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2 years ago

I'm having a bit of trouble finding out how to find the Domain of this function. Help please!

Mathematics

1 answer:

BabaBlast [244]2 years ago

6 0

In general, the domain is the set of all x-values for the graph.

The issue here is that this isn't the graph of a function. A function has at most one y-value for each x-value and this graph has an infinite number of y-values for the single x-value of 1.

So, either your teacher is wanting you to say the domain is {1}, because that's the only x-value used by the graph, or they're wanting you to say this is a trick question, because this isn't the graph of a function.

The range is the set of all y-values, which is -9<y<9, but again, do they intend this to be a trick question?

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Solve. 8x² + 5 = 35 Round to the nearest hundredth. Enter your answers in the boxes. The solutions are approximately and .

lora16 [44]

Answer:

x=1.94

x = - 1.94

Step-by-step explanation:

8x² + 5 = 35

Subtract 5 from each side

8x² + 5-5 = 35-5

8x² = 30

Divide each side by 8

8x² /8 = 30/8

x² = 15/4

Take the square root of each side

sqrt( x²) = ±sqrt(15/4)

x = ±sqrt(15/4)

x=1.93649

x = - 1.93649

To the nearest hundredth

x=1.94

x = - 1.94

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Find the middle numbers:

10, 12.

Find the number in between 10 and 12.

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3 years ago

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0.007407 to 1 significant figure

OLEGan [10]

Answer:

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3 years ago

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What is -5.25 to the nearest tenth?

kakasveta [241]

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3 years ago

Recall the equation for a circle with center (h, k) and radius r. At what point in the first quadrant does the line with equatio

lina2011 [118]

Answer:

The point of intersection is:

$\displaystyle \left(\frac{6\sqrt{5}}{5}, \frac{3\sqrt{5}}{5}+5\right)\approx \left(2.68, 6.34)$

Step-by-step explanation:

We want to find the point in QI at which the line with the equation:

$y=0.5x+5$

Intersect a circle with a radius of 3 and a center of (0, 5).

First, write the equation of a circle. The equation for a circle is given by:

$(x-h)^2+(y-k)^2=r^2$

Where (h, k) is the center and r is the radius.

Since our center is (0, 5), h = 0 and k = 5. The radius is 3. So, r = 3. Substitute:

$(x-0)^2+(y-5)^2=(3)^2$

Simplify:

$x^2+(y-5)^2=9$

At the point where the two equations intersect, its x-coordinate and y-coordinate will be the same. Therefore, we can substitute the equation of the line into the equation of the circle and solve for x. So:

$x^2+((0.5x+5)-5)^2=9$

Simplify:

$x^2+(0.5x)^2=9$

Square:

$x^2+0.25x^2=9$

Combine like terms:

$\displaystyle 1.25x^2=\frac{5}{4}x^2=9$

Solve for x:

$\displaystyle \begin{aligned} x^2&=\frac{36}{5} \\ x&=\pm\sqrt{\frac{36}{5}} \\ x&\Rightarrow \frac{6}{\sqrt{5}}=\frac{6\sqrt{5}}{5}\approx2.68\end{aligned}$

Note that since we are looking for the point of intersection in QI, x should be positive. So, we can ignore the negative answer.

To find the y-coordinate, substitute the x-value back into either equation. Using the linear equation:

$\displaystyle y=0.5\left(\frac{6\sqrt{5}}{5}\right)+5=\frac{3\sqrt{5}}{5}+5\approx 6.34$

So, the point of intersection in QI is:

$\displaystyle \left(\frac{6\sqrt{5}}{5}, \frac{3\sqrt{5}}{5}+5\right)\approx \left(2.68, 6.34)$

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2 years ago