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3 years ago

The coordinates of the vertices of AJKL are J(0, 2), K(3, 1), and L(1, – 5).

Mathematics

1 answer:

bonufazy [111]3 years ago

8 0

9514 1404 393

Answer:

JK slope: -1/3
KL slope: 3
JL slope: -7
is
two of the slopes have a product of -1

Step-by-step explanation:

The slope formula is useful.

m = (y2 -y1)/(x2 -x1)

JK slope = (1 -2)/(3 -1) = -1/3

KL slope = (-5-1)/(1-3) = -6/-2 = 3

JL slope = (-5-2)/(1-0) = -7

We observe that slopes (-1/3) and (3) have a product of -1, so those line segments are perpendicular. The triangle is a right triangle.

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Solve 2x + 5 = 12 please help

kherson [118]

Answer:

first method

subtract 5 from both side

2x + 5 -5 = 12 - 5

:. 2x = 7

:. X= 2

divide 2 from both side

X= 2x 7

2 2

:. X= 3.5

second method

2x = 12 - 5

2x = 7

X= 2 ÷ 7

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3 years ago

The sum and product of zeroes are 4 and 1 respectively then the quadratic equation is

Zinaida [17]

Answer:

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3 years ago

A car travelled 100 km with half the distance at 40 km/h and the other half at 80 km/h. Find the average speed of the car for th

Nana76 [90]

We must look at this question in steps
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50km / 40km/h = 1.25 hours

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6 0

3 years ago

Hi, can someone please help, and possibly explain the answer please.

mr_godi [17]

Answer:

$x=\frac{-(-2)\±\sqrt{(-2)^2-4(3)(0)} }{2(3)}$

Step-by-step explanation:

Quadratic formula: $x=\frac{-b\±\sqrt{b^2-4ac} }{2a}$ when the equation is $0=ax^2+bx+c$

The given equation is $1=-2x+3x^2+1$ . Let's first arrange this so its format looks like $y=ax^2+bx+c$ :

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Subtract 1 from both sides of the equation

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Now, we can easily identify 3 as a, -2 as b and 0 as c. Plug these into the quadratic formula:

$x=\frac{-b\±\sqrt{b^2-4ac} }{2a}\\x=\frac{-(-2)\±\sqrt{(-2)^2-4(3)(0)} }{2(3)}$

I hope this helps!

8 0

3 years ago

You have a 32-foot fence around a square garden. You paint 1/3 of one side of the fence. What fraction of the fence did you pain

dmitriy555 [2]

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8 0

3 years ago