All categories

4 years ago

PLZ ANSWER would be very helpful :)

Mathematics

1 answer:

Natalija [7]4 years ago

7 0

Https://photomath.net/en/

You might be interested in

I. There is a gap in the Potato Chips group

topjm [15]

Answer:

C) I and II

There is a gap in the Potato Chips group for the stem "14" and the Tortilla Chips group for the stem "12"

4 0

3 years ago

Prove that.........

tresset_1 [31]

Answer:

becauz the cos .1sj sin A the thing cos

8 0

4 years ago

There are 3 consecutive numbers whose sum is equal to the result of their multiplication. Which are they?

OleMash [197]

They are 1, 2 and 3.

Hope that helped ^^

7 0

3 years ago

Find the distance between the two points rounding to the nearest tenth (if

lana [24]

Answer:

5.7

Step-by-step explanation:

Creating the point into a triangle, both sides a 4.

As the Pythagorean theorem goes, a^2+b^2=c^2

4^2+4^2=

16+16=

The square root of 32 is about 5.7

The distance between the points is about 5.7 units

6 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