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3 years ago

If 6 pounds of strawberries cost $11.94 what is the cost of 8 pounds

1 answer:

3 0

First find the price of one pound. 11.94÷6=1.99 Once you find the price of 1 pound. You multiply the price of one pound 8 times. 1.99×8=15.92 Hope it helps!

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HELP!!!!!

antoniya [11.8K]

Answer: 82°

Step-by-step explanation:

<CDE = far arc - near arc / 2

i.e

<CDE = arcCBE - arcCE / 2

arcCE = 98°

arcCBE = 360° - 98° = 262°

<CDE = 262° - 98° / 2

<CDE = 164 / 2

<CDE = 82°

Therefore the measure of < CDE is 82°

8 0

3 years ago

Read 2 more answers

PLEASE ANSWER FAST!!

dem82 [27]

40.000 in 7 year the company deposit money

5 0

3 years ago

A rancher wishes to build a fence to enclose a rectangular pen having area 24 square yards. Along one side the fence is to be ma

Gnoma [55]

Answer:

The value is $C \approx \$76$

Step-by-step explanation:

From the question we are told that

The area of the rectangular pen is $A = 24 \ yard^2$

The cost of material used to make one side is $z = \$ 6$

The cost of material used to make the other sides is $r = \$ 3$

Now , the fence to be build around the rectangular pen has four sides, the first opposite sides are equal, let assume each of the to be x yard and the other opposite sides are also equal as well let assume of the to be y yard

So the cost is mathematically represented as

$C = zx + r (x + 2y )$

=> $C = 6x + 3(x + 2y)$

=> $C = 9x + 6y$

Now the area of the fence is mathematically represented as

$A = x* y = 24$

=> $y = \frac{24}{x}$

=> $C = 9x + 6[\frac{24}{x} ]$

=> $C = 9x + [\frac{144}{x} ]$

Now differentiating

$C' = 9 + 144* (-2) x^{-2}$

$C' = 9 - 288x^{-2}$

At minimum $C' = 0$

$9 - 288x^{-2} = 0$

$x^{-2} = 0.03125$

$x = \sqrt{\frac{1}{0.03125} }$

$x = 5.66$

Now substituting for x in the equation above to obtain minimum cost

$C = 9(5.66) + [\frac{144}{5.66} ]$

$C \approx \$76$

8 0

3 years ago

Write an equation for this trend line

jeka94

Answer:

y=-1/3x+8 i think

Step-by-step explanation:

3 0

3 years ago

A rectangular poster is to contain 722 square inches of print. the margins at the top and bottom of the poster are to be 2 inche

Delvig [45]

Let
x ----------> the height of the whole poster
y ----------> the width of the whole poster

We need to minimize the area A=x*y

we know that
(x-4)*(y-2)=722
(y-2)=722/(x-4)
(y)=[722/(x-4)]+2

so
A(x)=x*y--------->A(x)=x*{[722/(x-4)]+2}
Need to minimize this function over x > 4

find the derivative------> A1 (x)
A1(x)=2*[8x²-8x-1428]/[(x-4)²]

for A1(x)=0
8x²-8x-1428=0
using a graph tool
gives x=13.87 in

(y)=[722/(x-4)]+2

y=[2x+714]/[x-4]-----> y=[2*13.87+714]/[13.87-4]-----> y=75.15 in

the answer is
the dimensions of the poster will be
the height of the whole poster is 13.87 in
the width of the whole poster is 75.15 in

6 0

3 years ago