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3 years ago

What is the LCD of 1/6 and 3/10

1 answer:

6 0

The LCD is 30.

Think of it this way. You and I are on an assembly line checking i-pads. Your job is to quality check every 6th one and my job is to check every 10th one.
Here are the ones you will check:

6, 12, 18, 24, 30 and so on
Here are the ones I will check
10, 20 30 and so on.

Notice the first one we both check? #30 - that is the LCD of 6 and 10

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PLEASE HELP FAST :(

Ket [755]

Write the equation of the line passing through the points (-7,5) and (-5,9):
$y-5= \dfrac{9-5}{-5-(-7)} (x-(-7)) \\ y-5= \dfrac{4}{2}(x+7) \\ y-5=2(x+7)$ .
You also have another two points (-3,13) and (-1,17). Look whether coordinates of these points satisfy the line equation:

1. For (-3,13) you have
$13-5=2(-3+7) \\ \{8=8\}$ ;

2. For (-1,17) you have
$17-5=2(-1+7) \\ \{12=12\}$ .

Conclusion: All four points lie on the line y-5=2(x+7), so the relationship shown by the data is linear.

Answer: Correct choice is D.

4 0

2 years ago

Find the? inverse, if it? exists, for the given matrix. [4 3] [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

Someone please helpp meeee w this like i don’t understand

MrMuchimi

Answer:

Step-by-step explanation:

it cant pass since multiple points intersect

3 0

2 years ago

Suppose the line of best fit is being found for some data points that have an r-value of 0.657. If the standard deviation of the

oksian1 [2.3K]

We are given a line with the following data:

r-value = 0.657 (r)
standard deviation of x-coordinates = 2.445 (Sx)
standard deviation of y-coordinates = 9.902 (Sy)

We are asked to find the slope of the line up to 3 decimal places.

To find the slope of the line, based on the data that we have, we can use this formula:

slope, b = r * (Sy / Sx)

substitute the values to the formula:

b = 0.657 * ( 9.902 / 2.445 )

Solve for the b.

Therefore, the slope of the line is

b = 2.66078, round off to three decimal places:

b = 2.661 is the slope of the line.

5 0

3 years ago

Read 2 more answers

Question 13 of 20 :

grin007 [14]

Answer:

C. 70.4

Step-by-step explanation:

22% × 320 =

(22 ÷ 100) × 320 =

(22 × 320) ÷ 100 =

7,040 ÷ 100 =

70.4

8 0

3 years ago