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Yanka [14]
3 years ago
14

Seth solved a quadratic equation. His work is shown below, with Step 3 missing.

Mathematics
1 answer:
Lyrx [107]3 years ago
3 0

Answer:

|x - 1| = 6

Step-by-step explanation:

After step 2, we have that:

(x - 1)^2 = 36

and after step 4, we have the results:

x = -5 or x = 7

After step 2, what we can do in step 3 is: use square root in both sides of the equation:

|x - 1| = 6

This equation can be split in two:

x - 1 = 6 -> x = 7

x - 1 = -6 -> x = -5

So Seth could write this equation in step 3:

|x - 1| = 6

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Answer:

its b

Step-by-step explanation:

its b i for some reason got it right

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2 years ago
Patricia has a bank balance of −$54. If she deposits $100, what is her new balance?
Nadya [2.5K]

Answer:

Her new balance would be $46 all you have to do is take -54 and add 100 to it

Step-by-step explanation:

8 0
3 years ago
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12a = 48!

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3 years ago
Suppose you have a right triangle area marked in the ground with a base (8-h) meters long and a height h meters long. the hypote
Aleks [24]

Answer:

5.415m and 2.585m long

Step-by-step explanation:

For a right triangle

hyp^2 = opp^2 + adj^2 (Pythagoras theorem)

Given

hypotenuse = 6m

height(opposite) = h meters

Adjacent = (8-h)m

Substitute into the expression above;

6² = h²+(8-h)²

36 = h²+64-16h+h²

36 = 2h²-16h+64

2h²-16h+64-36 = 0

2h²-16h+28= 0

Divide through by 2

h²-8h+14 = 0

Using the general formula

h = 8±√8²-4(14)/2

h = 8±√64-56/2

h = 8±√8/2

h = 8±2.83/2

h = 8+2.83/2 and 8-2.83/2

h = 10.83/2 and 5.17/2

h = 5.415 and 2.585

hence the length of the height of the right triangle are 5.415m and 2.585m long

7 0
3 years ago
The coordinates of parallelogram UVWZ are U(a, 0), W(c - a, b), and Z(c, 0). Find the coordinates of V without using any new var
UkoKoshka [18]
Check the picture below.

so, as you can see, the UV segment is parallel to ZW, and therefore, they're the same slope, hmmm wait just a second, what is the slope of ZW anyway?

\bf \begin{array}{ccccccccc}
&&x_1&&y_1&&x_2&&y_2\\
%  (a,b)
&Z&(~ c &,& 0~) 
%  (c,d)
&W&(~ c-a &,& b~)
\end{array}
\\\\\\
% slope  = m
slope =  m\implies 
\cfrac{\stackrel{rise}{ y_2- y_1}}{\stackrel{run}{ x_2- x_1}}\implies \cfrac{b-0}{(c-a)-c}\implies \cfrac{b}{-a}

since now we know the ZW slope, we also know what is the slope for UV, thus,

\bf \begin{array}{ccccccccc}
&&x_1&&y_1&&x_2&&y_2\\
%  (a,b)
&U&(~ a &,& 0~) 
%  (c,d)
&V&(~ x &,& y~)
\end{array}
\\\\\\
% slope  = m
slope =  m\implies 
\cfrac{\stackrel{rise}{ y_2- y_1}}{\stackrel{run}{ x_2- x_1}}\implies \cfrac{y-0}{x-a}~~=~~\stackrel{ZW's~slope}{\cfrac{b}{-a}}
\\\\\\
\begin{cases}
y-0=b\implies \boxed{y=b}\\
-----------\\
x-a=-a\implies \boxed{x=0}
\end{cases}\qquad \qquad V~(0,b)

8 0
3 years ago
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