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3 years ago

Which One Doesn’t Belong: Comparing Speeds

Mathematics

1 answer:

TEA [102]3 years ago

4 0

Answer:

I think 3 minutes per mile. but dont quote me on it.

Step-by-step explanation: I think this because 3 minutes aint per mile.

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BRAINLIEST BRAINLIEST BRAINLIEST PLEASE ANDSER IT IS TIMED THANK YOU SO MUCCH

Doss [256]

Answer:

x=-b/2a

Step-by-step explanation:

hope that helps :)

mark as brainliest please :)

8 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

The sum of two numbers is 28. one number is 14 more. find the numbers

ruslelena [56]

The numbers are: "7 " and "21 " .
_______________________________________
Explanation:
_______________________________________
The numbers are: "x" and "x + 14" .

x + (x + 14) = 28 . Solve for "x" ; and then solve for "x + 14" .
_______________________________________
→ x + (x + 14) = 28 ;

Rewrite as:

→ x + x + 14 = 28 ;

→ 1x + 1x + 14 = 28 ;

→ 2x + 14 = 28 ;

Subtract "14" from each side of the equation;

→ 2x + 14 − 14 = 28 <span>− 14 ;

</span> → 2x = 14 ;

Divide EACH SIDE of the equation by "2" ;
to isolate "x" on one side of the equation; & to solve for "x" ;

→ 2x / 2 = 14 / 2 ;

→ x = 7 .
_________________________________________
So; one of the numbers is: " 7 " .

The other number is: " x + 14 " ; which equals: " 7 + 14 = 21".

The other number is: "21 " .
_________________________________________
The numbers are: "7 " and "21 " .
_________________________________________

5 0

3 years ago

Alex applies 350 n of force to move his stalled car 40 m how much work did alex do

Katyanochek1 [597]

Step-by-step explanation:

Work done = Force × displacement

= 350 × 40

=14000J

7 0

3 years ago

Convert to scientific notation: 0.0509.

fomenos

Answer:

5.09 × 10^-2

Step-by-step explanation:

You have to take the number go to the decimal and move it until you get to the first number and in this case, it is 5 put the decimal there. To write it you have to count the number of times you moved the decimal. when doing this it will always be a number times 10^ the number you moved the decimal.

5 0

3 years ago