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3 years ago

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Jake sold 42 tickets to the school fair and Jeanne sold 9 tickets. What is the ratio, in simplest form, of the number of tickets

Jeanne sold to the number of tickets Jake sold?

2 answers:

MrRissso [65]3 years ago

7 0

Answer:

Step-by-step explanation:

10

i think

nekit [7.7K]3 years ago

5 0

Answer:

3/14

Step-by-step explanation:

Jake sold 42 tickets while Jeanne sold 9 tickets. The questions is asking for the ratio in simplest form the number of tickets Jeanne sold to the number Jake sold. Since we know the amount they sold, let's put it into a ratio factor by Jeannes to Jakes.

9 : 42

To find the simplest form, we need to divide both numbers by their GCF (Greatest Common Factor), which would be 3.

$\frac{9}{3} : \frac{42}{3}$

3 /14

This means that the ratio from Jeanne's sakes to Jake's sales in simplest form is 3/14

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2 years ago

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Step-by-step explanation:

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3 years ago

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myrzilka [38]

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3 years ago

Read 2 more answers

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

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

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3 years ago

Which equation will solve the following word problem? One small pitcher and one large pitcher can hold 12 cups of water. If the

Inessa [10]

Answer:

Option (1).

Step-by-step explanation:

Amount of water hold by both the pitchers (Large + small) = 12 cups

Let the small pitcher holds the water = S cups

If the water hold by the large pitcher is twice the size of the small pitcher, then the amount of water hold by large pitcher = 2S

Total amount water hold by both = S + 2S

Equation for this situation will be,

S + 2S = 12

Therefore, Option (1) will be the answer.

7 0

3 years ago