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Mrac [35]
3 years ago
13

Adult tickets to space city amusement park cost x dollars. children's tickets cost y dollars. the garcia family bought 3 adult a

nd 1 child ticket for $164. the henson family bought 2 adult and 3 children's tickets for $174
Mathematics
1 answer:
3241004551 [841]3 years ago
8 0
A childs ticket = $28 and an adult = $45 
Total Cost to each family is the number of adult tickets bought multiplied by "x" PLUS the number of child tickets bought multiplied by "y" 
Henson's equation = 3x + y = 163 Garcia's equation = 2x + 3y = 174 
using the substitution method we need to express either x or y in terms of the other variable. In this example looking at the Henson equation it is very easy to express y in terms of x. 
Rewrite the Henson equation to make y the subject y = 163 - 3x Substitute this value (163 - 3x) for "y" in the Garcia equation which now becomes 
2x + 3(163 - 3x) = 174 Expand the bracket 2x + 489 - 9x = 174 -7x + 489 = 174 Add 7x and subtract 174 from both sides of the equation 315 = 7x 315/7 = x 45 = x An adult ticket costs $45 Substitute this back into the Henson equation 3 * 45 + y = 163 135 + y = 163 y = 163 - 135 = 28 A childs ticket costs $28 
Check in the Garcia equation 2 * 45 + 3 * 28 = 90 + 84 = 174 = CORRECT 
A childs ticket = $28 and an adult = $45
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Viefleur [7K]

Answer:

Hello There. ☆~\---___`:•€`___---/~☆ The correct answer is C. Both (1,3) and (-2,-4).

To check whether (-2,-4) is a solution of the equation let's substitute x= -2 and y= -4 into the equation:

-7x+3y=2

-7*(-2)+3*(-4)=2

14-12=2

2=2

Do the same thing with (1,3)

-7x + 3y = 2

-7 * 1 + 3 * 3 = 2

-7 + 9 = 2

2 = 2.

Hope It Helps!~ ♡

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8 0
3 years ago
Find the mass of the triangle with vertices (1; 0; 0), (0; 2; 0), (0; 1; 1) if the density function is given by (x; y; z)
Luda [366]

Answer:

=\frac{17}{3}

Step-by-step explanation:

Density of a function is \rho(x,y)=x^{2} +y^{2}

I have drawn the right angle triangle for visualization

equation for the line passing through (1,0) and (0,4) is

\frac{y}{4} =\frac{x-1}{-1}

y=4-4x

=$$\int_{0}^1\int_{0}^{4-4x} (x^{2}+y^{2})dydx$$

=$\int_{0}^{1}(x^{2}+\frac{y^{3} }{3})dx$  

=$\int_{0}^{1}(x^{2}(4-4x)+\frac{1}{3}(4-4x)^{3} )dx$

=$\int_{0}^{1}(\frac{-76}{3}x^{3} +68x^{2}-64x^{}+\frac{64}{3}  )dx$

==(\frac{-76}{3} \frac{x^{4} }{4})+\frac{68}{3}x^{3}-32x^{2}+\frac{64}{3}x^{})  

=\frac{17}{3}

3 0
3 years ago
Help me solve 1 <2n-5 <_7 also please explain
Naddik [55]
I'm not sure if that is supposed to be 7 or -7 If it's -7: 1<2n-5<-7 +5 +5 +5 Add 5 to all parts of the inequality get 2n by itself 6<2n<-2 /2 /2 /2 Divide all parts by 2 to isolate the variable 3
7 0
3 years ago
Multiply to -260 add to 7
djyliett [7]
What can you be more specific like a put it in a sentence. For example x*-260+7=?

6 0
3 years ago
Y=2(x+1)^2 has how many real roots
oee [108]

The equation y= 2(x+1)^2 has one real root and that is x=-1.

What is real roots of the equation?

    We are aware that when we resolve a linear or quadratic equation, we always arrive at the value variable of the equation, or, to put it another way, we always locate the equation's solution. This "solution" is what we refer to as the real roots. For instance, when the equation X^2-7x+12=0 is solved, the actual roots are 3 and 4.

Here given,

=> y = 2(x+1)^2

Take y=0 then,

=> 2(x+1)^2=0

=> (x+1)^2=0

=>(x+1)=0

=> x=-1

Hence the given equation has one real root and that is x=-1.

To learn more about real roots refer the below link

brainly.com/question/24147137

#SPJ1

8 0
1 year ago
Read 2 more answers
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