.

Step-by-step explanation:

600+470+170+430+300.

1970.

Now we have to divide 1,970 by 5 ( because we are given 5 data points)

The mean is:

1970/5=394

Variance: First, arrange the data from least to greatest.

s2 = 27130

Standard Deviation: 164.71187

3 0

3 years ago

Use triangle ABC drawn below & only the sides labeled. Find the side of length AB in terms of side a, side b & angle C o

Brrunno [24]

Answer:

$AB = \sqrt{a^2 + b^2-2abCos\ C}$

Step-by-step explanation:

Given:

The above triangle

Required

Solve for AB in terms of a, b and angle C

Considering right angled triangle BOC where O is the point between b-x and x

From BOC, we have that:

$Sin\ C = \frac{h}{a}$

Make h the subject:

$h = aSin\ C$

Also, in BOC (Using Pythagoras)

$a^2 = h^2 + x^2$

Make $x^2$ the subject

$x^2 = a^2 - h^2$

Substitute $aSin\ C$ for h

$x^2 = a^2 - h^2$ becomes

$x^2 = a^2 - (aSin\ C)^2$

$x^2 = a^2 - a^2Sin^2\ C$

Factorize

$x^2 = a^2 (1 - Sin^2\ C)$

In trigonometry:

$Cos^2C = 1-Sin^2C$

So, we have that:

$x^2 = a^2 Cos^2\ C$

Take square roots of both sides

$x= aCos\ C$

In triangle BOA, applying Pythagoras theorem, we have that:

$AB^2 = h^2 + (b-x)^2$

Open bracket

$AB^2 = h^2 + b^2-2bx+x^2$

Substitute $x= aCos\ C$ and $h = aSin\ C$ in $AB^2 = h^2 + b^2-2bx+x^2$

$AB^2 = h^2 + b^2-2bx+x^2$

$AB^2 = (aSin\ C)^2 + b^2-2b(aCos\ C)+(aCos\ C)^2$

Open Bracket

$AB^2 = a^2Sin^2\ C + b^2-2abCos\ C+a^2Cos^2\ C$

Reorder

$AB^2 = a^2Sin^2\ C +a^2Cos^2\ C + b^2-2abCos\ C$

Factorize:

$AB^2 = a^2(Sin^2\ C +Cos^2\ C) + b^2-2abCos\ C$

In trigonometry:

$Sin^2C + Cos^2 = 1$

So, we have that:

$AB^2 = a^2 * 1 + b^2-2abCos\ C$

$AB^2 = a^2 + b^2-2abCos\ C$

Take square roots of both sides

$AB = \sqrt{a^2 + b^2-2abCos\ C}$

6 0

3 years ago