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salantis [7]
3 years ago
14

Write the equation y= -1/3x -5 in standard form.

Mathematics
1 answer:
melomori [17]3 years ago
5 0

Answer:

Standard form is another way to write slope-intercept form (as opposed to y=mx+b). It is written as Ax+By=C. You can also change slope-intercept form to standard form like this: Y=-3/2x+3. Next, you isolate the y-intercept(in this case it is 3) like this: Add 3/2x to each side of the equation to get this: 3/2x+y=3.

Step-by-step explanation:

Standard form is another way to write slope-intercept form (as opposed to y=mx+b). It is written as Ax+By=C. You can also change slope-intercept form to standard form like this: Y=-3/2x+3. Next, you isolate the y-intercept(in this case it is 3) like this: Add 3/2x to each side of the equation to get this: 3/2x+y=3.

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The anwser is a hope it helps
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3 years ago
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I really am confused on how to write a proof.
kvasek [131]

Since ABCD is a parallelogram, the opposite sides will be parallel and equal,

\begin{gathered} AB=CD \\ BC=AD \end{gathered}

Consider that AC acts as a transversal to the parallel lines AB and CD, so we can write,

\begin{gathered} \angle CAD=\angle ACB\text{ (Alternate Interior Angles)} \\ BC=AD\text{ (Opposite sides of parallelogram)} \\ \angle ADB=\angle CBD\text{ (Alternate Interior Angles)} \end{gathered}

So by the ASA criteria, the triangle AED is congruent to the triangle CEB,

Then the corresponding parts of the triangles will be equal,

\begin{gathered} AE=CE \\ BE=DE \end{gathered}

Hence Proved.

8 0
1 year ago
A ball is thrown up in the air, and it's height (in feet) as a function of time (in seconds) can be written as h(t) = -16t2 + 32
olga2289 [7]

Answer:

1) Using properties of the quadratic equation:

Here the height equation is:

h(t) = -16t^2 + 32t + 6

We can see that the leading coefficient is negative, this means that the arms of the graph will open downwards.

Then, the vertex of the quadratic equation will be the maximum.

Remember that for a general quadratic equation:

a*x^2 + b*x + c = y

the x-value of the vertex is:

x = -b/2a

Then in our case, the vertex is at:

t = -32/(2*-16) = 1

The maximum height will be the height equation evaluated in this time:

h(1) = -16*1^2 + 32*1 + 6 = 22

The maximum height is 22ft

2) Second method, using physics.

We know that an object reaches its maximum height when the velocity is equal to zero (the velocity equal to zero means that, at this point, the object stops going upwards).

If the height equation is:

h(t) = -16*t^2 + 32*t + 6

the velocity equation is the first derivation of h(t)

Remember that for a function f(x) = a*x^n

we have that:

df(x)/dx = n*a*x^(n-1)

Then:

v(t) = dh(t)/dt = 2*(-16)*t + 32 + 0

v(t) = -32*t + 32

Now we need to find the value of t such that the velocity is equal to zero:

v(t) = 0 = -32*t + 32

       32*t = 32

            t = 32/32 = 1

So the maximum height is at t = 1

(same as before)

Now we just need to evaluate the height equation in t = 1:

h(1) = -16*1^2 + 32*1 + 6 = 22

The maximum height is 22ft

4 0
3 years ago
What is 18 2/3-3 5/8
xxTIMURxx [149]

Answer:

=15 1/24

Step-by-step explanation:

18 2/3−3 5/8

=15 1/24

5 0
3 years ago
How do I solve this problem?
Anvisha [2.4K]

Answer:

y+11=-\frac{1}{4}(x-4)

Step-by-step explanation:

since the equation is perpendicular to y=4x-2, m=-1/4 (negative reciprocal). (x_1,y_1)=(4,-11), plug all the values into the equation y- y_1 = m(x-x_1) and we get y+11=-\frac{1}{4}(x-4)

8 0
3 years ago
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