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3 years ago

Look at UVZ and WVZ in the image below.

Mathematics

1 answer:

baherus [9]3 years ago

4 0

Answer:

A Supplementery

Step-by-step explanation:

they show on usa test prep

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Choose yes or no to tell whether the expression represent a 20% discount off the price of an item that originally cost d dollars

Vaselesa [24]

Answer: B. d - 0.2

Step-by-step explanation: 20%= 20/100= 0.2

Which means = d-0.2

Hope this helped.

8 0

3 years ago

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The measure of an angle is 140°. What is the measure of its supplementary angle

Darina [25.2K]

Answer:

40 degrees

Step-by-step explanation:

Supplementary angles are angles in which their measures combined add up to 180 degrees. So we can set up an equation:

140+x=180

x=40

So the other supplementary angle is 40 degrees

7 0

3 years ago

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f. A fair coin is thrown in the air four times. If the coin lands with the head up on the first three tosses, what is the probab

vladimir2022 [97]

Answer:

D. 1/2

Step-by-step explanation:

Coin tosses are independent. Past results don't affect future probabilities. So the probability of getting heads on the fourth toss is still 1/2.

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3 years ago

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The Morris family bought 3 crates of eggs last week. Each crate had 18 eggs. Since then, they have eaten 25 of the eggs. How man

zzz [600]

Answer:

Step-by-step explanation:

3 x 8 = 54 - 25 = 29

5 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