Answer:
92 attendees had activity cards
Step-by-step explanation:
Let x be the number of students with activity cards. Then 130-x is the number without, and the total revenue is ...
7x +10(130 -x) = 1024
7x +1300 -10x = 1024 . . . . eliminate parentheses
-3x = -276 . . . . . . . . . . . . . collect terms; subtract 1300
x = 92 . . . . . . divide by 3
92 students with activity cards attended the dance.
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<em>Comment on the solution</em>
Often, you will see such a problem solved using two equations. For example, they might be ...
Let 'a' represent the number with an activity card; 'w' the number without. Then ...
- a+w = 130 . . . . the total number of students
- 7a +10w = 1024 . . . . the revenue from ticket sales
The problem statement asks for the value of 'a', so you want to eliminate w from these equations. You can do that using substitution. Using the first equation to write an expression for w, you have ...
w = 130-a
and making the substitution into the second equation gives ...
7a +10(130 -a) = 1024
This should look a lot like the equation we used above. There, we skipped the extra variable and went straight to the single equation we needed to solve.
In order to find height from where ball is dropped, you have to find height or h(t) when time or t is zero.So plug in t=0 into your quadratic equation:h(0) = -16.1(0^2) + 150h(0) = 0 +150h(0) = 150 ft is the height from where ball is dropped. When ball hits the ground, the height is zero. So plug in h(t) = 0 and solve for t.0 = -16.1t^2 + 15016.1 t^2 = 150t^2 = 150/16.1t = sqrt(150/16.1)t = ± 3.05Since time cannot be negative, your answer is positive solution i.e. t = 3.05
Answer:
B=-5
Step-by-step explanation:
Slope=rise/run
The line passes in
P1(-1,3)
and
P2(0,-2)
So slope=(3-(-2))/(-1-0)=5/-1=-5
Subtract 1111 from both sides
5{e}^{{4}^{x}}=22-115e4x=22−11
Simplify 22-1122−11 to 1111
5{e}^{{4}^{x}}=115e4x=11
Divide both sides by 55
{e}^{{4}^{x}}=\frac{11}{5}e4x=511
Use Definition of Natural Logarithm: {e}^{y}=xey=x if and only if \ln{x}=ylnx=y
{4}^{x}=\ln{\frac{11}{5}}4x=ln511
: {b}^{a}=xba=x if and only if log_b(x)=alogb(x)=a
x=\log_{4}{\ln{\frac{11}{5}}}x=log4ln511
Use Change of Base Rule: \log_{b}{x}=\frac{\log_{a}{x}}{\log_{a}{b}}logbx=logablogax
x=\frac{\log{\ln{\frac{11}{5}}}}{\log{4}}x=log4logln511
Use Power Rule: \log_{b}{{x}^{c}}=c\log_{b}{x}logbxc=clogbx
\log{4}log4 -> \log{{2}^{2}}log22 -> 2\log{2}2log2
x=\frac{\log{\ln{\frac{11}{5}}}}{2\log{2}}x=2log2
Answer= −0.171
Slope is 3/1 (3 simplified)
