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3 years ago

14

8=56 7=42 6=30 5=20 3=?

2 answers:

nalin [4]3 years ago

7 0

$8\rightarrow56=8\cdot7\\7\rightarrow42=7\cdot6\\6\rightarrow30=6\cdot5\\5\rightarrow20=5\cdot4\\\\4\rightarrow12=4\cdot3\\\\3\rightarrow6=3\cdot2\\\\Ans.\ 3\rightarrow6$

m_a_m_a [10]3 years ago

3 0

$3=6$

In this case, the pattern is the number, in this case, 3, multiplied by the number below it.

See?
$8\times7=56\\7\times6=42\\6\times5=30\\5\times4=20\\3\times2=6$

Hope this helped :)

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Define the variables set a system of linear questions,solve the system and check your solution.

LenaWriter [7]

7) Larger number= 30
Smaller number= 18

8) He bought 4 videogames and 11 movies.

3 0

3 years ago

Part of arianna’s earnings are based on a 5% commission rate. At the end of the day arianna earned 14.98 in commission. How much

Katarina [22]

Arianna sold for $ 299.6

<em><u>Solution:</u></em>

Given that, Part of Arianna’s earnings are based on a 5% commission rate

At the end of the day arianna earned 14.98 in commission

Therefore,

Commission earned = $ 14.98

Commission rate = 5 %

Let "x" be the amount that Arianna sell that day

Therefore, according to question,

5 % of amount earned = 14.98

5 % of "x" = 14.98

<em><u>Solve the above equation for "x"</u></em>

$5 \% \times x = 14.98\\\\\frac{5}{100} \times x = 14.98\\\\5x = 14.98 \times 100\\\\5x = 1498\\\\Divide\ both\ sides\ of\ equation\ by\ 5\\\\x = 299.6$

Thus arianna sold for $ 299.6

7 0

3 years ago

F(x) = 3x + 1 and f-1 =<br> then f(2)=

soldi70 [24.7K]

Answer:

f(x)= 3x + 1

f - 1 = 3x + 1-1

= 3x

f(x) -1= 3x

then f(2)= 3×2+1

=6+1

f(2)= 7.

f(-1)= 3×(-1) +1

= -3+1

f(-1)= -2

5 0

3 years ago

Use the variation of parameters method to solve the DE y"+y'- 2y=1

kotykmax [81]

Answer:

$y(t)\ =\ C_1e^{-2t}+C_2e^t\ -\ \dfrac{1}{2}$

Step-by-step explanation:

Given differential equation is

y"+y'-2y = 1

$=>\ (D^2\ +\ D\ -\ 2)y\ =\ 1$

Hence, the characteristics is

$D^2+D-2\ =\ 0$

$=>\ D^2\ +\ 2D\ -\ D\ -2\ =\ 0$

$=> (D+2)(D-1) = 0$

=> D = -2, 1

The general equation of the given differential equation is

$y_c(t)\ =\ C_1e^{-2t}+C_2e^t$

Let's consider that

$y_1(t)\ =\ e^{-2t}$ $y_2(t)\ =\ e^{t}$

$y'_1(t)\ =\ -2e^{-2t}$ $y'_2(t)\ =\ e^t$

g(t) = 1

Wronskian can be given by,

W = y_1(t)y'_2(t) - y_2(t)y'_1(t)

$=\ e^{-2t}.e^t\ -\ e^{t}.(-2e^{-2t})$

$=\ e^{-t}\ +\ 2e^{-t}$

$=\ 3e^{-t}$

Now, the particular integral can be given by

$y_p(t)\ = \ -y_1(t)\int{\dfrac{y_2(t).g(t)}{W}dt}\ +\ y_2(t)\int{\dfrac{y_1(t).g(t)}{W}dt}$

$=\ -e^{-2t}\int{\dfrac{e^t\times 1}{3e^{-t}}dt}\ +\ e^t\int{\dfrac{e^{-2t}\times 1}{3e^{-t}}dt}$

$=\ -e^{-2t}\int{\dfrac{e^{2t}}{3}dt}\ +\ e^t\int{\dfrac{e^{-t}}{3}dt}$

$=\ (-e^{-2t})(\dfrac{e^{2t}}{6})\ +\ (e^t)(\dfrac{e^{-t}}{-3})$

$=\ \dfrac{-1}{6}\ -\ \dfrac{1}{3}$

$=\ \dfrac{-3}{6}$

$=\ \dfrac{-1}{2}$

Now,

$y(t)\ =\ y_c(t)\ +\ y_p(t)$

$=\ C_1e^{-2t}+C_2e^t\ -\ \dfrac{1}{2}$

Hence, the complete solution of the given differential equation is

$y(t)\ =\ C_1e^{-2t}+C_2e^t\ -\ \dfrac{1}{2}$

5 0

3 years ago

How to solve and see if my answer choice is correct

joja [24]

the 3rd option is correct.

all the best.

3 0

3 years ago