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Sever21 [200]
3 years ago
7

Which triangle would be most helpful in finding the distance between the points (–4, 3) and (1, –2) on the coordinate plane? On

a coordinate plane, a triangle has points (negative 4, 3), (1, negative 2), (5, 3). On a coordinate plane, a triangle has points (negative 4, 2), (negative 1, 4), (1, negative 3). On a coordinate plane, a triangle has points (negative 4, 3), (negative 4, negative 2), (1, negative 2). On a coordinate plane, a line is drawn between points (negative 4, negative 3) and (negative 1, 4).
Mathematics
2 answers:
Rainbow [258]3 years ago
8 0

Answer:

The last one is the answer

Step-by-step explanation:

I got took the test. ( :

strojnjashka [21]3 years ago
7 0

Answer:

On a coordinate plane, a triangle has points (negative 4, 3), (1, negative 2), (5, 3).

Step-by-step explanation:

This coordinate above gives us the best range of values

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Which explicit formula best describes this sequence? (26, 32, 38, 44, 50, . . . )
Dominik [7]

The answer:

f(n) = 26+(n-1)

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3 years ago
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I need help with this??
Marina86 [1]

Select both the 2nd and 4th options!

3 0
3 years ago
How do you do this problem? I need to know how you found the answer.
Alexxandr [17]

to get the equation of a line, we simply need two points, say for the Red one ... notice in the graph the lines passes through (0,2) and (-1,6), so let's use those


\bf (\stackrel{x_1}{0}~,~\stackrel{y_1}{2})\qquad (\stackrel{x_2}{-1}~,~\stackrel{y_2}{6}) \\\\\\ slope = m\implies \cfrac{\stackrel{rise}{ y_2- y_1}}{\stackrel{run}{ x_2- x_1}}\implies \cfrac{6-2}{-1-0}\implies \cfrac{4}{-1}\implies -4 \\\\\\ \begin{array}{|c|ll} \cline{1-1} \textit{point-slope form}\\ \cline{1-1} \\ y-y_1=m(x-x_1) \\\\ \cline{1-1} \end{array}\implies y-2=-4(x-0) \\\\\\ y-2=-4x\implies \blacktriangleright y=-4x+2 \blacktriangleleft


now, for the Blue one, say let's use hmmm it passes through (0,2) and (1.6)


\bf (\stackrel{x_1}{0}~,~\stackrel{y_1}{2})\qquad (\stackrel{x_2}{1}~,~\stackrel{y_2}{6}) \\\\\\ slope = m\implies \cfrac{\stackrel{rise}{ y_2- y_1}}{\stackrel{run}{ x_2- x_1}}\implies \cfrac{6-2}{1-0}\implies \cfrac{4}{1}\implies 4 \\\\\\ \begin{array}{|c|ll} \cline{1-1} \textit{point-slope form}\\ \cline{1-1} \\ y-y_1=m(x-x_1) \\\\ \cline{1-1} \end{array}\implies y-2=4(x-0) \\\\\\ y-2=4x\implies \blacktriangleright y=4x+2 \blacktriangleleft

3 0
3 years ago
Simplify 2/3 (9x - 6)<br> A) 3x – 2<br> B) 6x – 4<br> C) 12x – 8<br> D) 18x – 12
Naily [24]
2/3 (9x - 6)

=2/3(9x) - 2/3(6)
= 6x - 4
answer
B) 6x – 4
8 0
3 years ago
A, b, c, d are the roots of 3x⁴-6x²+2=0. what is the value of a⁴+b⁴+c⁴+d⁴?
Pani-rosa [81]
Hello,

P(x)=x^4-6x²+2=(x-a)(x-b)(x-c)(x-d)
=x^4-(a+b+c+d)x^3+(ab+ac+ad+bc+bd+cd)x^2-(abc+abd+acd+bcd)x+abcd

==>
ab+ac+ad+bc+bd+cd=-6
abc+abd+acd+bcd=0
abcd=2
a+b+c+d=0 ==>(a+b+c+d)²=0=a²+b²+c²+d²+2(ab+ac+ac+bc+bd+cd)
==>a²+b²+c²+d²=0-2*(-6)=12




if a is a root P(a)=0==>a^4-6a²+2=0
if b is a root P(b)=0==>b^4-6b²+2=0
if c is a root P(a)=0==>c^4-6c²+2=0
if d is a root P(a)=0==>d^4-6d²+2=0

==>a^4+b^4+c^4+d^4-6(a²+b²+c²+d²)+4*2=0
==>a^4+b^4+c^4+d^4=-8+6*12=64


6 0
4 years ago
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