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TEA [102]
2 years ago
8

A recipe calls for 2 cups of flour and 3 cups of sugar. How many cups of flour would be needed if the amount of sugar were incre

ased to 12 cups?
Mathematics
2 answers:
Sergeu [11.5K]2 years ago
5 0

Answer:

It would be 8 cups of flour.

nika2105 [10]2 years ago
4 0

Answer:

8 cups

Step-by-step explanation:

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What is the answer to this inequality?<br> (x+7)&gt;80
Crank

Answer:

x > 73

Step-by-step explanation:

x+7 > 80

subtract 7 from both sides

x + 7 - 7 > 80 - 7

x > 73

-Chetan K

5 0
3 years ago
I need help fining X. I keep getting a negative number
KIM [24]

Answer:

3

Step-by-step explanation:

3 0
3 years ago
Read 2 more answers
WILL GIVE BRAINLIEST
Marizza181 [45]

Answer:

18 pots.

Step-by-step explanation:

10/2 to get one of the longs. 5(2) to get the other one. 10.

8/4 to get one of the wides. 2(2) to get the other one. 4.

There are 4 corners so there are 4 extra pots. 4.

Add the numbers: 4+4+10. 18

7 0
3 years ago
Read 2 more answers
The base of an aquarium with given volume V is made of slate and the sides are made of glass. If slate costs five times as much
Y_Kistochka [10]

Answer:

x = ∛ 2*V/5  

y = ∛ 2*V/5

h  = V/ ∛ 4*V²/25

Step-by-step explanation:

Dimensions of the aquarium base is  x*y

We call c₁ cost per unit area of the sides, then cost per unit area of slate is equal 5c₁.

let call h the height of the aquarium then volume of the aquarium is:

V = x*y*h      where   h =  V / x*y

As the base is a rectangular one there are 2 sides x*h .  and 2 sides  y*h

According to this:

Ct (cost of aquarium )  = cost of the base  + cost of the sides

cₐ  ( cost of the base) = 5*c₁*x*y

c₆ (cost of the sides ) = c₁*2*x*h   +   c₁*2*y*h

C(t)  =  5*c₁*x*y +2* c₁*x* V/x*y  +  2* c₁*y* V/x*y    or

C(t)  =  5*c₁*x*y  + 2*c₁*V/y   *2*c₁* V/x

Taking partial derivatives en x and y we have:

C´(x)  =  5*c₁*y - 2*c₁*V/x²

C´(y)  =  5*c₁*x - 2*c₁*V/y²

C´(x)  = C´(y)        ⇒  5*c₁*y - 2*c₁*V/x²  =   5*c₁*x - 2*c₁*V/y²

or    5*y - 2*V/x²  =   5*x - 2*V/y²

(5*y*x² - 2*V)/x²  = ( 5*y²x - 2*V) /y²

(5*y*x² - 2*V)*y²  = ( 5*y²x - 2*V)*x²

5*y³*x² - 2*V*y²  =  5*y²x³  - 2*V*x²

5*y³*x² - 5*y²x³  =  2*V * ( y² - x²)

by symmetry  x =  y

Then using x = y  and plugging that value on the derivatives

C´(x) =  5*c₁*y - 2*c₁*V/x²

C´(x) =  5*c₁*x - 2*c₁*V/x²

C´(x) = 0          ⇒     5*c₁*x - 2*c₁*V/x²  = 0

5*x  - 2*V/x² = 0      ⇒  5*x³ - 2*V = 0   ⇒   5*x³  = 2*V  ⇒ x³ = 2*V/5

x = ∛ 2*V/5       and   y = ∛ 2*V/5    and   h  =  V/ x*y    h  = V/ ∛ 4*V²/25

7 0
3 years ago
The International Space Station orbits at an altitude of about 250 miles about Earth’s surface. The radius of Earth is approxima
Hatshy [7]
In this case we are dealing with the pythagorean theorm involving right angled triangles. This theorm states that a^2 + b^2 = c^2 which means the square of the hypotenuse (side c, opposite the right angle) is equal to the square of the remaining two sides.

In this case we will say that a = 3963 miles which is the radius of the earth. c is equal to the radius of the earth plus the additional altitude of the space station which is 250 miles; therefore, c = 4213 miles. We must now solve for the value b which is equal to how far an astronaut can see to the horizon.

(3963)^2 + b^2 = (4213)^2
b^2 = 2,044,000
b = 1430 miles.

The astronaut can see 1430 miles to the horizon.
8 0
3 years ago
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