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Kazeer [188]
3 years ago
7

Trying to help my bro solve this Y= -2x+2

Mathematics
2 answers:
lesya [120]3 years ago
6 0
The slope-intercept form is <span><span>y=mx+b</span><span>y=mx+b</span></span>, where <span>mm</span> is the slope and <span>bb</span> is the y-intercept.<span><span>y=mx+b</span><span>y=mx+b</span></span>Find the values of <span>mm</span> and <span>bb</span> using the form <span><span>y=mx+b</span><span>y=mx+b</span></span>.<span><span>m=2</span><span>m=2</span></span><span><span>b=2</span><span>b=2</span></span>The slope of the line is the value of <span>mm</span>, and the y-intercept is the value of <span>bb</span>.Slope: <span>22</span>Y-Intercept: <span>2</span>
Alexus [3.1K]3 years ago
3 0
Are you trying to graph it because it is in y=mx+b form.

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It takes
kati45 [8]

Answer:

I don't really know if this is a trick question or not but, is 6363 minutes the answer?

Step-by-step explanation:

8 0
3 years ago
G(x) = -2x^3 – 15x^2 + 36x
shusha [124]

Consider the function G(x) = -2x^3 - 15x^2 + 36x. First, factor it:

G(x) = -2x^3 - 15x^2 + 36x=-x(2x^2+15x-36)=\\ \\=-x\cdot 2\cdot \left(x-\dfrac{-15-\sqrt{513}}{4}\right)\cdot \left(x-\dfrac{-15+\sqrt{513}}{4}\right).

The x-intercepts are at points \left(\dfrac{-15-\sqrt{513} }{4},0\right),\ (0,0),\ \left(\dfrac{-15+\sqrt{513} }{4},0\right).

1. From the attached graph you can see that

  • function is positive for x\in \left(-\infrty, \dfrac{-15-\sqrt{513} }{4}\right)\cup \left(0,\dfrac{-15+\sqrt{513} }{4}\right);
  • function is negative for x\in \left(\dfrac{-15-\sqrt{513} }{4},0\right)\cup \left(\dfrac{-15+\sqrt{513} }{4},\infty\right).

2. Since

G(-x) = -2(-x)^3 - 15(-x)^2 + 36(-x)=2x^3-15x^2-36x\neq G(x)\ \text{and }\neq -G(x) the function is neither even nor odd.

3. The domain is x\in (-\infty,\infty), the range is y\in (-\infty,\infty).

8 0
3 years ago
Read 2 more answers
A log is floating on swiftly moving water. A stone is dropped from rest from a 50.8-m-high bridge and lands on the log as it pas
TiliK225 [7]

Answer:

Horizontal distance between the log and the bridge when the stone is released = 17.24 m

Step-by-step explanation:

Height of bridge, h = 50.8 m

Speed of log = 5.36 m/s

We need to find the horizontal distance between the log and the bridge when the stone is released, for that first we need to find time taken by the stone to reach on top of log,

We have equation of motion. s = ut + 0.5 at²

       Initial velocity, u = 0 m/s

       Acceleration, a = 9.81 m/s²

       Displacement, s = 50.8 m

Substituting,

                  s = ut + 0.5 at²

                 50.8 = 0.5 x 9.81 x t²

                     t = 3.22 seconds,

So log travels 3.22 seconds at a speed of 5.36 m/s after the release of stone,

We have equation of motion. s = ut + 0.5 at²

       Initial velocity, u = 5.36 m/s

       Acceleration, a = 0 m/s²

       Time, t = 3.22 s

Substituting,

                  s = ut + 0.5 at²

                 s = 5.36 x 3.22 + 0.5 x 0 x 3.22²

                    s = 17.24 m

Horizontal distance between the log and the bridge when the stone is released = 17.24 m

3 0
3 years ago
If y = kx, where k is a constant,<br> and y = 30 when x = 6, what is<br> the value of y when x = 3?
Svetradugi [14.3K]

Answer:

Step-by-step explanation:

Actually it's 11 because 11 is technically the same when added to constant of x =6 = 11

7 0
3 years ago
Sjddkdkkdkdkdkdkdkdkkddkjsjssj
masya89 [10]

Answer:

PLESS WRITE A CORRECT QUESTION THAS I CAN HELP YOU

8 0
3 years ago
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