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zhuklara [117]
3 years ago
15

Mrs. Perry is taking a group of students to the Clark Planetarium where tickets cost $7 each. The bus will cost $50 and she has

a budget of $200. How many people can attend under these constraints?​
Mathematics
1 answer:
kenny6666 [7]3 years ago
7 0
Answer:
71 people can attend under these constraints

Explanation:

So you write the slope intercept form equation which is 7x+50=200
You got 7x because every tickets cost 7
You add 50 because of the bus
And you’re wondering how many people are able to go by just spending about $200

So you subtract 50 on both side and you are left with 7x=150 so you have to divide 7 on both side next and you get about 71.4... so basically just round it up to 71 people
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I need help please and thank you
dem82 [27]
So we see that the ratio between the two octagons is 7:1, since 28/4=7 so what we do next is multiply the values of the smaller octagon by 7. But that’s the long way. There’s actually a shortcut by multiplying the perimeter of the smaller octagon, 34, by 7. This in turn equals 238.
6 0
3 years ago
Read 2 more answers
A jar containing only nickels and dimes contains a total of 60 coins. The value of all the coins in the jar is $4.45. Solve by e
borishaifa [10]

Answer: 31 nickels and 29 dimes

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

Nickels (.05): x

Dimes (.10): y


 Value:     .05x + .10y = 4.45 →  -20(.05x + .10y = 4.45)   →   -x - 2y = -89

Quantity:        x  +    y  = 60   →       1(x  +    y  = 60)           →   <u> x  + y </u> = <u> 60 </u>

                                                                                                       -y  = -29

                                                                                                        y  =  29

Next, substitute "29" for "y" into either equation and solve for "x":

x +  y  = 60

x + 29 = 60

x          = 31


5 0
3 years ago
Use the discriminant of the following quadratic equation to state the number and type of solutions. 3x^(2)+4x-5=3x+2
krek1111 [17]
3x^(2)+4x-3x-5-2=0
3x^(2)+x-7=0
Discriminat= b^(2)-4ac= 1^(2)-4(3)(-7)= 1-12(-7)=1+84=85
The discrimination is bigger than 0 (delta>0) therefore there are 2 real solutions
4 0
3 years ago
It took my question down twice here's a real math question I guess
aliina [53]

Answer:

n-3

Interval notation: (-\infty, -10)\cup(-3,\infty)

Step-by-step explanation:

<u>First inequality:</u>

<u />n+8

Therefore, this inequality restricts:

n \in \mathrm{R};\: n

<u>Second inequality:</u>

< 8+n-3

Therefore, this inequality restricts:

n \in \mathrm{R};\: n>-3

Therefore, with both of these restrictions together, we have:

\fbox{$n \in \mathrm{R}; n-3$}\\\mathrm{or\:}\fbox{$n-3$}\\\mathrm{or\:}\fbox{$(-\infty, -10)\cup(-3,\infty)$}.

4 0
3 years ago
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Marina CMI [18]
I dont know the answer sorry am a biginner
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