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andreev551 [17]
3 years ago
13

If it's $5 for 4 pounds of apples, how much does 1 pound cost?

Mathematics
2 answers:
Gwar [14]3 years ago
6 0

Answer:

$1.25

Step-by-step explanation:

$5/4=$1.25

kodGreya [7K]3 years ago
4 0

80 cents

you divide 4 and 5 and get .8 so 80 cents

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842 rounded to the nearest ten
g100num [7]

Answer:

842 rounded to the nearest ten is 840

Step-by-step explanation:

6 0
3 years ago
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Does Emiliano need more than 1 stick of butter or less than 1 stick of butter for the corn bread and the apple muffins?
Nookie1986 [14]

Answer:

Emilio needs less than 1 stick of butter

Step-by-step explanation:

Emilio needs 1/2 stick of butter to make cornbread . He also needs 1/4 stick of butter to make apple muffins . Does Emilio need more than I stick of butter to make both recepes or less than 1 stick of butter ? How do you know?

Cornbread = 1/2 stick of butter

Apple muffins = 1/4 stick of butter

Total sticks of butter = cornbread + apple muffins

= 1/2 + 1/4

= (2+1) / 4

= 3/4 stick of butter

= 0.75 stick of butter

0.75 stick of butter is less than 1 stick of butter

Therefore,

Emilio needs less than 1 stick of butter

4 0
3 years ago
. A restaurant charges a $100 setup fee, plus $15 per guest, to host a private party. The table
netineya [11]

Answer:

42

Step-by-step explanation:

i said so

3 0
3 years ago
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99,000 = _____ tens<br>please answer
rusak2 [61]
The answer will be 9900. Hope it help!
5 0
3 years ago
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Find the dimensions of the open rectangular box of maximum volume that can be made from a sheet of cardboard 11 in. by 7 in. by
11111nata11111 [884]

Answer:

The dimension of the open rectangular box is 8.216\times 4.216\times 1.392.

The volume of the box is 8.217 cubic inches.

Step-by-step explanation:

Given : The open rectangular box of maximum volume that can be made from a sheet of cardboard 11 in. by 7 in. by cutting congruent squares from the corners and folding up the sides.

To find : The dimensions and the volume of the open rectangular box ?

Solution :

Let the height be 'x'.

The length of the box is '11-2x'.

The breadth of the box is '7-2x'.

The volume of the box is V=l\times b\times h

V=(11-2x)\times (7-2x)\times x

V=4x^3-36x^2+77x

Derivate w.r.t x,

V'(x)=4(3x^2)-2(36x)+77

V'(x)=12x^2-72x+77

The critical point when V'(x)=0

12x^2-72x+77=0

Solve by quadratic formula,

x=\frac{18+\sqrt{93}}{6},\frac{18-\sqrt{93}}{6}

x=4.607,1.392

Derivate again w.r.t x,

V''(x)=24x-72

Now, V''(4.607)=24(4.607)-72=38.568>0 (+ve)

V''(1.392)=24(1.392)-72=-38.592 (-ve)

So, there is maximum at x=1.392.

The length of the box is l=11-2x

l=11-2(1.392)=8.216

The breadth of the box is b=7-2x

b=7-2(1.392)=4.216

The height of the box is h=1.392.

The dimension of the open rectangular box is 8.216\times 4.216\times 1.392.

The volume of the box is V=l\times b\times h

V=8.216\times 4.216\times 1.392

V=48.217\ in.^3

The volume of the box is 8.217 cubic inches.

5 0
3 years ago
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