Please help with 19 and 20 and explain

A person wishes to mix coffee worth $6 per pound with coffee worth $3 per pound to get a 120 pound of a mixture worth $4 per pou

NikAS [45]

Answer:

$720 and $360

Step-by-step explanation:

8 0

3 years ago

Help me on this I really need it

CaHeK987 [17]

Answer:

Step-by-step explanation:

Given that sale of grills increase 6% per year

CUrrent sale of grills this year = 3300

So in the first year sales would increase by 3300(6%)

Total sales in the I year =3300+3300(6%)

This will be again increasing by 6% in II year and so on.

Hence we have

every year 3300 increases by 6% compoundly

So

No of sales in t year

= $3300(1+0.06)^t\\=3300(1.06^t)$

In 6th year sale

s(6) = $3300(1.06)^6\\=4681.11$

7 0

3 years ago

A plant grows the same amount every week. Which graph matches the situation described? b e .

dem82 [27]

Answer:

Option A

Step-by-step explanation:

Let the height of plant is 'b' units.

Graph representing the height of the plant will have y-intercept = b units

Since, the plant is growing at the same rate every week,

And growth of the plant is continuous.

Therefore, graph will be a straight line and continuous.

Since, the height of the plant is always increasing,

Slope of the line will be positive.

Option A will be the answer.

8 0

3 years ago

Describe the process of rewriting the expression Please Help

Mazyrski [523]

Answer:

$x^{\frac{21}{4} }$

Step-by-step explanation:

Given expression is:

$(\sqrt[8]{x^7} )^{6}$

First we will use the rule:

$\sqrt[n]{x} = x^{\frac{1}{n} }$

So for the given expression:

$\sqrt[8]{x^{7}}=(x^{7} )^{\frac{1}{8} }$

We will use tha property of multiplication on powers:

$=x^{7*\frac{1}{8} }$

$= x^{\frac{7}{8} }$

Applying the outer exponent now

$(x^{\frac{7}{8} })^6$

$= x^{\frac{7}{8}*6 } \\= x^{\frac{42}{8} }\\= x^{\frac{21}{4} }$

8 0

3 years ago

Consider the three points ( 1 , 3 ) , ( 2 , 3 ) and ( 3 , 6 ) . Let ¯ x be the average x-coordinate of these points, and let ¯ y

loris [4]

Answer:

$m=\dfrac{3}{2}$

Step-by-step explanation:

Given points are: ( 1 , 3 ) , ( 2 , 3 ) and ( 3 , 6 )

The average of x-coordinate will be:

$\overline{x} = \dfrac{x_1+x_2+x_3}{\text{number of points}}$

1) Finding $(\overline{x},\overline{y})$

Average of the x coordinates:

$\overline{x} = \dfrac{1+2+3}{3}$

$\overline{x} = 2$

Average of the y coordinates:

similarly for y

$\overline{y} = \dfrac{3+3+6}{3}$

$\overline{y} = 4$

2) Finding the line through $(\overline{x},\overline{y})$ with slope m.

Given a point and a slope, the equation of a line can be found using:

$(y-y_1)=m(x-x_1)$

in our case this will be

$(y-\overline{y})=m(x-\overline{x})$

$(y-4)=m(x-2)$

$y=mx-2m+4$

this is our equation of the line!

3) Find the squared vertical distances between this line and the three points.

So what we up till now is a line, and three points. We need to find how much further away (only in the y direction) each point is from the line.

Distance from point (1,3)

We know that when x=1, y=3 for the point. But we need to find what does y equal when x=1 for the line?

we'll go back to our equation of the line and use x=1.

$y=m(1)-2m+4$

$y=-m+4$

now we know the two points at x=1: (1,3) and (1,-m+4)

to find the vertical distance we'll subtract the y-coordinates of each point.

$d_1=3-(-m+4)$

$d_1=m-1$

finally, as asked, we'll square the distance

$(d_1)^2=(m-1)^2$

Distance from point (2,3)

we'll do the same as above here:

$y=m(2)-2m+4$

$y=4$

vertical distance between the two points: (2,3) and (2,4)

$d_2=3-4$

$d_2=-1$

squaring:

$(d_2)^2=1$

Distance from point (3,6)

$y=m(3)-2m+4$

$y=m+4$

vertical distance between the two points: (3,6) and (3,m+4)

$d_3=6-(m+4)$

$d_3=2-m$

squaring:

$(d_3)^2=(2-m)^2$

3) Add up all the squared distances, we'll call this value R.

$R=(d_1)^2+(d_2)^2+(d_3)^2$

$R=(m-1)^2+4+(2-m)^2$

4) Find the value of m that makes R minimum.

Looking at the equation above, we can tell that R is a function of m:

$R(m)=(m-1)^2+4+(2-m)^2$

you can simplify this if you want to. What we're most concerned with is to find the minimum value of R at some value of m. To do that we'll need to derivate R with respect to m. (this is similar to finding the stationary point of a curve)

$\dfrac{d}{dm}\left(R(m)\right)=\dfrac{d}{dm}\left((m-1)^2+4+(2-m)^2\right)$

$\dfrac{dR}{dm}=2(m-1)+0+2(2-m)(-1)$

now to find the minimum value we'll just use a condition that $\dfrac{dR}{dm}=0$

$0=2(m-1)+2(2-m)(-1)$

now solve for m:

$0=2m-2-4+2m$

$m=\dfrac{3}{2}$

This is the value of m for which the sum of the squared vertical distances from the points and the line is small as possible!

5 0

3 years ago