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2 years ago

How many child tickets were sold that day?

Mathematics

1 answer:

ExtremeBDS [4]2 years ago

3 0

$\huge\boxed{\fcolorbox{black}{red}{Answer}}$

first let's name a couples of variable

• the number of adults tickets sold: a

• the number of children tickets sold: c

From the problem we know

a + c = 128

and

$5.40c + $9.20a = $976.20

1) solve the equation to alpha

a+c-c = 128 -c

a+0=128-c

a=128-c

2) substitute (128 - c) for a in the second equation and solve to c

$5.40c + $9.20a = $976.20 become

$5.40c + $9.20(128 - c) = $976.20

$5.40c + ($9.20 × 128) - ($9.20 - c) = $976.20

$5.40c - $9.20c + $ 1177.6 = $976.20

($5.40 - $9.20)c +$1177.6 = $976.20

-$3.80c + $1177.6 = $9.76.20

-$3.80c + $1177.60 - $1177.60 = $976.20 - $1177.60

-$8.30c + 0 = $201.40

-$3.80c = - $201.40

-$3.80c. -$201.40

________. = _________

-$3.80. -$3.80

-$3.80c. -$201.40

________. = _________. - they are 4 cut the no

-$3.80. -$3.80

c = $201.40

________

3.80

c = 53

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2.the solution set of x +y =5 and x-y =3

Nataliya [291]

Answer:

Check the solution below

Step-by-step explanation:

2) Given the equation

x +y =5... 1 and

x-y =3 ... 2

Add both equations

x+x = 5+3

2x = 8

x = 8/2

x = 4

Substitute x = 4 into 1:

From 1: x+y = 5

4+y= 5

y = 5-4

y = 1

3) Given

x+3y =15 ... 1

2x+7y=19 .... 2

From 2: x = 15-3y

Substitute into 2

2(15-3y)+7y = 19

30-6y+7y = 19

30+y = 19

y = 19-30

y = -11

Substitute y=-11 into x = 15-3y

x =15-3(-11)

x = 15+33

x = 48

The solution set is (48, -11)

4) given

x/2 +y/3 =0 and x+2y=1

From 1

(3x+2y)/6 = 0

3x+2y = 0.. 3

x+2y= 1... 4

From 4: x = 1-2y

Substutute

3(1-2y) +2y = 0

3-6y+2y = 0

3 -4y = 0

4y = 3

y = 3/4

Since x = 1-2y

x = 1-2(3/4)

x = 1-3/2

x= -1/2

The solution set is (-1/2, 3/4)

5) Given

5.x=1/2 and y =x +1 then solution is

We already know the vkue of x

Get y

y= x+1

y = 1/2 + 1

y = 3/2

Hence the solution set is (1/2, 3/2)

6) Given

3x +y =5 and x -3y =5

From 3; x = 5+3y

Substitute into 1;

3(5+3y)+y = 5

15+9y+y = 5

10y = 5-15

10y =-10

y = -1

Get x;

x = 5+3y

x = 5+3(-1)

x = 5-3

x = 2

Hence two solution set is (2,-1)

6 0

2 years ago

Graph y > -1/3x+5<br><img src="https://tex.z-dn.net/?f=y%20%3E%20%20-%20%20%5Cfrac%7B1%7D%7B%203%20%7D%20x%20%2B%205" id="Tex

ad-work [718]

Explanation:

The function is $y>-\frac{1}{3} x+5$

To graph the function, let us find the x and y intercepts.

To find x-intercept, let us substitute y=0 in the function $y>-\frac{1}{3} x+5$

$\begin{aligned}0 &=-\frac{1}{3} x+5 \\-5 &=-\frac{1}{3} x \\x &=15\end{aligned}$

Thus, the x-intercept is $(15,0)$

To find the y-intercept, let us substitute x=0, we get,

$\begin{aligned}&y=-\frac{1}{3}(0)+5\\&y=5\end{aligned}$

Thus, the y-intercept is $(0,5)$

The graph has no asymptotes.

To plot the points in the graph, we need to substitute the values for x in the function $y>-\frac{1}{3} x+5$ , to find the y-values.

The points are $(-2,5.667),(-1,5.333),(1,4.667),(2,4.333),(3,4)$ . The image of the graph and table is attached below:

6 0

2 years ago

X intercept for 2x +5y = -6

Vlad1618 [11]

2x + 5y = -6
-5y -5y
___________
2x = -11y
__ ___
2 2
x = 5.5

8 0

3 years ago

Read 2 more answers

4 / 7 * 2 / 3 * *9/16

lara31 [8.8K]

$\bf \cfrac{4}{7}\cdot \cfrac{2}{3}\cdot \cfrac{9}{16}\implies \cfrac{4}{16}\cdot \cfrac{9}{3}\cdot \cfrac{2}{7}\implies \cfrac{1}{4}\cdot \cfrac{3}{1}\cdot \cfrac{2}{7}\implies \cfrac{1}{7}\cdot \cfrac{3}{1}\cdot \cfrac{2}{4}\implies \cfrac{1}{7}\cdot \cfrac{3}{1}\cdot \cfrac{1}{2}\\\\\\\cfrac{1\cdot 3\cdot 1}{7\cdot 1\cdot 2}\implies \cfrac{3}{14}$

4 0

3 years ago

Need help on all these questions!!

Yuki888 [10]

Okay, so YZ = 3 cm. You have XM correct. And YM = 0.5.

Now, you have the midpoint M at the correct spot.

Use Pythagorean's theorem o find the length of AB. a² + b² = c² a=6, b=8.

6² = 36 8² = 64 36 + 64 = 100 AB = 10!

If AB = 10 then AM = 5 MB also = 5

If B is the midpoint of AC, C would be 12 rows down from A, and 16 columns to the right. The last spot where the line intersects.

There are your answers!

5 0

2 years ago