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3 years ago

There are 32 female performers in a dance recital.The ratio of men to women in 3:8 how many men are in the dance recital

Mathematics

1 answer:

Levart [38]3 years ago

3 0

We have an equation: ?/32= 3/8

Cross multiply:
8*?= 3*32
⇒ ?= 3*32/8= 12

There are 12 men in the dance recital~

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Given the information in the diagram below, find the measure of angle N. Use RACE and complete sentences to show how you found t

Sonja [21]

Answer:

m<N = 76°

Step-by-step explanation:

Given:

∆JKL and ∆MNL are isosceles ∆ (isosceles ∆ has 2 equal sides).

m<J = 64° (given)

Required:

m<N

SOLUTION:

m<K = m<J (base angles of an isosceles ∆ are equal)

m<K = 64° (Substitution)

m<K + m<J + m<JLK = 180° (sum of ∆)

64° + 64° + m<JLK = 180° (substitution)

128° + m<JLK = 180°

subtract 128 from each side

m<JLK = 180° - 128°

m<JLK = 52°

In isosceles ∆MNL, m<MLN and <M are base angles of the ∆. Therefore, they are of equal measure.

Thus:

m<MLN = m<JKL (vertical angles are congruent)

m<MLN = 52°

m<M = m<MLN (base angles of isosceles ∆MNL)

m<M = 52° (substitution)

m<N + m<M° + m<MLN = 180° (Sum of ∆)

m<N + 52° + 52° = 180° (Substitution)

m<N + 104° = 180°

subtract 104 from each side

m<N = 180° - 104°

m<N = 76°

4 0

3 years ago

A collection of 20 coins made up of only nickels, dimes, and quarters has a total value of $3.35. If the dimes were nickels, the

Kay [80]

Answer: The required number of quarters in the collection is 11.

Step-by-step explanation: Given that a collection of 20 coins made up of only nickels, dimes and quarters has a total value of $3.35.

If the dimes were nickels, the nickels were quarters and the quarters were dimes, the collection of coins would have a total value of $2.75.

We are to find the number of quarters in the collection.

Let x, y and z represents the number of nickels, dimes and quarters respectively in the collection.

We will be using the following values of nickels, dimes and quarters in form of dollar :

1 nickel = $ 0.05, 1 dime = $ 0.10 and 1 quarter = $0.25.

Then, according to the given information, we have

$x+y+z=20\\\\\Rightarrow x=20-y-z~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~(i)\\\\0.05x+0.10y+0.25z=3.35\\\\\Rightarrow 5x+10y+25z=335\\\\\Rightarrow x+2y+5z=67~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~(ii)\\\\0.05y+0.25x+0.10z=2.75\\\\\Rightarrow 5y+25x+10z=275\\\\\Rightarrow y+2z+5x=55~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~(iii)$

Substituting the value of x from equation (i) in equations (ii) and (iii), we have

$(20-y-z)+2y+5z=67\\\\\Rightarrow y+4z=47\\\\\Rightarrow y=47-4z~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~(iv)$

and

$y+2z+5(20-y-z)=55\\\\\Rightarrow -4y-3z=-45\\\\\Rightarrow y=\dfrac{45-3z}{4}~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~(v)$

Comparing the values of y from equations (iv) and (v), we get

$47-4z=\dfrac{45-3z}{4}\\\\\Rightarrow 188-16z=45-3z\\\\\Rightarrow 16z-3z=188-45\\\\\Rightarrow 13z=143\\\\\Rightarrow z=\dfrac{143}{13}\\\\\Rightarrow z=11.$