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aalyn [17]
3 years ago
12

Please help i need verified please help

Mathematics
2 answers:
soldier1979 [14.2K]3 years ago
8 0
The first one is true
this is the equation: y=ax
you can put one of the points on the line as x and y. for example (2,8)
8= a \times 2 \\ 8 = 2a \\ a =  \frac{8}{2}  =  4
kramer3 years ago
5 0

Answer:

4

Step-by-step explanation:

Find the <em>rate</em><em> </em><em>of change</em><em> </em>[<em>slope</em>]:

[0, 0] and [2, 8]

OR

[1, 4] and [0, 0]

-y₁ + y₂\-x₁ + x₂ = m

\frac{-8 + 0}{-2 + 0} = \frac{-8}{-2} = 4 \\  \\ OR \\  \\ \frac{0 - 4}{0 - 1} = \frac{-4}{ -1}  = 4

I am joyous to assist you anytime.

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Triangle ABC is reflected across the y-axis and then translated down 1.5 units to form triangle A'B'C'. Determined if each state
Helga [31]

Answer:

They are all true.

Step-by-step explanation:

Rigid transformations, like reflections and translations, preserve angle measures and lengths.

8 0
3 years ago
While traveling to and from a certain destination, you realized increasing your speed by 40 mph saved 4 hours on your return. If
Luden [163]

Answer:

The speed driven while returning is 88 mph.

Step-by-step explanation:

We are given that while traveling to and from a certain destination, you realized increasing your speed by 40 mph saved 4 hours on your return.

Also, the total distance of the roundtrip was 420 miles.

Let the speed driven while returning be 'x mph' which means that the speed driven while going was '(x - 40) mph' because it has been given that while returning we have increased the speed by 40 mph.

As we know that the Speed-Distance-Time formula is given by;

                         \text {Speed} = \frac{\text{Distance}}{\text{Time}}   or  \text {Time} = \frac{\text{Distance}}{\text{Speed}}

So, according to the question;

\frac{420}{x-40} -\frac{420}{x} = 4 \text{ hours}       where Distance = 420 miles

\frac{420x-420(x-40)}{x(x-40)} = 4

\frac{420x-420x+16800}{x^{2} -40x} = 4

\frac{16800}{x^{2} -40x} = 4

4x^{2} -160x= 16800

4x^{2} -160x- 16800=0

x^{2} -40x- 4200=0

Now finding the roots of the above equation;

Here a = 1, b = -40 and c = -4200

D = b^{2} -4ac

   =  (-40)^{2} -4(1)(-4200) = 18400

Now, the roots of a quadratic equation is given by;

x = \frac{-b\pm \sqrt{D} }{2a}

x = \frac{-(-40)\pm \sqrt{18400} }{2\times 1}

So, the two roots of x are : x = \frac{40-\sqrt{18400} }{2}  and  x = \frac{40+\sqrt{18400} }{2}

Solving these two we get; x = -47.8  and  x = 87.8

Here we ignore the negative value of x, so the speed driven while returning is 87.8 ≈ 88 mph.

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