The unknown number is n. So n is the unknown variable in the equation.
In order to write an equation in which n is included we must know some dependence of n with other values.
We know that <span>the difference of three times a number and 44 is −19. So, we can write:
3*n-44=-19
3*n=-19+44
3*n=25
n=25/3
</span>
Step-by-step explanation:
Amelia got 56% of 75 so
(56/100)*75=(56*75)/100=4200/100=42
Hayden got 8/15 of 75 so
(8/15)*75=(8*75)/15=600/15=40
and Sophie that got 43 out of 75
so we see that Sophie got the highest points.
Step-by-step explanation:
For no 1
<em>2</em><em>5</em><em> </em><em>-</em><em> </em><em>3x </em><em>=</em><em> </em><em>4</em><em>0</em><em> </em>
<em>-</em><em> </em><em>3x </em><em>=</em><em> </em><em>4</em><em>0</em><em> </em><em>-</em><em> </em><em>2</em><em>5</em><em> </em>
<em>-</em><em> </em><em>3x </em><em>=</em><em> </em><em>1</em><em>5</em><em> </em>
<em> </em><em>-</em><em> </em><em>x </em><em>=</em><em> </em><em>1</em><em>5</em><em> </em><em>/</em><em> </em><em>3</em>
<em>Therefore </em><em> </em><em>x </em><em>=</em><em> </em><em>-</em><em> </em><em>5</em><em> </em>
<em>Now </em><em>for </em><em>no. </em><em> </em><em>2</em>
<em>1</em><em>/</em><em>3</em><em> </em><em>(</em><em> </em><em>x </em><em>-</em><em> </em><em>1</em><em>0</em><em>)</em><em> </em><em>=</em><em> </em><em>-</em><em> </em><em>4</em><em> </em>
<em>(</em><em> </em><em>x </em><em>-</em><em> </em><em>1</em><em>0</em><em> </em><em>)</em><em> </em><em>=</em><em> </em><em>-</em><em> </em><em>1</em><em>2</em><em> </em>
<em>x </em><em>=</em><em> </em><em>-</em><em> </em><em>1</em><em>2</em><em> </em><em>+</em><em> </em><em>1</em><em>0</em>
<em>Therefore </em><em> </em><em>x </em><em>=</em><em> </em><em>-</em><em> </em><em>2</em><em> </em>
<em>Hope </em><em>it </em><em>will </em><em>help </em><em>:</em><em>)</em>
6: 1,3,2,6
7: 1,7
Factors of 6 are 1,3,2,6 because 1x6 (6x1) and 3x2. (2x3)
Factors of 7 are 1,7 because seven is a odd number so you can only multiply it by one and itself. (1x7 or 7x1)
Answer: The required solution is
Step-by-step explanation: We are given to solve the following differential equation :
Let us consider that
be an auxiliary solution of equation (i).
Then, we have
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}.](https://tex.z-dn.net/?f=5m%5E2e%5E%7Bmt%7D%2B3me%5E%7Bmt%7D-2e%5E%7Bmt%7D%3D0%5C%5C%5C%5C%5CRightarrow%20%285m%5E2%2B3y-2%29e%5E%7Bmt%7D%3D0%5C%5C%5C%5C%5CRightarrow%205m%5E2%2B3m-2%3D0~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~%5B%5Ctextup%7Bsince%20%7De%5E%7Bmt%7D%5Cneq0%5D%5C%5C%5C%5C%5CRightarrow%205m%5E2%2B5m-2m-2%3D0%5C%5C%5C%5C%5CRightarrow%205m%28m%2B1%29-2%28m%2B1%29%3D0%5C%5C%5C%5C%5CRightarrow%20%28m%2B1%29%285m-1%29%3D0%5C%5C%5C%5C%5CRightarrow%20m%2B1%3D0%2C~~~~~5m-1%3D0%5C%5C%5C%5C%5CRightarrow%20m%3D-1%2C~%5Cdfrac%7B1%7D%7B5%7D.)
So, the general solution of the given equation is
Differentiating with respect to t, we get
According to the given conditions, we have
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}.](https://tex.z-dn.net/?f=y%5E%5Cprime%280%29%3D2.8%5C%5C%5C%5C%5CRightarrow%20-A%2B%5Cdfrac%7BB%7D%7B5%7D%3D2.8%5C%5C%5C%5C%5CRightarrow%20-5A%2BB%3D14%5C%5C%5C%5C%5CRightarrow%20-5A-A%3D14~~~~~~~~~~~~~~~~~~~~~~~~~~~%5B%5Ctextup%7BUisng%20equation%20%28ii%29%7D%5D%5C%5C%5C%5C%5CRightarrow%20-6A%3D14%5C%5C%5C%5C%5CRightarrow%20A%3D-%5Cdfrac%7B14%7D%7B6%7D%5C%5C%5C%5C%5CRightarrow%20A%3D-%5Cdfrac%7B7%7D%7B3%7D.)
From equation (ii), we get
Thus, the required solution is
