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3 years ago

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Lamont wants to buy raspberries and apples to make a fruit tart raspberries cost $3.50 per pound and apples cost $3.75 per pound

. How much dose he spend if he buys 3 pounds Aurora has been losing 2 pounds of apples ?how much does he spend if he buys X pounds of raspberries and Y pounds of apples ?

1 answer:

Monica [59]3 years ago

3 0

Complete question :

Lamont wants to buy raspberries and apples to make a fruit tart raspberries cost $3.50 per pound and apples cost $3.75 per pound. How much dose he spend if he buys 3 pounds raspberry and 2 pounds of apples ?how much does he spend if he buys X pounds of raspberries and Y pounds of apples ?

Answer:

$18.00 ;

3.50x + 3.75y

Step-by-step explanation:

Given that:

Cost per pound of raspberry = $3.50

Cost per pound of apple = $3.75

Cost of 2 pounds of raspberry = 3 * $3.50 = $10.50

Cost of 2 pounds of apple = 2 * $3.75 = $7.50

Total spent = $(10.50 + 7.50) = $18.00

Cost of X pounds of raspberry and Y pound of apple

.

(3.50 * X) + (3.75 * Y)

3.50x + 3.75y

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1. S(–4, –4), P(4, –2), A(6, 6) and Z(–2, 4) a) Apply the distance formula for each side to determine whether SPAZ is equilatera

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a) SPAZ is equilateral.

b) Diagonals SA and PZ are perpendicular to each other.

c) Diagonals SA and PZ bisect each other.

Step-by-step explanation:

At first we form the triangle with the help of a graphing tool and whose result is attached below. It seems to be a paralellogram.

a) If figure is equilateral, then SP = PA = AZ = ZS:

$SP = \sqrt{[4-(-4)]^{2}+[(-2)-(-4)]^{2}}$

$SP \approx 8.246$

$PA = \sqrt{(6-4)^{2}+[6-(-2)]^{2}}$

$PA \approx 8.246$

$AZ =\sqrt{(-2-6)^{2}+(4-6)^{2}}$

$AZ \approx 8.246$

$ZS = \sqrt{[-4-(-2)]^{2}+(-4-4)^{2}}$

$ZS \approx 8.246$

Therefore, SPAZ is equilateral.

b) We use the slope formula to determine the inclination of diagonals SA and PZ:

$m_{SA} = \frac{6-(-4)}{6-(-4)}$

$m_{SA} = 1$

$m_{PZ} = \frac{4-(-2)}{-2-4}$

$m_{PZ} = -1$

Since $m_{SA}\cdot m_{PZ} = -1$ , diagonals SA and PZ are perpendicular to each other.

c) The diagonals bisect each other if and only if both have the same midpoint. Now we proceed to determine the midpoints of each diagonal:

$M_{SA} = \frac{1}{2}\cdot S(x,y) + \frac{1}{2}\cdot A(x,y)$

$M_{SA} = \frac{1}{2}\cdot (-4,-4)+\frac{1}{2}\cdot (6,6)$

$M_{SA} = (-2,-2)+(3,3)$

$M_{SA} = (1,1)$

$M_{PZ} = \frac{1}{2}\cdot P(x,y) + \frac{1}{2}\cdot Z(x,y)$

$M_{PZ} = \frac{1}{2}\cdot (4,-2)+\frac{1}{2}\cdot (-2,4)$

$M_{PZ} = (2,-1)+(-1,2)$

$M_{PZ} = (1,1)$

Then, the diagonals SA and PZ bisect each other.

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