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Usimov [2.4K]
3 years ago
5

To make 16 hamburger patties that each weigh 1/3 pound, you need 16/3 pounds of beef. Write 16/3 as a mixed number.

Mathematics
2 answers:
Darina [25.2K]3 years ago
6 0
The answer would be 5 1/3.

solution:

if you were to divide 16 and 3, you wouldn’t have an exact whole number to divide it by. The closet would be 5. 5x3= 15 with a remainder of 1. You take the denominator and make it 1/3. Now, you have your whole number 5 and fraction 1/3. That would mean the answer is 5 1/3
Tanzania [10]3 years ago
3 0

Answer:

5 1/3

Step-by-step explanation:

fraction=division

16÷3= 5 R1

5 r1= 5 1/3

I hope this helps you learn how to change improper fractions to mixed numbers!

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Find the length of MA<br> What is the area of triangle MNT?
tia_tia [17]
The length is 7 and there is no N
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What is the range of the function f(x) = 2x +4 when the domain is {-2, 0, 3}?
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Apples cost $0.75 per pound and bananas cost $1.05 per pound
LuckyWell [14K]

Answer:

<em>The baker bought 8 apples and 4 bananas</em>

Step-by-step explanation:

<u>System of Equations</u>

Let's call:

a = pounds of apples

b = pounds of bananas

The baker bought a total of 12 pounds of apples and bananas, thus:

a + b = 12      [1]

Apple cost $0.75 per pound and each pound of bananas cost $1.05 per pound. Thus the total cost is 0.75a + 1.05b. We know he spent a total of $10.20, thus

0.75a + 1.05b = 10.20     [2]

Solving [1] for a:

a = 12 - b       [3]

Substituting in [2]:

0.75(12 - b) + 1.05b = 10.20

Operating

9 - 0.75b + 1.05b = 10.20

Simplifying:

0.30b = 10.20 - 9 = 1.20

Dividing by 0.30:

b = 1.20/0.30

b = 4

From [3]:

a = 12 - 4 = 8

a = 8

The baker bought 8 apples and 4 bananas

8 0
2 years ago
What is the area of this irregular shape and how do I work it out?
Brilliant_brown [7]

Answer:

17cm²

Step-by-step explanation:

i hope it's helpful

7 0
2 years ago
Let II be the tangent plane to the graph of f(x, y) = 8 – 2x^2 – 3y^2 at the point (1, 2,-6). Let S, x² + y^2 + z = 4 be another
stealth61 [152]

Let F(x,y,z)=f(x,y)-z. The tangent plane to f(x,y) at (1, 2, -6) has equation

\nabla F(1,2,-6)\cdot(x-1,y-2,z+6)=0

We have

\nabla F(x,y,z)=(-4x,-6y,-1)\implies\nabla F(1,2,-6)=(-4,-12,-1)

Then the tangent plane has equation

(-4,-12,-1)\cdot(x-1,y-2,z+6)=0\implies -4(x-1)-12(y-2)-(z+6)=0\implies 4x+12y+z=22

Let g(x,y)=4-x^2-y^2, and G(x,y,z)=g(x,y)-z. The tangent plane to S at a point (a,b,c) is

\nabla G(a,b,c)\cdot(x-a,y-b,z-c)=0

We have

\nabla G(x,y,z)=(-2x,-2y,-1)\implies \nabla G(a,b,c)=(-2a,-2b,-1)

so that this plane has equation

(-2a,-2b,-1)\cdot(x-a,y-b,z-c)=0\implies2ax+2by+z=2a^2+2b^2+c

In order for this plane to be parallel to the previous plane, we need to have

\begin{cases}2a=4\\2b=12\end{cases}\implies a=2,b=6\implies g(a,b)=c=-36

so the point we're looking for is (2, 6, -36).

6 0
3 years ago
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