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wariber [46]
3 years ago
10

55-8= 67+8= 99-4= 456-78= plz help me I need this done in one hour

Mathematics
2 answers:
bixtya [17]3 years ago
7 0
1. 47
2. 75
3.95
4.378

Hopefully I helped
kkurt [141]3 years ago
4 0
55 - 8 = 67 + 8 = 99 - 4 = 456 - 78
47 = 75 = 95 = 378
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Rectangle ABCD has vertices at (-7,-2); (1, -2); (1, -8); and (-7, -8) respectively. If GHJK is a similar rectangle where G(2, 5
Minchanka [31]

Answer:

J (6,2) K (2,2)

Step-by-step explanation:

Now distance of AB = [(y2-y1)^2 + (x2-x1)^2]^(1/2)

                                = 8 units

CB =  [(y2-y1)^2 + (x2-x1)^2]^(1/2)

= 6 units by using same formula

Now AB/CB must equal to GH/HJ as rectangles are similar

GH =  [(y2-y1)^2 + (x2-x1)^2]^(1/2)

= 4 units

so

8/6 = 4/HJ

So,

HJ = 3 units

Now if we see the coordinates given carefully, it is obvious that two perpendicular lines lie perfectly parallel to x and y coordinates in rectangle ABCD.  A is (-7,-2) and B is (1,-2) which means distance along y-axis doesn't change. Similarly for C (1,-8) and D (-7,-8), one can see that distance between y-axis doesn't change. So lines AB and CD of rectangle are parallel with x and AD and BC are parallel with y-axis.

In rectangle GHJK one can see that in given coordinates, G(2,5) and H(6,5), y coordinate is same so it is parallel to x axis. Now, HJ is perpendicular to GH so it must be parallel to y axis. It means if we know the lengths of sides we can easily determine unknown coordinates by simple addition and subtraction.

So,  we know HJ = 3 units

J is (6,2) since HJ is parallel to y axis so distance on x axis will remain unchanged and length of line HJ will effect distance of y axis.

Similarly K is (2,2) for the same reason.

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Hi, can you help to find (all the roots/zeros), please!!!
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Solution

Given the quadratic equation

x^2-2x-38=0

we need to find the zeros of the equation

To do that, we use the completing the square method

Step 1. Add 38 to both sides

\begin{gathered} \Rightarrow x^2-2x-38+38=38 \\  \\ \Rightarrow x^2-2x=38 \end{gathered}

Step 2: add the square of half of the coefficient of x to both sides

That is;

\begin{gathered} \Rightarrow x^2-2x+(\frac{1}{2}\cdot(-2))^2=38+(\frac{1}{2}\cdot(-2))^2 \\  \\ \Rightarrow x^2-2x+1=38+1 \\  \\ \Rightarrow x^2-2x+1=39 \\  \\ \Rightarrow(x-1)^2=39 \end{gathered}

Step 3: Simplify the above expression;

\begin{gathered} \Rightarrow x-1=\pm\sqrt[]{39} \\  \\ \Rightarrow x=1\pm\sqrt[]{39} \end{gathered}

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