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3 years ago

How do change slope-intercept form to standard form

2 answers:

7 0

Suppose you are given the slope-intercept form y = 5x - 1/16 and are to change it to standard form. Here's what you'd do:

1) multiply all 3 terms by 16 to eliminate the fraction 1/16:

16y = 80x - 1

Subtr. 16y from both sides: 0 = 80x - 16y - 1

This equation is in std. form.

olganol [36]3 years ago

6 0

Example. Convert y=54x+5 y = 5 4 x + 5 , graphed on the right, to standard form. The only fraction is $$ \frac { 5}{ \red 4} $$ so you can multiply everything by 4. Solve for the y-intercept (or "b" in the slope interceptequation) which is 20 in this example.

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The sum of 10 and a product of 7 and a number

NeX [460]

The answer to your question, would be 7x + 10.

Hopefully this helps!

8 0

4 years ago

Please help!!!!!! ill give you brainliest!

olga2289 [7]

Answer:

The team made x<u><</u>53 points total.

Step-by-step explanation:

1/4+1/10 is 7/20 or 0.35 in decimal form. If they both together scored 21 points then you can add another 7/20 (21 points) and have 14/20 (42 points). You have 6/20 more left. If you can solve what 1/20 could be then you would find the answer. Does this helps?

3 0

3 years ago

Read 2 more answers

Which equation is y = 2x2 – 8x + 9 rewritten in vertex form?

pishuonlain [190]

we have

$y=2x^{2} -8x+9$

Group terms that contain the same variable, and move the constant to the opposite side of the equation

$y-9=2x^{2} -8x$

Factor the leading coefficient

$y-9=2(x^{2} -4x)$

Complete the square. Remember to balance the equation by adding the same constants to each side

$y-9+8=2(x^{2} -4x+4)$

$y-1=2(x^{2} -4x+4)$

Rewrite as perfect squares

$y-1=2(x-2)^{2}$

$y=2(x-2)^{2} +1$ -----> equation in vertex form

therefore

the answer is the option C

$y=2(x-2)^{2} +1$

3 0

3 years ago

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For each vector field f⃗ (x,y,z), compute the curl of f⃗ and, if possible, find a function f(x,y,z) so that f⃗ =∇f. if no such f

butalik [34]

$\vec f(x,y,z)=(2yze^{2xyz}+4z^2\cos(xz^2))\,\vec\imath+2xze^{2xyz}\,\vec\jmath+(2xye^{2xyz}+8xz\cos(xz^2))\,\vec k$

Let

$\vec f=f_1\,\vec\imath+f_2\,\vec\jmath+f_3\,\vec k$

The curl is

$\nabla\cdot\vec f=(\partial_x\,\vec\imath+\partial_y\,\vec\jmath+\partial_z\,\vec k)\times(f_1\,\vec\imath+f_2\,\vec\jmath+f_3\,\vec k)$

where $\partial_\xi$ denotes the partial derivative operator with respect to $\xi$ . Recall that

$\vec\imath\times\vec\jmath=\vec k$

$\vec\jmath\times\vec k=\vec i$

$\vec k\times\vec\imath=\vec\jmath$

and that for any two vectors $\vec a$ and $\vec b$ , $\vec a\times\vec b=-\vec b\times\vec a$ , and $\vec a\times\vec a=\vec0$ .

The cross product reduces to

$\nabla\times\vec f=(\partial_yf_3-\partial_zf_2)\,\vec\imath+(\partial_xf_3-\partial_zf_1)\,\vec\jmath+(\partial_xf_2-\partial_yf_1)\,\vec k$

When you compute the partial derivatives, you'll find that all the components reduce to 0 and

$\nabla\times\vec f=\vec0$

which means $\vec f$ is indeed conservative and we can find $f$ .

Integrate both sides of

$\dfrac{\partial f}{\partial y}=2xze^{2xyz}$

with respect to $y$ and

$\implies f(x,y,z)=e^{2xyz}+g(x,z)$

Differentiate both sides with respect to $x$ and

$\dfrac{\partial f}{\partial x}=\dfrac{\partial(e^{2xyz})}{\partial x}+\dfrac{\partial g}{\partial x}$

$2yze^{2xyz}+4z^2\cos(xz^2)=2yze^{2xyz}+\dfrac{\partial g}{\partial x}$

$4z^2\cos(xz^2)=\dfrac{\partial g}{\partial x}$

$\implies g(x,z)=4\sin(xz^2)+h(z)$

Now

$f(x,y,z)=e^{2xyz}+4\sin(xz^2)+h(z)$

and differentiating with respect to $z$ gives

$\dfrac{\partial f}{\partial z}=\dfrac{\partial(e^{2xyz}+4\sin(xz^2))}{\partial z}+\dfrac{\mathrm dh}{\mathrm dz}$

$2xye^{2xyz}+8xz\cos(xz^2)=2xye^{2xyz}+8xz\cos(xz^2)+\dfrac{\mathrm dh}{\mathrm dz}$

$\dfrac{\mathrm dh}{\mathrm dz}=0$

$\implies h(z)=C$

for some constant $C$ . So

$f(x,y,z)=e^{2xyz}+4\sin(xz^2)+C$

3 0

4 years ago

The period of a function is 4 pi how many cycles of the function occur in a horizontal length of 12 pi

eduard

Answer:

3 cycles

Step-by-step explanation:

in one period, you have one full cycle

because the period for this function is 4pi, you have one cycle per 4pi

if you have a total of 12pi, you would have 3 cycles (12pi/4pi=3)

hope this helps!

3 0

3 years ago

Read 2 more answers