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3 years ago

5

at the movie theater chid admission is 5.70 and adult admission is 9.90 on Sunday 154 tickets were sold for a total sales price

of 1125.60 how many child tickets were sold that day

1 answer:

saw5 [17]3 years ago

6 0

Answer:

94 Child tickets were sold.

Step-by-step explanation:

Let x represent child tickets

Let y represent adult tickets

Cost Equation: 5.70x + 9.75y = 1306.10

Quantity Equation: x + y = 173

Multiply the Quantity Equation by -5.70

5.70x + 9.75y = 1306.10

-5.70x - 5.70y = -986.10

4.05y = 320.00

÷4.05 ÷4.05

y = 79

x + y = 173

x + 79 = 173

x = 94

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Black_prince [1.1K]

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Step-by-step explanation:

the triangles are in AA similarity.

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3 years ago

There are 34 blocks in each box how many blocks are in 54 boxes

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4 years ago

Read 2 more answers

State the number of possible triangles that can be formed using the given measurements.

romanna [79]

Answer: 39) 1 40) 2

41) 1 42) 0

<u>Step-by-step explanation:</u>

39) ∠A = ? ∠B = ? ∠C = 129°

a = ? b = 15 c = 45

Use Law of Sines to find ∠B:

$\dfrac{\sin B}{b}=\dfrac{\sin C}{c} \rightarrow\quad \dfrac{\sin B}{15}=\dfrac{\sin 129}{45}\rightarrow \quad \angle B=15^o\quad or \quad \angle B=165^o$

If ∠B = 15°, then ∠A = 180° - (15° + 129°) = 36°

If ∠B = 165°, then ∠A = 180° - (165° + 129°) = -114°

Since ∠A cannot be negative then ∠B ≠ 165°

∠A = 36° ∠B = 15° ∠C = 129° is the only valid solution.

40) ∠A = 16° ∠B = ? ∠C = ?

a = 15 b = ? c = 19

Use Law of Sines to find ∠C:

$\dfrac{\sin A}{a}=\dfrac{\sin C}{c} \rightarrow\quad \dfrac{\sin 16}{15}=\dfrac{\sin C}{19}\rightarrow \quad \angle C=20^o\quad or \quad \angle C=160^o$

If ∠C = 20°, then ∠B = 180° - (16° + 20°) = 144°

If ∠C = 160°, then ∠B = 180° - (16° + 160°) = 4°

Both result with ∠B as a positive number so both are valid solutions.

Solution 1: ∠A = 16° ∠B = 144° ∠C = 20°

Solution 2: ∠A = 16° ∠B = 4° ∠C = 160°

41) ∠A = ? ∠B = 75° ∠C = ?

a = 7 b = 30 c = ?

Use Law of Sines to find ∠A:

$\dfrac{\sin A}{a}=\dfrac{\sin B}{b} \rightarrow\quad \dfrac{\sin A}{7}=\dfrac{\sin 75}{30}\rightarrow \quad \angle A=13^o\quad or \quad \angle A=167^o$

If ∠A = 13°, then ∠C = 180° - (13° + 75°) = 92°

If ∠A = 167°, then ∠C = 180° - (167° + 75°) = -62°

Since ∠C cannot be negative then ∠A ≠ 167°

∠A = 13° ∠B = 75° ∠C = 92° is the only valid solution.

42) ∠A = ? ∠B = 119° ∠C = ?

a = 34 b = 34 c = ?

Use Law of Sines to find ∠A:

$\dfrac{\sin A}{a}=\dfrac{\sin B}{b} \rightarrow\quad \dfrac{\sin A}{34}=\dfrac{\sin 119}{34}\rightarrow \quad \angle A=61^o\quad or \quad \angle A=119^o$

If ∠A = 61°, then ∠C = 180° - (61° + 119°) = 0°

If ∠A = 119°, then ∠C = 180° - (119° + 119°) = -58°

Since ∠C cannot be zero or negative then ∠A ≠ 61° and ∠A ≠ 119°

There are no valid solutions.

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3 years ago

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3 years ago

Find an equation of the tangent to the curve at the given point by both eliminating the parameter and without eliminating the pa

konstantin123 [22]

Answer:

$y=2x-8$

Step-by-step explanation:

The given parametric equation is;

$x=6+ln(t),y=t^2+3$

<h3><u>BY ELIMINATING THE PARAMETER</u></h3>

To eliminate the parameter we make $t$ the subject in one equation and put it inside the other.

We make $t$ the subject in $x=6+ln(t)$ because it is easier.

$\Rightarrow x-6=ln(t)$

$\Rightarrow {e}^{x-6}=e^{ln(t)}$

$\Rightarrow {e}^{x-6}=t$

Or

$t={e}^{x-6}$

We now substitute this into $y=t^2+3$ .

This gives us;

$y=(e^{x-6})^2+3$ .

$\Rightarrow y=e^{2(x-6)}+3$ .

We have now eliminated the parameter.

The equation of the tangent at (6,4) is given by;

$y-y_1=m(x-x_1)$

where the gradient function is given by;

$\frac{dy}{dx}=2e^{2(x-6)}$

We substitute $x=6$ into the gradient function to obtain the gradient.

$\Rightarrow m=2e^{2(6-6)}$

$\Rightarrow m=2e^0$

$\Rightarrow m=2$

The equation of the tangent becomes

$y-4=2(x-6)$

We simplify to obtain

$y=2x-12+4$

$y=2x-8$

<h3><u>WITHOUT ELIMINATING THE PARAMETER</u></h3>

The given parametric equation is;

$x=6+ln(t),y=t^2+3$

For $x=6+ln(t)$

$\frac{dx}{dt}=\frac{1}{t}$

For $y=t^2+3$

$\frac{dy}{dt}=2t$

The slope is given by;

$\frac{dy}{dx}=\frac{\frac{dy}{dt} }{\frac{dx}{dt} }$

$\frac{dy}{dx}=\frac{2t }{\frac{1}{t} }$

$\frac{dy}{dx}=2t^2$

At the point, (6,4), we plug in any of the values into the parametric equation and find the corresponding value for $t$ .

Notice that

When $x=6$ , $6=6+\ln(t)$

$6-6=\ln(t)$

$0=\ln(t)$

$e^0=e^\ln(t)$

$1=t$

when $y=4$ , $4=t^2+3$

$4-3=t^2$

$1=t^2$

$t=\pm1$

But the slope is the same when we plug in any of these values for t.

$\frac{dy}{dx}=2(\pm1)^2=2$

The equation of the tangent becomes

$y-4=2(x-6)$

We simplify to obtain

$y=2x-12+4$

$y=2x-8$

7 0

3 years ago