Part A: How tall is the pole?

Mary used Delicious (d) and Golden Delicious (g) apples to make homemade applesauce. Delicious apples are $0.75 each and Golden

alina1380 [7]

Answer:

Mary bought 9 Golden Delicious apples

Step-by-step explanation:

This can be solved by setting up a system of equations. Based on the information given, the equations would be set up as follows.

$.75d+1.25g=21.00\\d+g=22$

I prefer to solve using substitution, but elimination can be used as well

$.75d+1.25g=21.00\\g=22-d$

We can now plug in this g value to the first equation

$.75d + 1.25 (22-d)=21.00\\.75d + 27.5-1.25d=21\\27.5-0.5d=21\\-.5d=-6.5\\d=13$

This means mary bought 13 delicious apples. She bought 22 apples in total, so the total number minus the amount of delicious apples will give us the amount of golden delicious apples.

22 - 13 = g

9 = g

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Let me know if you have any questions! Spread the love.

3 0

2 years ago

Bananas cost $0.54 per lb and grapes cost$1.28 per lb. Leanne bought 2.6 lb of bananas and 3.1 lb of grapes .How much did she pa

slamgirl [31]

So all u have to do is just add $1.28 and $0.54

5 0

3 years ago

What is the slope of a line parallel to the line whose equation is 3x – 4y = 8

liq [111]

Hi there! The answer is C.

We can find the slope of this line by making y the subject of the equation.

$3x - 4y = 8 \\ - 4y = - 3x + 8 \\ y = \frac{ - 3}{ - 4} x + \frac{8}{ - 4} \\ y = \frac{3}{4} x - 2$

Now we can see that the slope of this line (which is positioned in front of x) is 3/4. Therefore, the slope of any line parallel to this one is 3/4 as well. The answer is C.

7 0

3 years ago

Solve for x and y A=3y B=105 C=9x+5 D=2x+10

Leto [7]

Where’s the equation?

6 0

2 years ago

Read 2 more answers

Use Lagrange multipliers to find the maximum and minimum values of (i) f(x,y)-81x^2+y^2 subject to the constraint 4x^2+y^2=9. (i

sp2606 [1]

i. The Lagrangian is

$L(x,y,\lambda)=81x^2+y^2+\lambda(4x^2+y^2-9)$

with critical points whenever

$L_x=162x+8\lambda x=0\implies2x(81+4\lambda)=0\implies x=0\text{ or }\lambda=-\dfrac{81}4$

$L_y=2y+2\lambda y=0\implies2y(1+\lambda)=0\implies y=0\text{ or }\lambda=-1$

$L_\lambda=4x^2+y^2-9=0$

If $x=0$ , then $L_\lambda=0\implies y=\pm3$ .
If $y=0$ , then $L_\lambda=0\implies x=\pm\dfrac32$ .
Either value of $\lambda$ found above requires that either $x=0$ or $y=0$ , so we get the same critical points as in the previous two cases.

We have $f(0,-3)=9$ , $f(0,3)=9$ , $f\left(-\dfrac32,0\right)=\dfrac{729}4=182.25$ , and $f\left(\dfrac32,0\right)=\dfrac{729}4$ , so $f$ has a minimum value of 9 and a maximum value of 182.25.

ii. The Lagrangian is

$L(x,y,z,\lambda)=y^2-10z+\lambda(x^2+y^2+z^2-36)$

with critical points whenever

$L_x=2\lambda x=0\implies x=0$ (because we assume $\lambda\neq0$ )

$L_y=2y+2\lambda y=0\implies 2y(1+\lambda)=0\implies y=0\text{ or }\lambda=-1$

$L_z=-10+2\lambda z=0\implies z=\dfrac5\lambda$

$L_\lambda=x^2+y^2+z^2-36=0$

If $x=y=0$ , then $L_\lambda=0\implies z=\pm6$ .
If $\lambda=-1$ , then $z=-5$ , and with $x=0$ we have $L_\lambda=0\implies y=\pm\sqrt{11}$ .

We have $f(0,0,-6)=60$ , $f(0,0,6)=-60$ , $f(0,-\sqrt{11},-5)=61$ , and $f(0,\sqrt{11},-5)=61$ . So $f$ has a maximum value of 61 and a minimum value of -60.

5 0

3 years ago