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slavikrds [6]
4 years ago
14

The x-coordinate of an ordered pair is more than half the value of its corresponding y-coordinate. If the x-coordinate of the or

dered pair is 10, what is the largest whole number value for the corresponding y-coordinate?
Mathematics
2 answers:
lisabon 2012 [21]4 years ago
7 0

Answer:

19 :)

Step-by-step explanation:

Vladimir79 [104]4 years ago
4 0

since half of 18 is 9 and the number more than 18 does not have half less than x.

i guess it is 18.

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Which graph represents 4x + 7y < -21
allsm [11]

Answer:

C

Step-by-step explanation:

Since the sign is not \leq, and it is <, then you'd choose the option with the dotted line showing it isn't inclusive.

3 0
4 years ago
When a fraction of 17 is taken away from 17 , what remains exceeds one-third of seventeen by six.
Andrew [12]

Answer:

The fraction is less than \frac{17}{18}.

Step-by-step explanation:

Let the fraction is \frac{1}{n}.

Given that when the fraction of 17 is subtracted from 17 itself, then the remaining will be greater than one-third of 17 divided by six.

Therefore, by the given condition,  

17 - \frac{17}{n} > \frac{1}{3}(\frac{17}{6} )

⇒ 17 - \frac{17}{n} > \frac{17}{18}

⇒ 1 - \frac{1}{n} > \frac{1}{18}

⇒ \frac{1}{n} < 1 - \frac{1}{18}

⇒ \frac{1}{n} < \frac{17}{18}

Therefore, the fraction is less than \frac{17}{18}. (Answer)

5 0
3 years ago
Pleasepleasepleaseplease help me
Tanzania [10]

Answer:

(\frac{f}{g} )(x)=5-x

Step-by-step explanation:

If f(x)=25-x^2 and g(x)=x+5

Then (\frac{f}{g} )(x)=\frac{25-x^2}{x+5}

We can factor out the numerator to get

(\frac{f}{g} )(x)=\frac{(5-x)(5+x)}{x+5}

And thanks to the communicative property of addition, 5+x=x+5

This means that the x+5 on both the top and the bottom cancel out. This leaves you with

(\frac{f}{g} )(x)=5-x

7 0
4 years ago
Analyze Nick's solution. Is he correct? If not, what was
NemiM [27]

Answer: No, the power multiplied to 8.64 should have an exponent of 0. B pretty much

Step-by-step explanation: just got that question trust me it’s right.

7 0
3 years ago
Read 2 more answers
A piece of wire of length 56 is​ cut, and the resulting two pieces are formed to make a circle and a square. Where should the wi
Llana [10]

Answer:

1) Cut the wire at a distance of 31.36 and draw this length into a square and the remaining into a circle to minimize the area.

2) To maximize the area do not cut the wire but make the whole wire into a circle.

Step-by-step explanation:

let the wire be cut at a distance of x

We make a square of this wire

For the remaining length of (56 - x) we make a circle

Thus

Area of square = Area_{square}(side)^{2}\\\\=(x/4)^{2}

Similarly the area of the circle equals

A_{circle}=\frac{(56-x)^{2}}{4\pi }

Thus summing the areas we get

A=\frac{x^{2}}{16}+\frac{(56-x)^{2}}{4\pi }

1) to find the critical points we differentiate the given area with respect to 'x'

Thus we have

\frac{dA}{dx}=\frac{d(\frac{x^{2}}{16}+\frac{(56-x)^{2}}{4\pi })}{dx}\\\\0=\frac{x}{8}-\frac{(56-x)}{2\pi }\\\\\frac{x}{8}+\frac{x}{2\pi }=\frac{28}{\pi }\\\\\therefore x=31.36

Thus the length of the square wire should be x = 31.36

The length of the circular portion should be 56 - 31.36 = 24.64.

These lengths shall give the minimum combined area of the 2 figures.

2) Since the given function is a quadratic thus with the graph attached below we can see the maximum area occurs if all the wire is to be made into a circle thus to maximize the area the wire shall not be cut and wholly shaped into a circle.

4 0
3 years ago
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