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2 years ago

6

A family of 6 is going to the fair. They have a coupon for $1.50 off each ticket. If they pay for $46.50 for all their tickets,

what equation could be used to solve for how much each ticket cost without the coupon?

2 answers:

Karolina [17]2 years ago

8 0

Answer:

6x=46.50

Step-by-step explanation:

rusak2 [61]2 years ago

3 0

Answer:

(1.50x6)+46.50

Step-by-step explanation:

You might be interested in

8(5021)(3054)+(-8)(-5021)(-3054)+3

kifflom [539]

Answer:

3

Step-by-step explanation:

first two terms cancel eachother out!

5 0

2 years ago

Read 2 more answers

Five swimmers entered a competition. Four of the swimmers have had their turns. Their scores are 10.8s, 10.75s, 10.79s and 10.81

Shalnov [3]

Answer:

10.74s or faster (lower)

Step-by-step explanation:

The lowest time out of all of them is 10.75, so to all the 5th swimmer needs to do to win is to undercut the swimmer who got 10.75s by a 100th of a second. i.e 10.74s or faster (lower) would win.

6 0

2 years ago

Find the measures of the three angles, in radians, of the triangle with the given vertices: d(1,1,1), e(1,−5,2), and f(−2,2,7).

Oduvanchick [21]

Consider triangle DEF with vertices D(1,1,1), E(1,-5,2) and F(-2,2,7).

1. Find

$\overrightarrow{DE}=(1-1,-5-1,2-1)=(0,-6,1),\\ \\\overrightarrow{DF}=(-2-1,2-1,7-1)=(-3,1,6).$

Then

$\cos \angle D=\dfrac{0\cdot (-3)+(-6)\cdot 1+1\cdot 6}{\sqrt{0^2+(-6)^2+1^2}\cdot \sqrt{(-3)^2+1^2+6^2}}=\dfrac{0}{\sqrt{37} \cdot \sqrt{46} }=0.$

2. Find

$\overrightarrow{ED}=(1-1,1-(-5),1-2)=(0,6,-1),\\ \\\overrightarrow{EF}=(-2-1,2-(-5),7-2)=(-3,7,5).$

Then

$\cos \angle E=\dfrac{0\cdot (-2)+6\cdot 7+(-1)\cdot 5}{\sqrt{0^2+6^2+(-1)^2}\cdot \sqrt{(-3)^2+7^2+5^2}}=\dfrac{37}{\sqrt{37} \cdot \sqrt{83} }=\sqrt{\dfrac{37}{83}}.$

3. Find

$\overrightarrow{FE}=(1-(-2),-5-2,2-7)=(3,-7,-5),\\ \\\overrightarrow{FD}=(1-(-2),1-2,1-7)=(3,-1,-6).$

Then

$\cos \angle F=\dfrac{3\cdot 3+(-7)\cdot (-1)+(-5)\cdot (-6)}{\sqrt{3^2+(-7)^2+(-5)^2}\cdot \sqrt{3^2+(-1)^2+(-6)^2}}=\dfrac{46}{\sqrt{83} \cdot \sqrt{46} }=\sqrt{\dfrac{46}{83}}.$

4.

$\angle E=\arccos0=\dfrac{\pi}{2},\\ \\
\angle D=\arccos\letf(\sqrt{\dfrac{37}{83}}\right)\approx 0.27\pi,\\ \\
\angle F=\arccos\letf(\sqrt{\dfrac{46}{83}}\right)\approx 0.23\pi.$

3 0

1 year ago

The map needs to be reproduced at a different scale so that it has an area of 6 square inches and can fit in a brochure at what

Ilia_Sergeevich [38]

Answer:

4 inches

Step-by-step explanation:

The computation of the scale that should be required to reproduced that fitted on the brochure is

Given that

Length of rectangular park = 4 inches

And the Breadth of rectangular park = 6 inches

As we know that

Area of rectangular park = length × breadth

= 4×6

= 24 square inches

Now the scale that required to reproduced is

= 24 ÷ 6

= 4 inches

5 0

2 years ago

HELP PLS>>> The vector (3, - 7) ) is rotated by an angle of (3pi)/4 radians and then reflected across the y-axis. If th

horsena [70]

Answer:

$a \approx -2.826$ , $b \approx 7.072$

Step-by-step explanation:

First, the vector must be transformed into its polar form:

$r = \sqrt{3^{2}+ (-7)^{2}}$

$r \approx 7.616$

$\theta \approx 2\pi - \tan^{-1}\left(\frac{7}{3} \right)$

$\theta \approx 1.629\pi$

Let assume that vector is rotated counterclockwise. The new angle is:

$\theta' = \theta + \frac{3\pi}{4}$

$\theta' = 2.379\pi$

Which is coterminal with $\theta'' = 0.379\pi$ . The reflection across y-axis is:

$\theta''' = \pi - \theta''$

$\theta''' = 0.621\pi$

The equivalent vector in rectangular coordinates is:

$a = 7.616\cdot \cos 0.621\pi$

$a \approx -2.826$

$b = 7.616\cdot \sin 0.621\pi$

$b \approx 7.072$

7 0

2 years ago