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olasank [31]
3 years ago
13

A convenience store pays a farmer $1.25 per pineapple. If it costs the farmer $0.15 in seeds, $0.25 in fertilizer, and $0.25 in

forgone output to grow each pineapple, the value added by the farmer to each pineapple is:
Mathematics
1 answer:
maw [93]3 years ago
6 0

Answer:

$0.60

Step-by-step explanation:

The costs are:

$0.15 + $0.25 = $0.40

He sells it for $1.25.

The value added is:

$1.25 - $0.40 = $0.85

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Two gears have a ratio of 8:3.
34kurt

Step-by-step explanation:

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6 0
3 years ago
Find the value of xx and yy in the parallelogram below.
user100 [1]

Answer: x = 33, y = 95

Step-by-step explanation:

The defintion of a parallelogram states that opposite sides are equal.

Therefore, 3x - 4 = 65 and y - 4 = 91.

We can now solve these equations.

3x = 69

x = 33

--------------------

y = 95

7 0
2 years ago
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Enter T or F depending on whether the statement is true or not. (You must enter T or F -- True and False will not work.) If it i
slava [35]

Answer:

See below

Step-by-step explanation:

If a is divisible by 3 then a is divisible by 9

<u>FALSE </u>

Counter-example

6 is divisible by 3 but not by 9

The subtraction of 2 rational numbers is rational.

<u>TRUE </u>

<em>Proof </em>

If a, b are two rational numbers

\large a=\frac{p}{q}\;b=\frac{r}{s}

for some integers p, q, r, s.

Then

\large a-b=\frac{p}{q}-\frac{r}{s}=\frac{ps-qr}{qs}

since ps-qr and qs are integers, a-b is rational

A sufficient condition for an integer to be divisible by 8 is that it is divisible by 2

<u>FALSE </u>

<em>Counter-example </em>

4 is divisible by 2 but not by 8

A sufficient condition for an integer to be divisible by 6 is that it is divisible by 2

<u>FALSE </u>

<em>Counter-example </em>

4 is divisible by 2 but not by 6

If a is divisible by 9 then a is divisible by 3.

<u>TRUE </u>

<em>Proof</em>

If a is divisible by 9, then a = 9k for some integer k, but 9=3*3, so a = 3*(3k).

Since 3k is integer, a is divisible also by 3.

The product of 2 consecutive integers is even.

<u>TRUE </u>

<em>Proof </em>

Let p, q be two consecutive integers, then either p is even or odd.

Suppose first p is even. Then

p = 2n and q = 2n+1, so p*q=2n(2n+1)=2n*2n+2n=2(n*2n+n)

since (n*2n+n) is integer p*q is even.

Suppose now p is odd

p = 2n+1 q = 2n+2, then

p*q=(2n+1)(2n+2)=2n*2n+2*2n+2n+2=2(n*2n+2n+n+1)

since (n*2n+2n+n+1) is integer p*q is also even.

Answer the following questions:

30 division 3 =

10

30 mod 3 =

0 (the remainder when dividing 30 by 3)

-26 division 5 =

-5 (plus remainder -1)

-26 mod 5 =

-1 (the remainder when dividing -26 by 5)

28 division 4 =

7

28 mod 4 =

0

-29 division 10 =

-2 (plus remainder -9)

-29 mod 10 =

-9

24 division 9 =

2 (plus remainder 6)

24 mod 9 =

6

-28 division 6=

-4 (plus remainder -4)

-28 mod 6 =

-4

965255471 mod 101 =

87 (the remainder when dividing 965255471 by 101)

630153353 mod 101 =

11 (the remainder when dividing 630153353 by 101)

8 0
4 years ago
Please help me Please please please help me
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It’s because that ❣️ is 1000
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3 years ago
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If f(x)=x^2-45, and f(2a)= -20, what is the value of a
Klio2033 [76]
Hope this helps!!! :)

5 0
3 years ago
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