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3 years ago

Mathematics, Help, not sure about my answer on this one

Mathematics

1 answer:

olga nikolaevna [1]3 years ago

3 0

Answer:

F. 2 1/9

Step-by-step explanation:

Step 1: Substitute q

$r = 3(\frac{1}{3^n} )+2$

When you plug in <em>n</em> = 1, 2, 3:

n(1) = 3

n(2) = 2.11111

n(3) = 2.3333

So our answer is F.

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mixas84 [53]

Answer:

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Step-by-step explanation:

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2 years ago

What is the value of x?

gtnhenbr [62]

Value of x is used to consider unknown value. The letter “x” is commonly used in algebra to indicate an unknown value. It is referred to as a “variable” or, in some cases, a “unknown.” In x + 2 = 7, x is a variable

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2 years ago

Read 2 more answers

Consider the system of equations.

Harman [31]

The solution to the system of equations x + 2y = 1 and -3x-2y = 5 is:

x = -3, y = 2

The given system of equations:

x + 2y = 1............(1)

-3x - 2y = 5..........(2)

This can be written in matrix form as shown:

$\left[\begin{array}{ccc}1&2\\-3&-2\end{array}\right] \left[\begin{array}{ccc}x\\y\end{array}\right] = \left[\begin{array}{ccc}1\\5\end{array}\right]$

Find the determinant of $\left[\begin{array}{ccc}1&2\\-3&-2\end{array}\right]$

$\triangle = 1(-2) - 2(-3)\\\triangle = -2+6\\\triangle = 4$

$\triangle_x = \left[\begin{array}{ccc}1&2\\5&-2\end{array}\right]\\\triangle_x = 1(-2)-2(5)\\\triangle_x = -2-10\\\triangle_x =-12$

$\triangle_y = \left[\begin{array}{ccc}1&1\\-3&5\end{array}\right]\\\triangle_y = 1(5)-1(-3)\\\triangle_y = 5 + 3\\\triangle_y =8$

$x = \frac{\triangle_x}{\triangle} \\x = \frac{-12}{4} \\x = -3$

$y = \frac{\triangle_y}{\triangle} \\y = \frac{8}{4} \\y = 2$

The solution to the system of equations x + 2y = 1 and -3x-2y = 5 is:

x = -3, y = 2

Learn more here: brainly.com/question/4428059

3 0

2 years ago

Help me. I don't understand. Please help me find X and IK

sertanlavr [38]

Answer:

$x=4\\\\y=6\\\\IK=47$

Step-by-step explanation:

Two Tangent Theorem: Tangents which meet at same point are equal in length.

Here tangents at J and H meet at I.

$Hence\ HI=IJ\\\\y^2-10=4y+2\\\\y^2-4y-12=0\\\\y^2-6y+2y-12=0\\\\y(y-6)+2(y-6)=0\\\\(y-6)(y+2)\\\\y\ can\ not\ be\ negative\ as\ in\ that\ case\ IJ\ will\ be\ negative.\\\\Hence\ y=6\\\\IJ=4y+2=4\times 6+2=26\\\\\\\\Tangents\ at\ J\ and\ L\ meet\ at\ K.\ Use\ two\ tangent\ theorem.\\\\JK=KL\\\\5x+1=6x-3\\\\6x-5x=1+3\\\\x=4\\\\JK=5x+1=5\times 4+1=21\\\\\\IK=IJ+JK=26+21\\\\IK=47$

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3 years ago

If a circle has a radius length of 2x units, what is the circle's circumference?

Brut [27]

Answer: 12.56

Step-by-step explanation:

If I remember correctly, the equation for circumference is C=2r*pi (R representing radius). For this problem, if 2 is your radius 2R would equal 4. Take this number and multiply it by 3.14 (pi) to get your final answer.

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3 years ago