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nasty-shy [4]
3 years ago
9

A trucker buys crates of apples and pears to sell at a Farmer's Market. The apples cost $6 per crate and the pears cost $5.50 pe

r crate. Each crate weights 25 pounds. If the truck can carry 5000 pounds and the trucker has $1180 to spend, how many crates of each kind of fruit can he buy? Write your answer as an ordered pair (x,y).
Mathematics
1 answer:
ivolga24 [154]3 years ago
4 0

Answer:

He can buy 160 crates of apples and 40 crates of pears. (160, 40).

Step-by-step explanation:

From the information given, you can write the following equations:

25x+25y=5000 (1)

6x+5.50y=1180 (2), where:

x is the number of crates of apples

y is the number of crates of pears

First, you can solve for x in (1):

25x=5000-25y

x=(5000/25)-(25y/25)

x=200-y (3)

Then, you can replace (3) in (2) and solve for y:

6(200-y)+5.50y=1180

1200-6y+5.50y=1180

1200-1180=6y-5.50y

20=0.5y

y=20/0.5

y=40

Finally, you can replace the value of y in (3) to find the value of x:

x=200-y

x=200-40

x=160

According to this, the answer is that he can buy 160 crates of apples and 40 crates of pears. (160, 40).

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baherus [9]

Answer:

Total amount of money Sabra spends on both clubs = 20.50 + 10m

Amount paid to join the cooking club = $8.50

Monthly payment at the cooking club = $6.25

Payment for m months at the cooking club = 6.25m

Total money paid at the cooking club = 8.50 + 6.25m

Amount paid to join the movie club = $12

Monthly payment at the movie club = $3.75

Payment for m months at the movie club = 3.75m

Total money paid at the movie club = 12 + 3.75m

Total amount of money Sabra spends on both clubs = Total money paid at the cooking club + Total money paid at the movie club

Total amount of money Sabra spends on both clubs = 8.50 + 6.25m + 12 + 3.75m

Total amount of money Sabra spends on both clubs = 20.50 + 10m

Step-by-step explanation:

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3 years ago
Which fraction is equivilent to -2/3
Vikki [24]

Answer:

-4/6, -6/9,- 8/12,-10/15

Step-by-step explanation:

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3 years ago
1) Use Newton's method with the specified initial approximation x1 to find x3, the third approximation to the root of the given
neonofarm [45]

Answer:

Check below, please

Step-by-step explanation:

Hello!

1) In the Newton Method, we'll stop our approximations till the value gets repeated. Like this

x_{1}=2\\x_{2}=2-\frac{f(2)}{f'(2)}=2.5\\x_{3}=2.5-\frac{f(2.5)}{f'(2.5)}\approx 2.4166\\x_{4}=2.4166-\frac{f(2.4166)}{f'(2.4166)}\approx 2.41421\\x_{5}=2.41421-\frac{f(2.41421)}{f'(2.41421)}\approx \mathbf{2.41421}

2)  Looking at the graph, let's pick -1.2 and 3.2 as our approximations since it is a quadratic function. Passing through theses points -1.2 and 3.2 there are tangent lines that can be traced, which are the starting point to get to the roots.

We can rewrite it as: x^2-2x-4=0

x_{1}=-1.1\\x_{2}=-1.1-\frac{f(-1.1)}{f'(-1.1)}=-1.24047\\x_{3}=-1.24047-\frac{f(1.24047)}{f'(1.24047)}\approx -1.23607\\x_{4}=-1.23607-\frac{f(-1.23607)}{f'(-1.23607)}\approx -1.23606\\x_{5}=-1.23606-\frac{f(-1.23606)}{f'(-1.23606)}\approx \mathbf{-1.23606}

As for

x_{1}=3.2\\x_{2}=3.2-\frac{f(3.2)}{f'(3.2)}=3.23636\\x_{3}=3.23636-\frac{f(3.23636)}{f'(3.23636)}\approx 3.23606\\x_{4}=3.23606-\frac{f(3.23606)}{f'(3.23606)}\approx \mathbf{3.23606}\\

3) Rewriting and calculating its derivative. Remember to do it, in radians.

