Problem 3: Let x = price of bag of pretzels Let y = price of box of granola bars
We have Lesley's purchase: 4x+2y=13.50
And Landon's: 1x+5y=17.55
We can use the elimination method. Let's negate Landon's purchase by multiplying by -1. -1x-5y=-17.55
We add this four times to Lesley's purchase to eliminate the x variable.
2y-20y=13.50-70.2
-18y=-56.7
y = $3.15 = Price of box of granola bars
Plug back into Landon's purchase to solve for pretzels.
x+5*3.15=17.55
x+15.75=17.55
x = $1.80 = price of bag of pretzels
Problem 4.
Let w = number of wood bats sold
Let m = number of metal bats sold
From sales information we have: w + m = 23
24w+30m=606
Substitution works well here. Solve for w in the first equation, w = 23 - m, and plug this into the second.
24*(23-m)+30m=606
552-24m+30m=606
6m=54
m=9 = number of metal bats sold
Therefore since w = 23-m, w = 23-9 = 14. 14 wooden bats were sold.
solution:
Z1 = 5(cos25˚+isin25˚)
Z2 = 2(cos80˚+isin80˚)
Z1.Z2 = 5(cos25˚+isin25˚). 2(cos80˚+isin80˚)
Z1.Z2 = 10{(cos25˚cos80˚ + isin25˚cos80˚+i^2sin25˚sin80˚) }
Z1.Z2 =10{(cos25˚cos80˚- sin25˚sin80˚+ i(cos25˚sin80˚+sin25˚cos80˚))}
(i^2 = -1)
Cos(A+B) = cosAcosB – sinAsinB
Sin(A+B) = sinAcosB + cosAsinB
Z1.Z2 = 10(cos(25˚+80˚) +isin(25˚+80˚)
Z1.Z2 = 10(cos105˚+ isin105˚)
The answer for the exercise shown above is the second option, which is: <span>
Maximum: 32°; minimum: −8°; period: 10 hours. The explanation is shown below:
</span> You can make a graph of the function given in the problem above: f(t)=20Sin(π/5t)+12.
As you can see in the graph, the maximum point is at 32 over the y-axis, and the minimum is at -8.
The lenght of the repeating pattern of the function (Its period) is 10.
10 is the answer
why are you struggling
so simple
<h2>
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Answer:
A) y=-6x+1
Step-by-step explanation:
m=(y2-y1)/(x2-x1)
m=(-17-(-5))/(3-1)
m=(-17+5)/2
m=-12/2
m=-6
y-y1=m(x-x1)
y-(-5)=-6(x-1)
y+5=-6(x-1)
y=-6x+6-5
y=-6x+1
