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3 years ago

11xto the 7th power ×6x to the 5th power

Mathematics

1 answer:

solong [7]3 years ago

6 0

The first one is 19,487,171 and the second is 7,776

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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

3 years ago

The length of a rectangle is 1 ft less than double the width, and the area of the rectangle is 21 ft^2. Find the dimensions of t

ICE Princess25 [194]

Let the width be w, then length = 2w - 1

Area = length times width = w(2w - 1) = 21.

Thus, 2w^2 - w = 21 or 2w^2 - w - 21 = 0

Solving the quadratic equation gives that w = 3.5.

Therefore, width is 3.5 ft and length is 2(3.5) - 1 = 6 ft.

4 0

3 years ago

A shipping service restricts the dimensions of the boxes it will ship for a certain type of service. The restriction states that

Igoryamba

Answer:

w =< 70

(width is less or equal to 70 inches)

Step-by-step explanation:

Let l = length, w = width, h = height

Restrictions given in this question:

'sum of perimeter of the base and the height cannot exceed 130 inches'

perimeter of the base is 2 width and 2 length of the box

perimeter = 2w + 2l

Therefore, inequality involves here is

2w + 2l + h =< 130

(Note that =< here means less or equal)

Then a new condition given with

height, h = 60 in

and length is 2.5 times the width

l = 2.5w

Substitute this new condition into the equation will give us the following:

2w + 2(2.5w) + 60 =< 130

2w + 5w + 60 =< 130

7w + 60 =< 130

7w =< 130-60

7w =< 70

w =< 10

4 0

3 years ago

What is the quotient of 3.201 x 109 and 4.85 x 103 expressed in scientific notation?

Ray Of Light [21]

Answer:

=3.201x109

=4.85x103

Step-by-step explanation:

5 0

3 years ago

The amount of money spent on textbooks per year for students is approximately normal.

Ostrovityanka [42]

Answer:a

$336.04 < \mu < 443.96$

The margin of error will increase

The margin of error will decreases

The 99% confidence interval is $0.4107 < p < 0.4293$

Step-by-step explanation:

From the question we are told that

The sample size $n = 19$

The sample mean is $\= x = \$\ 390$

The standard deviation is $\sigma = \$ \ 120$

Given that the confidence level is 95% then the level of significance is mathematically represented as

$\alpha = 100 - 95$

$\alpha = 5 \%$

$\alpha = 0.05$

Next we obtain the critical value of $\frac{\alpha }{2}$ from the normal distribution table

$Z_{\frac{\alpha }{2} } = 1.96$

The margin of error is mathematically represented as

$E = Z_{\frac{\alpha }{2} } * \frac{\sigma}{\sqrt{n} }$

=> $E = 1.96 * \frac{120}{\sqrt{19} }$

=> $E = 53.96$

The 95% confidence interval is

$\= x - E < \mu < \= x + E$

=> $390 - 53.96 < \mu < 390 - 53.96$

=> $336.04 < \mu < 443.96$

When the confidence level increases the $Z_{\frac{\alpha }{2} }$ also increases which increases the margin of error hence the confidence level becomes wider