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Daniel [21]
3 years ago
9

Choose the most appropriate strategy and then use it to solve the quadratic equation.

Mathematics
1 answer:
dsp733 years ago
7 0

 (x-8) ^ 2 = 121
 
(x-8) = + / - root (121)
 x = 8 +/- root (121)
 The solutions are:
 
x1 = 8 + root (121)
 
x2 = 8 - root (121)
 2a ^ 2 = 8a-6
 2a ^ 2-8a + 6 = 0
 a ^ 2-4a + 3 = 0
 (a-1) (a-3) = 0
 The solutions are:
 
a1 = 1
 
a2 = 3
 x ^ 2 + 12x + 36 = 4
 x ^ 2 + 12x + 36-4 = 0
 x ^ 2 + 12x + 32 = 0
 (x + 4) (x + 8) = 0
 The solutions are:
 
x1 = -8
 
x2 = -4
 x ^ 2-x + 30 = 0
 
x = (- b +/- root (b ^ 2 - 4 * a * c)) / 2 * a
 x = (- (- 1) +/- root ((- 1) ^ 2 - 4 * (1) * (30))) / 2 * (1)
 x = (1 +/- root (1 - 120))) / 2
 x = (1 +/- root (-119))) / 2
 x = (1 +/- root (119) * i)) / 2
 The solutions are:
 
x1 = (1 + root (119) * i)) / 2
 
x2 = (1 - root (119) * i)) / 2
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90 plus what equals 395
rodikova [14]
305 plus 90= 395. Found by doing 395-90
5 0
3 years ago
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Will give brainliest
Anna [14]

Answer: 23.5

Step-by-step explanation: Divide 47.00 by 2 and you get 23.5

hope this helps

7 0
2 years ago
The coordinates of rhombus ABCD are A(–4, –2), B(–2, 6), C(6, 8), and D(4, 0). What is the area of the rhombus? Round to the nea
a_sh-v [17]
Check the picture below.

so the rhombus has the diagonals of AC and BD, now keeping in mind that the diagonals bisect each, namely they cut each other in two equal halves, let's find the length of each.

\bf ~~~~~~~~~~~~\textit{distance between 2 points}
\\\\
A(\stackrel{x_1}{-4}~,~\stackrel{y_1}{-2})\qquad 
C(\stackrel{x_2}{6}~,~\stackrel{y_2}{8})\qquad \qquad 
%  distance value
d = \sqrt{( x_2- x_1)^2 + ( y_2- y_1)^2}
\\\\\\
AC=\sqrt{[6-(-4)]^2+[8-(-2)]^2}\implies AC=\sqrt{(6+4)^2+(8+2)^2}
\\\\\\
AC=\sqrt{10^2+10^2}\implies AC=\sqrt{10^2(2)}\implies \boxed{AC=10\sqrt{2}}\\\\
-------------------------------

\bf ~~~~~~~~~~~~\textit{distance between 2 points}
\\\\
B(\stackrel{x_1}{-2}~,~\stackrel{y_1}{6})\qquad 
D(\stackrel{x_2}{4}~,~\stackrel{y_2}{0})\qquad \qquad BD=\sqrt{[4-(-2)]^2+[0-6]^2}
\\\\\\
BD=\sqrt{(4+2)^2+(-6)^2}\implies BD=\sqrt{6^2+6^2}
\\\\\\
BD=\sqrt{6^2(2)}\implies \boxed{BD=6\sqrt{2}}

that simply means that each triangle has a side that is half of 10√2 and another side that's half of 6√2.

namely, each triangle has a "base" of 3√2, and a "height" of 5√2, keeping in mind that all triangles are congruent, then their area is,

\bf \stackrel{\textit{area of the four congruent triangles}}{4\left[ \cfrac{1}{2}(3\sqrt{2})(5\sqrt{2}) \right]\implies 4\left[ \cfrac{1}{2}(15\cdot (\sqrt{2})^2) \right]}\implies 4\left[ \cfrac{1}{2}(15\cdot 2) \right]
\\\\\\
4[15]\implies 60

7 0
2 years ago
Read 2 more answers
If f(x)= -3x -6and f(n)=9, what is the value of n?
Alchen [17]
Try to use photo math it gives you the work and answer
7 0
3 years ago
One side of the triangle is 8 in, and the other two sides are equal to a positive whole number of inches. What is the smallest p
frutty [35]
In a triangle the sum of two sides is always greater than the third side.

If one side = 8 in and other two sides are equal to a positive whole number, then the smallest possible sum of the other two sides = 9 in ⇒
<span>Smallest possible perimeter = 8 + 9 = 17 in</span>
6 0
2 years ago
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