5\cos(x)-x-1=0 \:and f'(x)=-5\sin(x)-1

x_{1}=1\\x_{2}=1-\frac{f(1)}{f'(1)}=1.13471\\x_{3}=1.13471-\frac{f(1.13471)}{f'(1.13471)}\approx 1.13060\\x_{4}=1.13060-\frac{f(1.13060)}{f'(1.13060)}\approx 1.13059\\x_{5}= 1.13059-\frac{f( 1.13059)}{f'( 1.13059)}\approx \mathbf{ 1.13059}

For the second root, let's try -1.5

x_{1}=-1.5\\x_{2}=-1.5-\frac{f(-1.5)}{f'(-1.5)}=-1.71409\\x_{3}=-1.71409-\frac{f(-1.71409)}{f'(-1.71409)}\approx -1.71410\\x_{4}=-1.71410-\frac{f(-1.71410)}{f'(-1.71410)}\approx \mathbf{-1.71410}\\

For x=-3.9, last root.

x_{1}=-3.9\\x_{2}=-3.9-\frac{f(-3.9)}{f'(-3.9)}=-4.06438\\x_{3}=-4.06438-\frac{f(-4.06438)}{f'(-4.06438)}\approx -4.05507\\x_{4}=-4.05507-\frac{f(-4.05507)}{f'(-4.05507)}\approx \mathbf{-4.05507}\\

5) In this case, let's make a little adjustment on the Newton formula to find critical numbers. Remember their relation with 1st and 2nd derivatives.

x_{n+1}=x_{n}-\frac{f'(n)}{f''(n)}

f(x)=x^6-x^4+3x^3-2x

\mathbf{f'(x)=6x^5-4x^3+9x^2-2}

\mathbf{f''(x)=30x^4-12x^2+18x}

For -1.2

x_{1}=-1.2\\x_{2}=-1.2-\frac{f'(-1.2)}{f''(-1.2)}=-1.32611\\x_{3}=-1.32611-\frac{f'(-1.32611)}{f''(-1.32611)}\approx -1.29575\\x_{4}=-1.29575-\frac{f'(-1.29575)}{f''(-4.05507)}\approx -1.29325\\x_{5}= -1.29325-\frac{f'( -1.29325)}{f''( -1.29325)}\approx  -1.29322\\x_{6}= -1.29322-\frac{f'( -1.29322)}{f''( -1.29322)}\approx  \mathbf{-1.29322}\\

For x=0.4

x_{1}=0.4\\x_{2}=0.4\frac{f'(0.4)}{f''(0.4)}=0.52476\\x_{3}=0.52476-\frac{f'(0.52476)}{f''(0.52476)}\approx 0.50823\\x_{4}=0.50823-\frac{f'(0.50823)}{f''(0.50823)}\approx 0.50785\\x_{5}= 0.50785-\frac{f'(0.50785)}{f''(0.50785)}\approx  \mathbf{0.50785}\\

and for x=-0.4

x_{1}=-0.4\\x_{2}=-0.4\frac{f'(-0.4)}{f''(-0.4)}=-0.44375\\x_{3}=-0.44375-\frac{f'(-0.44375)}{f''(-0.44375)}\approx -0.44173\\x_{4}=-0.44173-\frac{f'(-0.44173)}{f''(-0.44173)}\approx \mathbf{-0.44173}\\

These roots (in bold) are the critical numbers

3 0
3 years ago
If h(x) = (f o g)(x) and h(x) = sqrt(x+5), find g(x) if f(x) = sqrt(x+2)
agasfer [191]

Answer:

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Step-by-step explanation:

<h3>Given</h3>
  • h(x) = (f ο g)(x)
  • h(x) = \sqrt{x+5}
  • f(x) = \sqrt{x+2}
<h3>To find</h3>
  • g(x)
<h3>Solution</h3>

<u>We know that:</u>

  • (f ο g)(x) = f(g(x))

<u>Substitute x with g(x) and solve for g(x):</u>

  • \sqrt{x+5} = \sqrt{g(x)+2}
  • x + 5 = g(x) + 2
  • x + 3 = g(x)
  • g(x) = x + 3
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Answer:

x=5

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Step-by-step explanation:

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