All categories

2 years ago

Isabella is cooking a meal for 14 friends. Each person gets

Mathematics

1 answer:

algol132 years ago

8 0

I’m pretty sure that says 6 cups of vegetables so 84 cups but if you meant 1/6 of a cup then 7/3 cups or 2/1/3

You might be interested in

NEED HELP ASAP!! (15 points)

LiRa [457]

Answer:

Step-by-step explanation:

a b c

13 10 z

169 100 z 269 16.4 hope this help's

3 0

2 years ago

Lynette buys candy at the grocery store. she gives half of the pieces to her son and 2 pieces to her daughter. if she has 4 piec

nevsk [136]

Let Lynette has originally candies in hand = x

She gave half to her son means x/2 given to son.

2 candies gave to daughter and left with 4.

x - x/2 - 2 = 4

adding 2 both sides

x - x/2 - 2 + 2 = 4 + 2

x - x/2 = 6

x /2 = 6

x = 12

Means she have 12 candies originally : Answer

Hope it will help :)

5 0

2 years ago

2×6 <br>euruuryduruhdhdhd

Pachacha [2.7K]

2 x 6 = 12

Hope it helps :)

5 0

2 years ago

Read 2 more answers

Dr. Clarence invested $5000 in an account that draws 3.2% interest, compounded annually. What is the total value of the account

DanielleElmas [232]

Given that the amount invested was compounded annually, the total value of the investment after 5 years will be:
FV=P(1+r/100)^n
where:
FV=future value
p=principle
r=rate
n=time
hence
FV=5000(1+3.2/100)^5
FV=$5,852.865

3 0

3 years ago

Find the directional derivative of at the point (1, 3) in the direction toward the point (3, 1). g

Anika [276]

Complete Question:

Find the directional derivative of g(x,y) = $x^2y^5$ at the point (1, 3) in the direction toward the point (3, 1)

Answer:

Directional derivative at point (1,3), $D_ug(1,3) = \frac{162}{\sqrt{8} }$

Step-by-step explanation:

Get $g'_x$ and $g'_y$ at the point (1, 3)

g(x,y) = $x^2y^5$

$g'_x = 2xy^5\\g'_x|(1,3)= 2*1*3^5\\g'_x|(1,3) = 486$

$g'_y = 5x^2y^4\\g'_y|(1,3)= 5*1^2* 3^4\\g'_y|(1,3)= 405$

Let P = (1, 3) and Q = (3, 1)

Find the unit vector of PQ,

$u = \frac{\bar{PQ}}{|\bar{PQ}|} \\\bar{PQ} = (3-1, 1-3) = (2, -2)\\{|\bar{PQ}| = \sqrt{2^2 + (-2)^2}\\$

$|\bar{PQ}| = \sqrt{8}$

The unit vector is therefore:

$u = \frac{(2, -2)}{\sqrt{8} } \\u_1 = \frac{2}{\sqrt{8} } \\u_2 = \frac{-2}{\sqrt{8} }$

The directional derivative of g is given by the equation:

$D_ug(1,3) = g'_x(1,3)u_1 + g'_y(1,3)u_2\\D_ug(1,3) = (486*\frac{2}{\sqrt{8} } ) + (405*\frac{-2}{\sqrt{8} } )\\D_ug(1,3) = (\frac{972}{\sqrt{8} } ) + (\frac{-810}{\sqrt{8} } )\\D_ug(1,3) = \frac{162}{\sqrt{8} }$

8 0

2 years ago