<img src="https://tex.z-dn.net/?f=%5Csqrt%7B0.81%7D" id="TexFormula1" title="\sqrt{0.81}" alt="\sqrt{0.81}" align="absmiddle" cl

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galben [10]

2 years ago

ass="latex-formula">

Mathematics

1 answer:

Mandarinka [93]2 years ago

3 0

Answer: .9

Step-by-step explanation:

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Generate the first 5 terms of this sequence:

AURORKA [14]

<span>We are given f(1) = 0 and f(2) = 1. Going forward, the term is the sum of the two previous terms. f(3) = f(2) + f(1) = 0 + 1 = 1. f(4) = f(3) + f(2) = 1 + 1 = 2 f(5) = f(4) + f(3) = 2 + 1 = 3 This matches answer B.</span>

4 0

2 years ago

Find an equation for the linear function g(x) which is perpendicular to the line 5x-2y=6 and intersects the line 5x-2y=6 at x=12

exis [7]

Answer:

≡

Step-by-step explanation:

8 0

2 years ago

yan [13]

Answer:

7,854,000

because the 308 people would have to be at least 500 to round up to 7,855,000

5 0

1 year ago

What type of polyhedra is this ?????<br>

eimsori [14]

Answer: Cube

Its faces are all squares

Step by step:

6 0

2 years ago

Need help with homework

motikmotik

We have two points describing the diameter of a circumference, these are:

$\begin{gathered} A=(-12,-4) \\ B=(-4,-10) \end{gathered}$

Recall that the equation for the standard form of a circle is:

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

Where (h,k) is the coordinate of the center of the circle, to find this coordinate, we find the midpoint of the diameter, that is, the midpoint between points A and B.

For this we use the following equation:

$M=(\frac{x_1+x_2_{}_{}}{2},\frac{y_1+y_2}{2})$

Now, we replace and solve:

$\begin{gathered} M=(\frac{-12+(-4)}{2},\frac{-4+(-10)}{2} \\ M=(\frac{-12-4}{2},\frac{-4-10}{2}) \\ M=(\frac{-16}{2},\frac{-14}{2}) \\ M=(-8,-7) \end{gathered}$

The center of the circle is (-8,-7), so:

$\begin{gathered} h=-8 \\ k=-7 \end{gathered}$

On the other hand, we must find the radius of the circle, remember that the radius of a circle goes from the center of the circumference to a point on its arc, for this we use the following equation:

$r^2=\Delta x^2+\Delta y^2$

In this case, we will solve the delta with the center coordinate and the B coordinate.

$\begin{gathered} r^2=((-4)-(-8))^2+((-10)-(-7)) \\ r^2=(-4+8)^2+(-10+7)^2 \\ r^2=4^2+(-3)^2 \\ r^2=16+9 \\ r^2=25 \\ r=5 \end{gathered}$

Therefore, the equation for the standard form of a circle is:

$\begin{gathered} (x-(-8))^2+(y-(-7))^2=25 \\ (x+8)^2+(y+7)^2=25 \end{gathered}$

In conclusion, the equation is the following:

$(x+8)^2+(y+7)^2=25$

4 0

2 months ago