How to solve step by step : ax2-3bxy-axy+3by2 (the 2's are squares)

Two numbers have a sum of 1,212 and a difference of 518. What are the two numbers?

sashaice [31]

Answer:

347 and 865

Step-by-step explanation:

Let the two number be x and y

Therefore, from the given information:

x + y = 1212

x - y = 518

Rewrite both equations to make x the subject:

x = 1212 - y

x = 518 + y

Equate equations and solve for y:

1212 - y = 518 + y

694 = 2y

y = 347

Substitute found value of y into one of the equations and solve for x:

x = 518 + 347

x = 865

Therefore, the two numbers are 347 and 865

5 0

3 years ago

Skylar is going to invest $69,000 and leave it in an account for 6 years. Assuming the

SSSSS [86.1K]

$~~~~~~ \textit{Continuously Compounding Interest Earned Amount} \\\\ A=Pe^{rt}\qquad \begin{cases} A=\textit{accumulated amount}\dotfill & \$105000\\ P=\textit{original amount deposited}\dotfill & \$69000\\ r=rate\to r\%\to \frac{r}{100}\\ t=years\dotfill &6 \end{cases}$

$105000=69000e^{\frac{r}{100}\cdot 6}\implies \cfrac{105000}{69000}=e^{\frac{3r}{50}}\implies \cfrac{7}{46}=e^{\frac{3r}{50}} \\\\\\ \log_e\left( \cfrac{35}{23} \right)=\log_e\left( e^{\frac{3r}{50}} \right)\implies \ln\left( \cfrac{35}{23} \right)=\cfrac{3r}{50}\implies 50\cdot \ln\left( \cfrac{35}{23} \right)=3r \\\\\\ \cfrac{50\cdot \ln\left( \frac{35}{23} \right)}{3}=r\implies 6.9976\approx r\implies \stackrel{\%}{7.00}\approx r$

8 0

3 years ago

Enter the missing numbers in the boxes to complete the table of equivalent ratios of time to distance.

andre [41]

Answer: Here is the complete table, with the filled in values:
______________________________________________________________
Time (h) Distance (mi)
  3 2
9 6
12 8
18   12
___________________________________________________

Explanation:
___________________________________________________
Let us begin by obtaining the "?" value; that is, the "distance" (in "mi.") ;
when the time (in "h") is "18" ;
___________________________________________________

12/8 = 18/?

Note: "12/8 = (12÷4) / (8÷4) = 3/2 ;

Rewrite: 3/2 = 18/? ; cross-multiply: 3*? = 2 * 18 ;
3*? = 36 ;

Divide each side by "3" ;
The "?" = 36/3 = 12 ;

So, 12/8 = 18/12 ;

The value: "12" takes the place for the "?" in the table for "distance (in "mi.);
when the "time" (in "h") is "18".
__________________________________________________________
Now, let us obtain the "? " value for the "distance" (in "mi.");
when the "time" (in "h") is: "9" .

12/8 = 9/? ; Solve for "?" ;

We know (see aforementioned) that "12/8 = 3/2" ;

So, we can rewrite: 3/2 = 9/? ; Solve for "?" ;

Cross-multiply: 3* ? = 2* 9 ; 3* ? = 18 ;
Divide each side by "3" ;
to get: "6" for the "?" value.

When the time (in "h") is "9", the distance (in "mi.") is "6" .
____________________________________________________
Now, to solve the final "?" value in the table given.

9/6 = ?/2 ; Note: We get the "6" from our "calculated value" (see above problem).

9/6 = (9÷3) / (6÷3) = 3/2 ;

So, we know that the "?" value is: "3" .

Alternately: 9/6 = ?/2 ;

Cross-multiply: 6*? = 2*9 ; 6 * ? = 18 ; Divide each side by "6" ;
to find the value for the "?" ;
"?" = 18/6 = "3" .

When the "distance" (in "mi.") is: "2" ; the time (in "h") is: "3" .
____________________________________________________
Here is the complete table—with all the values filled in:
____________________________________________________

<span>Time (h) Distance (mi)
____________________________________________________
3 2
9 6
12 8
18   12
____________________________________________________</span>

6 0

4 years ago

Simplify the expression:<br> 9–7k+9k–k

vodomira [7]

9+k I am pretty sure

8 0

3 years ago

The number of messages that arrive at a Web site is a Poisson distributed random variable with a mean of 6 messages per hour. Ro

QveST [7]

Answer:

a) There is a 16.0623% probability that 5 messages are received in 1 hour.

b) There is a 11.5880% probability that 10 messages are received in 1.5 hours.