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2 years ago

Please help with these

Mathematics

1 answer:

PtichkaEL [24]2 years ago

6 0

Re ask.. the photo is not visible

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When would a set of data be described as variable?

valina [46]

I guess when the data isn't consitant

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3 years ago

On Sunday Lindsay rode her bikeb3 times as far as she did on Saturday. On Saturday she rode 3 miles. How far did she ride on Sun

Naddik [55]

Answer:

9 miles

Step-by-step explanation:

3 times 3 is 9

7 0

2 years ago

At which points does the graph of f(x)=2x−4 cross the x-axis and y-axis?

blagie [28]

crosses x-axis at (2, 0 ) and y-axis at (0, - 4 )

To find where the graph crosses the x and y axes ( intercepts )

• let x = 0, in the equation for y- intercept

• let y = 0, in the equation for x- intercept

x = 0 : y = 0 - 4 = - 4 ⇒ (0, - 4 )

y = 0 : 2x - 4 = 0 ⇒ 2x = 4 ⇒ x = 2 ⇒ (2, 0 )

6 0

2 years ago

If the endpoints of the diameter of a circle are (−8, −6) and (−4, −14), what is the standard form equation of the circle?

kondaur [170]

Equation of the circle is $(x+6)^{2}+(y+10)^{2}=20$ .

Solution:

The endpoints of the diameter of a circle are (–8, –6) and (–4, –14).

Center of the circle = Mid point of the diameter

Mid point formula:

$$P(x, y)=\left(\frac{x_{1}+x_{2}}{2}, \frac{y_{1}+y_{2}}{2}\right)$

Here, $x_1=-8, y_1=-6, x_2=-4, y_2=-14$

$$P(x, y) =\left(\frac{-8-4}{2}, \frac{-6-14}{2}\right)$

$$P(x, y) =\left(\frac{-12}{2}, \frac{-20}{2}\right)$

$$P(x, y) =(-6, -10)$

Center of the circle = (–6, –10)

Radius is the distance between center and any endpoint of the diameter.

To calculate the radius using distance formula.

$r=\sqrt{\left(x_{2}-x_{1}\right)^{2}+\left(y_{2}-y_{1}\right)^{2}}$

Here, $x_1=-6, y_1=-10, x_2=-8, y_2=-6$

$r=\sqrt{\left(-8-(-6)\right)^{2}+\left(-6-(-10)}\right)^{2}}$

$r=\sqrt{(-8+6)^{2}+(-6+10)^{2}}$

$r=\sqrt{(-2)^{2}+(4)^{2}}$

$r=\sqrt{20}$ units

The standard form of the equation of a circle is

$(x-a)^{2}+(y-b)^{2}=r^{2}$ , where (a, b) are center and r is the radius.

Here, center = (–6, –10) and $r=\sqrt{20}$

$(x-(-6))^{2}+(y-(-10))^{2}={(\sqrt{20})} ^{2}$

$(x+6)^{2}+(y+10)^{2}=20$

Equation of the circle is $(x+6)^{2}+(y+10)^{2}=20$ .

4 0

2 years ago

Sketch the graph of and identify the axis of symmetry

kaheart [24]

The axis of symmetry is found within the set of parenthesis with the x. If our h value of the vertex is -4, then the axis of symmetry is x = -4. D is that choice. Cannot graph it here, but your vertex is sitting at (-4, 4), it's an upside down parabola, and some other points on this graph are (-5, 0), (-3, 0), (-6, -12), (-2, -12). You could graph it using those points and the vertex without a problem, I'm sure.

6 0

2 years ago

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