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3 years ago

The terms shown below are the first three terms of a geometric sequence whose common ratio is positive.

Mathematics

1 answer:

vichka [17]3 years ago

3 0

Answer:

a. 4

b. 5

Step-by-step explanation:

a. r = √(112/7) =√16 =4

b. 4x+8 = 28

4x = 28-8

4x = 20

x= 5

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Find the factors of function f, and use them to complete this statement.

Crank

Answer:

f(x)= 2x^4 - x^3=24 -18x^2 + 9x= 45 45-24=21

Step-by-step explanation:

just multipkye that

4 0

3 years ago

Read 2 more answers

Write the solution to the given inequality in interval notation.

Mazyrski [523]

Answer:

$(-\infty,4)\cup (5,\infty)$

Step-by-step explanation:

From the number line, the solution to the inequality is x<4 or x>5.

We can write x<4 in interval notation as (-∞,4) and x>5 as (5,∞).

The "or" represents the union of the two intervals.

Therefore the solution to the given inequality in interval notation is:

$(-\infty,4)\cup (5,\infty)$

The third choice is the correct answer.

3 0

4 years ago

Pls answer asap!<br> Nonsense - report<br> Good - brainliest

iren [92.7K]

The solution is

a) 1.412

b) 0.662

What is the place value chart?

A place value chart is a table that is used to determine the value of each digit in a number based on its numerical position. We place the given number in the place value chart to examine the place value of each digit in order to precisely identify the worth of distinct digits in a number.

Given data ,

Let the number be A = 4.236 / 3

So , the number A = 1.412

So , in order to place the number in place value chart

The digit 1 is in ones place

The digit 4 is in tenths place

The digit 1 is in hundredths place

The digit 2 is in thousandths place

Let the number be B = 1.324 / 2

So , the number is B = 0.662

So , in order to place the number in place value chart

The digit 0 is in ones place

The digit 6 is in tenths place

The digit 6 is in hundredths place

The digit 2 is in thousandths place

Hence , the place value chart of numbers 1.412 and 0.662 are solved

To learn more about place value chart click :

brainly.com/question/10759386

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6 0

1 year ago

What are the coordinates of the image of point B, after the segment has been dilated by a scale factor of 3 with a center of dil

shepuryov [24]

Answer: (–9, 9)

Step-by-step explanation:

if the original point (x,y) gets dilated by a scale factor 'k' with a center of dilation at the origin, then

The coordinates of the image point are (kx, ky).

Given: The coordinates of line segment A B are A(-6,8) and B(-3,3).

then , the coordinates of B after dilation by scale factor of 3 with a center of dilation at the origin,

$(-3,3)\to(3(-3),3(3))\\\\\Rightarrow\ (-3,3)\to(-9,9)$

Hence, the coordinates of the image of point B, after the segment has been dilated by a scale factor of 3 with a center of dilation at the origin = (–9, 9).

8 0

3 years ago

Use the following matrices, A, B, C and D to perform each operation.

wolverine [178]

Step-by-step explanation:

$\bold{40.}\\\\A+B=\left[\begin{array}{ccc}3&1\\5&7\end{array}\right] +\left[\begin{array}{ccc}4&1\\6&0\end{array}\right] =\left[\begin{array}{ccc}3+4&1+1\\5+6&7+0\end{array}\right]=\left[\begin{array}{ccc}7&1\\11&7\end{array}\right]$

$\bold{41.}\\\\B-A=\left[\begin{array}{ccc}4&1\\6&0\end{array}\right]-\left[\begin{array}{ccc}3&1\\5&7\end{array}\right]=\left[\begin{array}{ccc}4-3&1-1\\6-5&0-7\end{array}\right]=\left[\begin{array}{ccc}1&0\\1&-7\end{array}\right]$

$\bold{42.}\\\\3C=3\left[\begin{array}{ccc}-2&3&1\\-1&0&4\end{array}\right]=\left[\begin{array}{ccc}(3)(-2)&(3)(3)&(3)(1)\\(3)(-1)&(3)(0)&(3)(4)\end{array}\right]=\left[\begin{array}{ccc}-6&9&3\\-3&0&12\end{array}\right]$

$\bold{43.}\\\\CD=\left[\begin{array}{ccc}-2&3&1\\-1&0&4\end{array}\right]\left[\begin{array}{ccc}-2&3&4\\0&-2&1\\3&4&-1\end{array}\right]\\\\=\left[\begin{array}{ccc}(-2)(-2)+(3)(0)+(1)(3)&(-2)(3)+(3)(-2)+(1)(4)&(-2)(4)+(3)(1)+(1)(-1)\\(-1)(-2)+(0)(0)+(4)(3)&(-1)(3)+(0)(-2)+(4)(4)&(-1)(4)+(0)(1)+(4)(-1)\end{array}\right]\\\\=\left[\begin{array}{ccc}4+0+1&-6-6+4&-8+3-1\\2+0+12&-3+0+16&-4+0-4\end{array}\right]\\\\=\left[\begin{array}{ccc}5&-8&-6\\14&13&-8\end{array}\right]$

$\bold{44.}\\\\2D+3C=2\left[\begin{array}{ccc}-2&3&4\\0&-2&1\\3&4&-1\end{array}\right]+3\left[\begin{array}{ccc}-2&3&1\\-1&0&4\end{array}\right]\\\\=\left[\begin{array}{ccc}(2)(-2)&(2)(3)&(2)(4)\\(2)(0)&(2)(-2)&(2)(1)\\(2)(3)&(2)(4)&(2)(-1)\end{array}\right]+\left[\begin{array}{ccc}(3)(-2)&(3)(3)&(3)(1)\\(3)(-1)&(3)(0)&(3)(4)\end{array}\right]\\\\=\left[\begin{array}{ccc}-4&6&8\\0&-4&2\\6&8&-2\end{array}\right]+\left[\begin{array}{ccc}-6&9&3\\-3&0&12\end{array}\right]$

$\large\bold{You\ can\ not\ add\ matrices\ of\ different\ dimensions!!!}$

4 0

3 years ago