Solve A=1/2h(b1+b2) Solve for b1

- 2 + 0.8) ÷ (- 1.3) = <img src="https://tex.z-dn.net/?f=%20%28%20-%202%20%2B%200.8%29%20%5Cdiv%20%28%201%7D%5E%7B2%7D%20%20-

Nookie1986 [14]

(-2 + 0.8) ÷ (1²-1.3) =

(-1.2) ÷ (1-1.3)

(-1.2) ÷ (-0.3)

= 4

6 0

3 years ago

One positive number is 14 less than another positive number. If the reciprocal of the smaller number is added to five times the

Kipish [7]

Let's call the two numbers $s$ and $l$ .

Given these variables, we can say: $s = l - 14$ , based on the first sentence in the problem.

Also, remember that the reciprocal of a number is simply 1 divided by the number. Thus, we can say that:

$\dfrac{1}{s} + 5\Big( \dfrac{1}{l} \Big) = \dfrac{1}{4}$

To solve, we can simply substitute $l -14$ in for $s$ in the second equation and solve.

$\dfrac{1}{l - 14} + \dfrac{5}{l} = \dfrac{1}{4}$

Set up

$\dfrac{l}{l(l - 14)} + \dfrac{5(l - 14)}{l(l - 14)} = \dfrac{1}{4}$

Get terms on the left side to a common denominator for easier addition

$\dfrac{l + 5l - 70}{l(l - 14)} = \dfrac{1}{4}$

Simplify

$4(6l - 70) = l(l - 14)$

Cross multiplication ( $\frac{a}{b} = \frac{c}{d} \Rightarrow ad = bc$ )

$24l - 280 = l^2 - 14l$

Simplify

$l^2 - 38l + 280 = 0$

Subtract $24l - 280$ from both sides of the equation

$(l - 10)(l - 28) = 0$

Factor left side of the equation

$l = 10, 28$

Zero Product Property

Now, notice that we have found two solutions, but the problem is only asking for one. This likely means that one of our solutions is extraneous. Let's take a look. Remember that the smaller positive number is equal to 14 less than the larger number. However,

$s = 10 - 14 = -4$ ,

Since $s$ is not positive in this case, $l = 10$ is not a solution.

Thus, $l = 28$ is our only solution. In this case,

$s = 28 - 14 = 14$ ,

which means that the smaller number is 14 and the larger number is 28.

5 0

2 years ago

Please help! Thank you!

KIM [24]

The correct answer is: [B]: " (2, 5) ".
__________________________________________
Given:
__________________________________________
-5x + y = -5 ;
  -4x + 2y = 2 .
___________________________________________
Consider the first equation:
___________________________
-5x + y = -5 ; ↔ y + (-5x) = -5 ;

↔ y - 5x = -5 ; Add "5x" to each side of the equation; to isolate "y" on one side of the equation; and to solve in terms of "y".
_____________________________________________
   y - 5x + 5x = -5 + 5x

   y = -5 + 5x ; ↔ y = 5x - 5 ;
____________________________________________
Now, take our second equation:
______________________________
     -4x + 2y = 2 ; and plug in "(5x - 5)" for "y" ; and solve for "x" :
_____________________________________________________
         -4x + 2(5x - 5) = 2 ;
______________________________________________________
Note, 2(5x - 5) = 2(5x) - 2(5) = 10x - 10 ;
__________________________________________
So:   -4x + 10x - 10 = 2 ;

On the left-hand side of the equation, combine the "like terms" ;

-4x +10x = 6x ; and rewrite:

6x - 10 = 2 ;

Now, add "10" to each side of the equation:

6x - 10 + 10 = 2 + 10 ;

to get:

           6x = 12 ; Now, divide EACH side of the equation by "6" ; to isolate "x" on one side of the equation; and to solve for "x" ;

           6x/6 = 12 / 6 ;

                 x = 2 ;
_________________________________
Now, take our first given equation; and plug our solved value for "x" ; which is "2" ; and solve for "y" ;
_____________________________________
-5x + y = -5 ;

-5(2) + y = -5 ;

-10 + y = -5 ; ↔
                              y - 10 = -5 ;

Add "10" to each side of the equation; to isolate "y" on one side of the equation; and to solve for "y" ;

         y - 10 + 10 = -5 + 10 ;

                   y = 5 .
_____________________________
So, we have, x = 2 ; and y = 5 .
____________________________
Now, let us check our work by plugging in "2" for "x" and "5" for "y" in BOTH the original first and second equations:
______________________________
first equation:

-5x + y = -5 ;

-5(2) + 5 =? -5?

-10 + 5 =? -5 ? YES!
______________________
second equation:

-4x + 2y = 2 ;

-4(2) + 2(5) =? 2 ?

-8 + 10 =? 2 ? Yes!
_______________________________________________________
So, the answer is:
___________________________________________________________
          x = 2 , y = 5 ; or, "(2, 5)" ; which is: "Answer choice: [B] " .
___________________________________________________________

3 0

3 years ago

Which point is the solution to the following system of equations? x2 + y2 = 13 2x - y = 4

Anit [1.1K]

There are two solutions.
.. (3, 2)
.. (0.2, -3.6)

_____
If you're choosing possibilities from a list, trying them in the equations usually gives quick results.

4 0

3 years ago

Read 2 more answers

What is the mean of the values in the dot plot? Enter your answer in the box.

Alona [7]

Answer:45.3333333333333333333333333333333333333333333...

Step-by-step explanation: ad all of them together (272) divide them by 6 =45.33...

8 0

2 years ago

Read 2 more answers