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2 years ago

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You are checking the intermediate results of a phone app that calculates the weight of an object in kilograms given the weight i

n pounds. The particular calculation you are checking, shown below, is for an input weight of 142 pounds and outputs 29.3 kilograms. What is the first step that has an error, and why is it an error?

1 answer:

Illusion [34]2 years ago

4 0

The wrong step is step 4, because you only need one conversion.

The formula to convers pounds to kilograms is

$\text{weight in pounds} = 2.2 \times \text{weight in kilograms}$

Obviously, the inverse formula is

$\text{weight in kilograms} = \cfrac{\text{weight in pounds}}{2.2}$

So, you're given an input of 142 pounds. To convert it in kilograms, plug that value into out formula:

$\text{weight in kilograms} = \cfrac{142}{2.2} \approx 64.5$

and this is the answer you're looking for, so you need no more steps.

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Marty and Ethan both wrote a function, but in different ways.

marishachu [46]

<em><u>Question:</u></em>

Marty and Ethan both wrote a function, but in different ways.

Marty

y+3=1/3(x+9)

Ethan

x y

-4 9.2

-2 9.6

0 10

2 10.4

Whose function has the larger slope?

1. Marty’s with a slope of 2/3

2. Ethan’s with a slope of 2/5

3. Marty’s with a slope of 1/3

4. Ethan’s with a slope of 1/5

<em><u>Answer:</u></em>

Marty’s with a slope of 1/3 has the larger slope

<em><u>Solution:</u></em>

<em><u>Given that Marty equation is:</u></em>

$y + 3 = \frac{1}{3}(x+9)$

<em><u>The point slope form is given as:</u></em>

$y - y_1 = m(x-x_1)$

Where, "m" is the slope of line

On comapring both equations,

$m = \frac{1}{3}$

<em><u>Ethan wrote a function:</u></em>

Consider any two values from the table we have;

(0, 10) and (2, 10.4)

<em><u>The slope is given by formula:</u></em>

$m = \frac{y_2-y_1}{x_2-x_1}$

From above two points,

$(x_1, y_1) = (0, 10)\\\\(x_2, y_2) = (2, 10.4)$

Therefore,

$m = \frac{10.4-10}{2-0}\\\\m = \frac{0.4}{2} \\\\m = 0.2$

Thus we get,

$\text{Slope of Ethan} < \text{Slope of Marty}$

Therefore, Marty’s with a slope of 1/3 has the larger slope

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2 years ago

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Which of the following systems of Inequalities would produce the region indicated on the graph below?

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The line rising to the right has equation y = x + 2
and the other one has equation y = -x + 4

the lines are solid so the inequalities will be <= or >=

in both case the shading is below so the sign is <=

also x and Y values are both positive
so the answer is choice C

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The volume of air inside a rubber ball with radius r can be found using the function V(r) = four-thirds pi r cubed. What does V

aleksklad [387]

Answer:

The answer is explained below

Step-by-step explanation:

Given that The volume of air inside a rubber ball with radius r can be found using the function V(r) = $\frac{4}{3}\pi r^3$ , this means that the volume of the air inside the rubber ball is a function of the radius of the rubber ball, that is as the radius of the rubber ball changes, also the volume of the ball changes.

As seen from the function, the radius is directly proportional to the volume of the ball, if the radius increases, the volume also increases.

$V(\frac{5}{7} )$ is equal to the volume of the ball when the radius of the ball is $\frac{5}{7}$ . Therefore:

$V(\frac{5}{7} )=\frac{4}{3} \pi *(\frac{5}{7} )^3=1.53\ unit^3$

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For the binomial expansion of (x + y)^10, the value of k in the term 210x 6y k is a) 6 b) 4 c) 5 d) 7

Sloan [31]

Answer:

a) 6

Step-by-step explanation:

Expanding the polynomial using the formula:

$(x+y)^n=\sum_{k=0}^n \binom{n}{k} x^{n-k} y^k$

Also

$\binom{n}{k}=\frac{n!}{(n-k)!k!}$

I think you mean $210x^6y^4$

We can deduce that this term will be located somewhere in the middle. So I will calculate $k= 5; k=6 \text{ and } k =7$ .

For $k=5$

$\binom{10}{5} (y)^{10-5} (x)^{5}=\frac{10!}{(10-5)! 5!}(y)^{5} (x)^{5}= \frac{10 \cdot 9 \cdot 8 \cdot 7 \cdot 6 \cdot 5! }{5! \cdot 5 \cdot 4 \cdot 3 \cdot 2 \cdot 1 } \\ =\frac{30240}{120} =252 x^{5} y^{5}$

Note that we actually don't need to do all this process. There's no necessity to calculate the binomial, just $x^{n-k} y^k$

For $k=6$

$\binom{10}{6} \left(y\right)^{10-6} \left(x\right)^{6}=\frac{10!}{(10-6)! 6!}\left(y\right)^{4} \left(x\right)^{6}=210 x^{6} y^{4}$

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40 degrees might be use the question box

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