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tatiyna
3 years ago
15

What is the sum of an infinite geometric series if the first term is 81 and the common ratio is 2⁄3?

Mathematics
1 answer:
Marrrta [24]3 years ago
4 0

C

the sum to infinity of a geometric series  = \frac{a}{1-r}

where a is the first term and r the common ratio

= \frac{81}{\frac{1}{3} } = 81 × 3 = 243



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3 years ago
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Please help I am terrible at algebra
Marina CMI [18]

Answer:

y = (x-0)^2 + (-5)  ⇒ y = x^2 - 5

Step-by-step explanation:

The general vertex form of the parabola  y = a(x - h)² + k

Where (h,k) is the coordinates of the vertex.

As shown at the graph the vertex of the parabola is the point (0, -5)

So,

y = a(x-0) + (-5)

y = ax^2 - 5

To find substitute with another point from the graph like (1,-4)

So, at x = 1 ⇒ y = -4

-4 = a * 1^2 - 5

a = -4 + 5 = 1

<u>So, the equation of the given parabola is ⇒ y = x^2 - 5</u>

6 0
3 years ago
Classify each as an expression or an equation.
fenix001 [56]

Answer:

1. equation

2. equation

3. expression

4. expression

explanation is that an equation has an equal sign and an expression doesnt .  

5. equation

Step-by-step explanation:

6 0
3 years ago
Find the? inverse, if it? exists, for the given matrix.<br><br> [4 3]<br><br> [3 6]
True [87]

Answer:

Therefore, the inverse of given matrix is

=\begin{pmatrix}\frac{2}{5}&-\frac{1}{5}\\ -\frac{1}{5}&\frac{4}{15}\end{pmatrix}

Step-by-step explanation:

The inverse of a square matrix A is A^{-1} such that

A A^{-1}=I where I is the identity matrix.

Consider, A = \left[\begin{array}{ccc}4&3\\3&6\end{array}\right]

\mathrm{Matrix\:can\:only\:be\:inverted\:if\:it\:is\:non-singular,\:that\:is:}

\det \begin{pmatrix}4&3 \\3&6\end{pmatrix}\ne 0

\mathrm{Find\:2x2\:matrix\:inverse\:according\:to\:the\:formula}:\quad \begin{pmatrix}a\:&\:b\:\\ c\:&\:d\:\end{pmatrix}^{-1}=\frac{1}{\det \begin{pmatrix}a\:&\:b\:\\ c\:&\:d\:\end{pmatrix}}\begin{pmatrix}d\:&\:-b\:\\ -c\:&\:a\:\end{pmatrix}

=\frac{1}{\det \begin{pmatrix}4&3\\ 3&6\end{pmatrix}}\begin{pmatrix}6&-3\\ -3&4\end{pmatrix}

\mathrm{Find\:the\:matrix\:determinant\:according\:to\:formula}:\quad \det \begin{pmatrix}a\:&\:b\:\\ c\:&\:d\:\end{pmatrix}\:=\:ad-bc

4\cdot \:6-3\cdot \:3=15

=\frac{1}{15}\begin{pmatrix}6&-3\\ -3&4\end{pmatrix}

=\begin{pmatrix}\frac{2}{5}&-\frac{1}{5}\\ -\frac{1}{5}&\frac{4}{15}\end{pmatrix}

Therefore, the inverse of given matrix is

=\begin{pmatrix}\frac{2}{5}&-\frac{1}{5}\\ -\frac{1}{5}&\frac{4}{15}\end{pmatrix}

4 0
4 years ago
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Svet_ta [14]

Step-by-step explanation:

first you do 3×1 get your answer and then × it by 2 then your answer you get you × it by 3 what you get - by 2 what you get you - by 1 and you get your answer hope I helped

3 0
3 years ago
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