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4 years ago

A caterer prepares three times as many pizzas as she usually prepares for a large party.

Mathematics

2 answers:

VMariaS [17]4 years ago

8 0

Answer: There are 45 party guests who will be served pizzas.

Step-by-step explanation:

Since we have given that

Number of pizzas usually prepare by the caterer = 5

Part of pizza will eat by each party guest is given by

$\frac{1}{3}$

As we have given that a caterer prepares 3 times as many pizzas as she usually prepares for a large party,

So, it becomes

$\text{Number of pizzas prepare by carterer this time}=5\times 3=15\\$

Let the number of party guests be x

According to question, it becomes,

Number of party guests will get the pizza serve is given by

$\frac{1}{3}\times x=15\\\\x=15\times 3\\\\x=45$

Hence, there are 45 party guests who will be served pizzas.

anastassius [24]4 years ago

8 0

3 x 5 pizzas = 15 pizzas
1/3 of pizzas x people = 15 pizzas
people = 15 pizzas / (1/3) = 45

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Which symbol correctly compares the two measurementsone pound 16 oz

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Answer:

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Step-by-step explanation:

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3 years ago

A local theater charges $7.50 for adult tickets and $5.00 for discount tickets. The theater needs to

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3 years ago

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Subtract. (2/9v)−(−1/3v+5/9) A. 5/9v+5/9 B. 1/9v+5/9 C. 5/9v−1/3 D. 5/9v−5/9

Alexeev081 [22]

D. (5/9v - 5/9)

Step-by-step explanation:

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4 years ago

(1/1+sintheta)=sec^2theta-secthetatantheta pls help me verify this

Xelga [282]

Answer:

See Below.

Step-by-step explanation:

We want to verify the equation:

$\displaystyle \frac{1}{1+\sin\theta} = \sec^2\theta - \sec\theta \tan\theta$

To start, we can multiply the fraction by (1 - sin(θ)). This yields:

$\displaystyle \frac{1}{1+\sin\theta}\left(\frac{1-\sin\theta}{1-\sin\theta}\right) = \sec^2\theta - \sec\theta \tan\theta$

Simplify. The denominator uses the difference of two squares pattern:

$\displaystyle \frac{1-\sin\theta}{\underbrace{1-\sin^2\theta}_{(a+b)(a-b)=a^2-b^2}} = \sec^2\theta - \sec\theta \tan\theta$

Recall that sin²(θ) + cos²(θ) = 1. Hence, cos²(θ) = 1 - sin²(θ). Substitute:

$\displaystyle \displaystyle \frac{1-\sin\theta}{\cos^2\theta} = \sec^2\theta - \sec\theta \tan\theta$

Split into two separate fractions:

$\displaystyle \frac{1}{\cos^2\theta} -\frac{\sin\theta}{\cos^2\theta} = \sec^2\theta - \sec\theta\tan\theta$

Rewrite the two fractions:

$\displaystyle \left(\frac{1}{\cos\theta}\right)^2-\frac{\sin\theta}{\cos\theta}\cdot \frac{1}{\cos\theta}=\sec^2\theta - \sec\theta \tan\theta$

By definition, 1 / cos(θ) = sec(θ) and sin(θ)/cos(θ) = tan(θ). Hence:

$\displaystyle \sec^2\theta - \sec\theta\tan\theta \stackrel{\checkmark}{=} \sec^2\theta - \sec\theta\tan\theta$

Hence verified.

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