MATH SOLVE

2 months ago

Q:
# 2For question 1, find the x- and y-intercept of the line.1. -10x+5y = 40 (1 point)x-intercept is 5; y-intercept is -10.x-intercept is 8; y-intercept is -4.x-intercept is -10;y-intercept is 5.x-intercept is -4;y-intercept is 8.For question 2, find the x- and y-intercept of the line.2. 5x + 4y= 80 (1 point)x-intercept is 4; y-intercept is 5.x-intercept is 20; y-intercept is 16.x-intercept is 5; y-intercept is 4.x-intercept is 16; y-intercept is 20.Write y=-x+4 in standard form using integers.(1 point)-x-by= 24-x+6y= 24Ox+6y=46x-y= 244. The grocery store sells kumquats for $4.75 a pound and Asian pears for $2.25 a pound. Write an equation (1 point)in standard form for the weights of kumquats k and Asian pears p that a customer could buy with $22.4.75k +2.25p = 224.75k = 2.25p +224.75 +2.25=k4.75p +2.25k = 225. Graph the equation. (1 point)

Accepted Solution

A:

Answer:Part 1) x-intercept is -4;y-intercept is 8Part 2) x-intercept is 16; y-intercept is 20Part 3) [tex]x+6y=24[/tex]Part 4)Part a) [tex]4.75k+2.25p=22[/tex]Part b) The graph in the attached figureStep-by-step explanation:Part 1) Find the x- and y-intercept of the linewe have[tex]-10x+5y = 40[/tex]we know thatThe x-intercept is the value of x when the value of y is equal to zeroso For y=0[tex]-10x+5(0) = 40[/tex][tex]-10x=40[/tex][tex]x=-4[/tex]The y-intercept is the value of y when the value of x is equal to zeroso For x=0[tex]-10(0)+5y = 40[/tex][tex]5y = 40[/tex][tex]y=8[/tex]thereforex-intercept is -4;y-intercept is 8Part 2) Find the x- and y-intercept of the linewe have[tex]5x+4y=80[/tex]we know thatThe x-intercept is the value of x when the value of y is equal to zeroso For y=0[tex]5x+4(0)=80[/tex][tex]5x=80[/tex][tex]x=16[/tex]The y-intercept is the value of y when the value of x is equal to zeroso For x=0[tex]5(0)+4y=80[/tex] [tex]4y=80[/tex][tex]y=20[/tex] thereforex-intercept is 16; y-intercept is 20Part 3) Write y=-(1/6)x+4 in standard form using integers.we know thatThe equation of a line in standard form is equal to[tex]Ax+By=C[/tex]whereA is a positive integerB and C are integerswe have[tex]y=-\frac{1}{6}x+4[/tex]Multiply both sides by 6 to remove the fraction[tex]6y=-x+24[/tex]Adds x both sides[tex]x+6y=24[/tex]Part 4) The grocery store sells kumquats for $4.75 a pound and Asian pears for $2.25 a pound. Part a) Write an equation in standard form for the weights of kumquats k and Asian pears p that a customer could buy with $22Part b) Graph the equationPart a) Letk -----> the number of pounds of kumquats boughtp ----> the number of pounds of Asian pears bough we know thatThe number of pounds of kumquats bought (k) multiplied by it cost of $4.75 a pound plus the number of pounds of Asian pears bough (p) multiplied by it cost of $2.25 a pound must be equal to $22so[tex]4.75k+2.25p=22[/tex]Part b) Graph the equationTo graph the line find out the interceptsLetk the first coordinate of the pointp the second coordinate of the pointThe k-intercept is the value of k when the value of p is equal to zeroso For p=0[tex]4.75k+2.25(0)=22[/tex][tex]4.75k=22[/tex][tex]k=4.63[/tex]soThe k-intercept is the point (4.63,0)The p-intercept is the value of p when the value of k is equal to zeroso For k=0[tex]4.75(0)+2.25p=22[/tex][tex]2.25p=22[/tex][tex]p=9.78[/tex]soThe k-intercept is the point (0,9.78)using a graphing toolPlot the intercepts and join the points to graph the linesee the attached figureRemember that the weight cannot be a negative number