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3 years ago

8

Harry and Jenny are going to the movies. They each bought a medium popcorn, and Harry got a small soft drink. Jenny had a $4 cou

pon as a discount, and Harry paid the rest, which came to $23. A movie ticket costs $8, and a medium popcorn costs $4.50. What was the cost for a soft drink?

2 answers:

Rudik [331]3 years ago

8 0

Answer:

2$

Step-by-step explanation:

2 medium popcorns = 9$

2 Movie tickets = 16$

One coupon = -4$

This is equal to the cost without the soft drink(21$)

23$ - 21$ = 2$

Alisiya [41]3 years ago

7 0

Answer:

$2

Step-by-step explanation:

It was originally $27 but, Jenny used a $4 coupon. Each movie ticket costs $8, each popcorn cost $4.50.

1. Start from the cost of each item.

8+8=16 (cost of the tickets)
4.50+4.50=9 (cost of the popcorn) .

2. Add the totals together.

16+9+ 25

Now subtract 25 from the original amount.

27-25=2

It costs 2 dollars for a soft drink.

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4 years ago

Based on the information given say whether or not △ABC∼△FED. Explain your reasoning.

GalinKa [24]

Answer:

Yes, △ABC ∼ △FED by AA postulate.

Step-by-step explanation:

Given:

Two triangles ABC and FED.

m∠A = m∠B

m∠C = m∠A + 30°

m∠E = m∠F = $x$

m∠D = $2x-20$ °.

Now, let m∠A = m∠B = $y$

So, m∠C = m∠A + 30° = $y+30$

Now, sum of all interior angles of a triangle is 180°. Therefore,

m∠A + m∠B + m∠C = 180

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Therefore, m∠A = 50°, m∠B = 50° and m∠C = m∠A + 30° = 50 + 30 = 80°.

Now, consider triangle FED,

m∠D+ m∠E + m∠F = 180

$2x-20+x+x=180\\4x=180+20\\4x=200\\x=\frac{200}{4}=50$

Therefore, m∠F = 50°

m∠E = 50° and

m∠D = $2x-20=2(50)-20=100-20=80\°$

So, both the triangles have congruent corresponding angle measures.

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3 years ago

Write out the first four terms of the series to show how the series starts. Then find the sum of the series or show that it dive

Nostrana [21]

Answer:

The first four terms of the series are

$(9+3),(\frac97+\frac35),(\frac9{7^2}+\frac3{5^2}),(\frac9{7^3}+\frac3{5^3})$

$\sum_{n=0}^\infty \frac9{7^n}+\frac{3}{5^n}$ = 14.25

Step-by-step explanation:

We know that

Sum of convergent series is also a convergent series.

We know that,

$\sum_{k=0}^\infty a(r)^k$

If the common ratio of a sequence |r| <1 then it is a convergent series.

The sum of the series is $\sum_{k=0}^\infty a(r)^k=\frac{a}{1-r}$

Given series,

$\sum_{n=0}^\infty \frac9{7^n}+\frac{3}{5^n}$

$=(9+3)+(\frac97+\frac35)+(\frac9{7^2}+\frac3{5^2})+(\frac9{7^3}+\frac3{5^3})+.......$

The first four terms of the series are

$(9+3),(\frac97+\frac35),(\frac9{7^2}+\frac3{5^2}),(\frac9{7^3}+\frac3{5^3})$

Let

$S_n=\sum_{n=0}^\infty \frac{9}{7^n}$ and $t_n=\sum_{n=0}^\infty \frac{3}{5^n}$

Now for $S_n$ ,

$S_n=9+\frac97+\frac{9}{7^2}+\frac9{7^3}+.......$

$=\sum_{n=0}^\infty9(\frac 17)^n$

It is a geometric series.

The common ratio of $S_n$ is $\frac17$

The sum of the series

$S_n=\sum_{n=0}^\infty \frac{9}{7^n}$

$=\frac{9}{1-\frac17}$

$=\frac{9}{\frac67}$

$=\frac{9\times 7}{6}$

=10.5

Now for $t_n$

$t_n= 3+\frac35+\frac{3}{5^2}+\frac3{5^3}+.......$

$=\sum_{n=0}^\infty3(\frac 15)^n$

It is a geometric series.

The common ratio of $t_n$ is $\frac15$

The sum of the series

$t_n=\sum_{n=0}^\infty \frac{3}{5^n}$

$=\frac{3}{1-\frac15}$

$=\frac{3}{\frac45}$

$=\frac{3\times 5}{4}$

=3.75

The sum of the series is $\sum_{n=0}^\infty \frac9{7^n}+\frac{3}{5^n}$

= $S_n+t_n$

=10.5+3.75

=14.25

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4 years ago

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Answer:

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Step-by-step explanation:

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3 years ago