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goldenfox [79]
3 years ago
10

Can someone help me figure this out

Mathematics
1 answer:
Setler79 [48]3 years ago
5 0

Answer:

<h2>See the explanation.</h2>

Step-by-step explanation:

It is told that, 330 students that are surveyed, are 9th graders.

Hence, 0.55 represents 330 students, that is 1 represents \frac{330}{0.55} = 600 students.

The number of 9th grade students whose favorite elective is music is represented by 0.18.

1 ≡ 600

0.18 ≡600\times0.18 = 108.

The percentage of students whose favorite elective is home economics and are also in 10th grade is \frac{0.06}{1} \times100 = 6%.

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Variables on both sides<br> 4x + 9 = 2x + 13
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There are a total of 15 bicycles and tricycles.There are 37 wheels all together if we count them up. How many tricycles and how
m_a_m_a [10]

Answer: there are eight bicycles and 7 tricycles.

Step-by-step explanation:

Let x represent the number of bicycles that are there.

Let y represent the number of tricycles that are there.

There are a total of 15 bicycles and tricycles. This means that

x + y = 15

A bicycle has 2 wheels and a tricycle has 3 wheels. There are 37 wheels all together if we count them up. This means that

2x + 3y = 37- - - - - - - - - - - - - 1

Substituting x = 15 - y into equation 1, it becomes

2(15 - y) + 3y = 37

30 - 2y + 3y = 37

- 2y + 3y = 37 - 30

y = 7

x = 15 - y = 15 - 7

x = 8

7 0
3 years ago
How to find (c,d,e)<br>Please do a step by step working.Thank you.
Natasha2012 [34]
(a)
The inverse is when you swap the variables and solve for y.
g(t) = 2t - 1 (Note: g(t) represents y)
rewrite as: y = 2t - 1
swap the variables: t = 2y - 1
solve for y: t + 1 = 2y
                   \frac{t + 1}{2} = y
Answer for (a): g^{-1}(t) =  \frac{t + 1}{2}

(b)
Same steps as part (a) above:
h(t) = 4t + 3
rewrite as: y = 4t + 3
swap the variables: t = 4y + 3
solve for y: y =\frac{t - 3}{4}

Answer for (b): h^{-1}(t) = \frac{t - 3}{4}

(c)
g^{-1} ( h^{-1}(t)) =  g^{-1} (\frac{t - 3}{4})
replace all t's in the g^{-1}(t) equation with \frac{t - 3}{4}
 g^{-1} (\frac{t - 3}{4}) = \frac{ \frac{t-3}{4} + 1}{2}
= \frac{ \frac{t-3}{4} +  \frac{4}{4}}{2} = \frac{ \frac{t - 3 + 4}{4}}{2} = \frac{ \frac{t + 1}{4}}{2} =  \frac{t + 1}{8}
Answer for (c): g^{-1} ( h^{-1}(t)) = \frac{t + 1}{8}

 (d)
h(g(t)) = h(2t - 1) = 4(2t - 1) + 3 = 8t - 4 + 3 = 8t - 1
Answer for (d): h(g(t)) = 8t - 1

(e)
h(g(t)) = 8t - 1
   y = 8 t - 1
   t = 8y - 1
  t + 1 = 8y
\frac{t + 1}{8} = y
Answer for (e): inverse of h(g(t)) = \frac{t + 1}{8}
 
























8 0
3 years ago
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