Loading DocumentName: BIRE ELVIS KOLBIRE
STUDENTS ID: ADS24B00134Y
Solution
1(A).
Let S= local Sales
P= price (Ghc)
M= slope
M = (P2−P1S2−S1)
Given two points (10, 200) and (9, 250)
M= 9−10250−200= −50
The equation of a straight line
y= mx + b.
Y= Sales(S)
x=p
S=mP+b200=−50(10)+bb=200+500b=700The expresion on a linear is S=−50P+b
(B) Price of toy=ghc7.50Substitute the value of price into the equationS=−50(7.50)+700S=−375+700S=325. The number if sales is 325(C). when price of toy=ghc15Substitute the value of price into the expressionS=−50(15)+700=−750+700S=−50The Sales in the week will be −50(D). we observe that the weekly sales of B increases as the prices becomes lower while the sales value of ghc15decreaseswhen prices of the item toy in been increase to ghc15 in C.This will totally mean that, the low the prices of your goods the more sales you make and the more your prices are expensive to buythe less sales you will make in a week.xx(2) ax+b2x=c3By using properties,of exponents and logarithmsTake natural logarithms of boths sizes.(ab2)x+=c3xln(ab2)=3ln(c)⇒ln(ab2)xln(ab2) =In(ab2)3ln(c)x=ln(ab2)3ln(c)(b) f(x)= x3−5x2+4g(x)=7x2−5x−6p(x)=3x2+2x2−x+14f(x)+g(x)−(px)4(x3−5x2+4)+(7x2−5x−6)−(3x3+2x2−x+1)4x3−20x2+16+7x2−5x−6−3x3−2x2+x−1group like terms 4x3−3x3−20x2+7x2−2x2−5x+x+16−6−1x3−15x2−4x+9(c)y=mx+b⇒ 5x+6y=206y=−5x+20⇒ y=−65+310slope of first line is m1+−65for second slope18x−15y=17−15y=17−18x−1515(y)=−1517−15−18xy=56x−1517 the second slope of the linem2=56 Multiply the product to find if it is −1m1⋅m2−65⋅56=−1 since the product is −1, the line18x−15y=17 AND 5x+6y=20are at right angles −−5x−6=
D). If there are 100 students and food provisions for 20 days, we can calculate amount of food provision as the product of the number of students and the number of days:
Total provisions= number of students × number of days=100 × 20
Total provisions =2000
If 25 more students join the class then the total students = 100+ 25= 125
To find out how the provisions will last after the increased number of students we divided the total provisions by the new number of students.
New number of days= 2000 ÷ 125= 16 days
As the number of students increases while the total provisions remain constant then the number of days the provisions will last for reduces.
This emphasizes the importance of proper planning and adjustments in resource allocation when the size of a group changes.
(E)2−1y43y−618(y31y)−2 ⇒21y43y−618(y31(2)y21)21y43+−6y34y28⇒21y127y34+28 2y127 y388⇒2y127y388⇒ y388×y1272=y38×y13716=y41316(f). x−√x−3=5add both,sides by √x−3X=5+√x−3x2=(5+√x−3)2(a+b)2=a2+2ab+b2x2=25+10√x−3+x−3x2=x+22+10√x−3x2−x−22=10√x−310x2−x−22=√x−33102(x2−x−10)2=(x−3)2x4−2x3−3x2+44x2−44x+484=x−3x4−2x3+41x2−44x+484=x−3x4−2x3+41x2−44x+484=100x−300x4−2x3+41x2−144x+784=0)