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maria [59]
3 years ago
14

Andrea is selling candles as a fundraiser. she spent $50 on supplies for making the candles. she plans to sell the candles for $

10 each. her profit can be modeled by c(x)=10x-50. what is the domain and range of the function
Mathematics
1 answer:
Luda [366]3 years ago
8 0
In order for Andrea to make profit, the value of the C in the given function must not be 0 or less than zero. So, let's find the value of x when C is zero.

C = 0 = 10x - 50
x = 5

Hence, the number of candles sold represented by x must be more than 5 (< 5). Therefore, the domain is (5,+∞). When the value of x is more than 5, the C or the range would be greater than 0. Thus, the range is (0,+∞).
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Write an expression that is equivalent to -2/5(15 - 20d +50)<br> Explain please!:)
valina [46]

Answer:

-26 + 4/5d

Step-by-step explanation:

-2/5 ( 15 - 2d + 50)

= -2/5 ( 65 - 2d)

= -2/5 * 65 - (-2/5) * 2d

= -26 - (-4/5d)

= -26 + 4/5d

3 0
3 years ago
Determine the value of x.
harina [27]

Answer:

2

Step-by-step explanation:

7 0
3 years ago
What is 25/100 and 350/1000 simplified?
sergey [27]
25/100 = 1/4 (divide both numerator and denominator by 25)

350/1000 = 35/100 = 7/20.
4 0
3 years ago
2x + y - z = -8
Andreyy89

Answer:

Multiply row 1 by \frac{1}{2}.

Step-by-step explanation:

The augmented matrix of the system of linear equation is described below:

\left[\begin{array}{cccc}2&1&-1&-8\\0&2&3&-6\\-\frac{1}{2} &1&1&-4\end{array}\right]

Where a_{11} = 2, if we need to create a_{11} = 1, we need to multiply row 1 by \frac{1}{2}, that is to say:

\left[\begin{array}{cccc}1&\frac{1}{2} &-\frac{1}{2} &-4\\0&2&3&-6\\-\frac{1}{2} &1&1&-4\end{array}\right]

Hence, the correct answer is: Multiply row 1 by \frac{1}{2}.

5 0
3 years ago
A bucket that weighs 6 lb and a rope of negligible weight are used to draw water from a well that is 80 ft deep. The bucket is f
zaharov [31]

Answer:

3200 ft-lb

Step-by-step explanation:

To answer this question, we need to find the force applied by the rope on the bucket at time t

At t=0, the weight of the bucket is 6+36=42 \mathrm{lb}

After t seconds, the weight of the bucket is 42-0.15 t \mathrm{lb}

Since the acceleration of the bucket is the force on the bucket by the rope is equal to the weight of the bucket.

If the upward direction is positive, the displacement after t seconds is x=1.5 t

Since the well is 80 ft deep, the time to pull out the bucket is \frac{80}{2}=40 \mathrm{~s}

We are now ready to calculate the work done by the rope on the bucket.

Since the displacement and the force are in the same direction, we can write

W=\int_{t=0}^{t=36} F d x

Use x=1.5 t and F=42-0.15 t

W=\int_{0}^{36}(42-0.15 t)(1.5 d t)

=\int_{0}^{36} 63-0.225 t d t

=63 \cdot 36-0.2 \cdot 36^{2}-0=3200 \mathrm{ft} \cdot \mathrm{lb}

=\left[63 t-0.2 t^{2}\right]_{0}^{36}

W=3200 \mathrm{ft} \cdot \mathrm{lb}

4 0
3 years ago
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