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3 years ago

What is c3 - 1 when c = 4

Mathematics

2 answers:

den301095 [7]3 years ago

6 0

The answer to your question is 11.
4x3=12-1=11

weeeeeb [17]3 years ago

4 0

You just have to substitute
c3-1
4*3-1
12-1
=11

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P divided by 1 2/3 = 2 1/5

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Find the distance between the points. (Round your answer to two decimal places. (9.7, - 2.8), (- 3.2, 8.8)

Dmitry [639]

Answer:

Rounding it to two decimal places, we get distance, $d=17.35$

Step-by-step explanation:

Given:

The two points are $(9.7, -2.8)\textrm{ and }(-3.2, 8.8)$

The distance between the two points can be obtained using the distance formula which is given as:

$d=\sqrt{(x_{2}-x_{1})^2+(y_{2}-y_{1})^2}$

Here, for the points, $(9.7, -2.8)\textrm{ and }(-3.2, 8.8)$

$x_{1}=9.7,x_{2}=-3.2,y_{1}=-2.8,y_{2}=8.8$

Therefore, the distance between the points is:

$d=\sqrt{(-3.2-9.7)^2+(8.8-(-2.8))^2}\\d=\sqrt{(-12.9)^2+(8.8+2.8)^2}\\d=\sqrt{(12.9)^2+(11.6)^2}\\d=\sqrt{166.41+134.56}\\d=\sqrt{300.97}=17.348$

Rounding it to two decimal places, we get $d=17.35$

7 0

3 years ago

Solve 5y'' + 3y' – 2y = 0, y(0) = 0, y'(0) = 2.8 y(t) = 0 Preview

mario62 [17]

Answer: The required solution is

$y(t)=-\dfrac{7}{3}e^{-t}+\dfrac{7}{3}e^{\frac{1}{5}t}.$

Step-by-step explanation: We are given to solve the following differential equation :

$5y^{\prime\prime}+3y^\prime-2y=0,~~~~~~~y(0)=0,~~y^\prime(0)=2.8~~~~~~~~~~~~~~~~~~~~~~~~(i)$

Let us consider that

$y=e^{mt}$ be an auxiliary solution of equation (i).

Then, we have

$y^prime=me^{mt},~~~~~y^{\prime\prime}=m^2e^{mt}.$

Substituting these values in equation (i), we get

$5m^2e^{mt}+3me^{mt}-2e^{mt}=0\\\\\Rightarrow (5m^2+3y-2)e^{mt}=0\\\\\Rightarrow 5m^2+3m-2=0~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~[\textup{since }e^{mt}\neq0]\\\\\Rightarrow 5m^2+5m-2m-2=0\\\\\Rightarrow 5m(m+1)-2(m+1)=0\\\\\Rightarrow (m+1)(5m-1)=0\\\\\Rightarrow m+1=0,~~~~~5m-1=0\\\\\Rightarrow m=-1,~\dfrac{1}{5}.$

So, the general solution of the given equation is

$y(t)=Ae^{-t}+Be^{\frac{1}{5}t}.$

Differentiating with respect to t, we get

$y^\prime(t)=-Ae^{-t}+\dfrac{B}{5}e^{\frac{1}{5}t}.$

According to the given conditions, we have

$y(0)=0\\\\\Rightarrow A+B=0\\\\\Rightarrow B=-A~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~(ii)$

and

$y^\prime(0)=2.8\\\\\Rightarrow -A+\dfrac{B}{5}=2.8\\\\\Rightarrow -5A+B=14\\\\\Rightarrow -5A-A=14~~~~~~~~~~~~~~~~~~~~~~~~~~~[\textup{Uisng equation (ii)}]\\\\\Rightarrow -6A=14\\\\\Rightarrow A=-\dfrac{14}{6}\\\\\Rightarrow A=-\dfrac{7}{3}.$

From equation (ii), we get

$B=\dfrac{7}{3}.$

Thus, the required solution is

$y(t)=-\dfrac{7}{3}e^{-t}+\dfrac{7}{3}e^{\frac{1}{5}t}.$

7 0

3 years ago

The width of a rectangle is the length minus 3 units.The area of the rectangle is 10 units. What is the width, in units, of the

labwork [276]

Answer:

2 units

Step-by-step explanation:

It's because the only whole number factor pair to 10 is 2x5 and luckily, it turned out that 5-3=2, so there are no decimals or fractions.

5 0

3 years ago