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3 years ago

4a²-8ab+4b=????????????

Mathematics

1 answer:

GuDViN [60]3 years ago

3 0

Step-by-step explanation:

4a² - 8ab + 4b²

= (2a)² - 2 * 2a * 2b + (2b) ²

= (2a - 2b) ²

Hope it helps :)

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2x+3y=1 and y=-2 - 9 solve using linear combination method. Please explain steps.

kaheart [24]

Answer:

$x=17$

Step-by-step explanation:

$2x+3y=1$

$y=-2-9$

In order to combine these two equations, an idea you need to keep in mind is finding a way of setting these equations as equal to each other. I saw that each equation shared a common value, $y$ . In this case, we need to isolate $y$ in the first equation so that both equations $=y$ .

$2x+3y=1$

$3y=-2x+1$

$\frac{3y}{3}=\frac{-2x+1}{3}$

$y=-\frac{2}{3}x+\frac{1}{3}$

With this, we now know that both $-2-9$ and $-\frac{2}{3}x+\frac{1}{3}$ are equal to $y$ , so we can set them equal to each other.

$y=-\frac{2}{3}x+\frac{1}{3}$

$y=-11$

$-\frac{2}{3}x+\frac{1}{3}=-11$

Reply to this if anything I'm saying or doing is confusing in any way, or if you find a mistake. :) Solve for $x$ .

$-\frac{2}{3}x+\frac{1}{3}=-11$

$-\frac{2}{3}x=-11-\frac{1}{3}$

$-\frac{2}{3}x =-\frac{33}{3}-\frac{1}{3}$

$-\frac{2}{3}x =-\frac{34}{3}$

$-\frac{2}{3}x(-\frac{3}{2})=-\frac{34}{3}(-\frac{3}{2})$

$x=\frac{102}{6}=17$

$x=17$

Hopefully this answer is correct AND makes sense in terms of how I achieved it. Again, reply to this with any questions or mistakes I made and I'll do my best to answer or fix them.

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3 years ago

Which equation in standard form has a graph that passes through the point (-4 , 2) and has a slope of 9/2?

malfutka [58]

Answer is <span>C. 9x-2y=-40

hope that helps

because y = 9/2x +20 then 2y = 9x + 40 so 9x -2y = -40
</span>

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3 years ago

Use implicit differentiation to find an equation of the tangent line to the curve at the given point. x2/3 + y2/3 = 4 (−3 3 , 1)

vovikov84 [41]

Answer with Step-by-step explanation:

We are given that an equation of curve

$x^{\frac{2}{3}}+y^{\frac{2}{3}}=4$

We have to find the equation of tangent line to the given curve at point $(-3\sqrt3,1)$

By using implicit differentiation, differentiate w.r.t x

$\frac{2}{3}x^{-\frac{1}{3}}+\frac{2}{3}y^{-\frac{1}{3}}\frac{dy}{dx}=0$

Using formula : $\frac{dx^n}{dx}=nx^{n-1}$

$\frac{2}{3}y^{-\frac{1}{3}}\frac{dy}{dx}=-\frac{2}{3}x^{-\frac{1}{3}}$

$\frac{dy}{dx}=\frac{-\frac{2}{3}x^{-\frac{1}{3}}}{\frac{2}{3}y^{-\frac{1}{3}}}$

$\frac{dy}{dx}=-\frac{x^{-\frac{1}{3}}}{y^{-\frac{1}{3}}}$

Substitute the value x= $-3\sqrt3,y=1$

Then, we get

$\frac{dy}{dx}=-\frac{(-3\sqrt3)^{-\frac{1}{3}}}{1}$

$\frac{dy}{dx}=-(-3^{\frac{3}{2}})^{-\frac{1}{3}}=-\frac{1}{-(3)^{\frac{3}{2}\times \frac{1}{3}}}=\frac{1}{\sqrt3}$

Slope of tangent=m= $\frac{1}{\sqrt3}$

Equation of tangent line with slope m and passing through the point $(x_1,y_1)$ is given by

$y-y_1=m(x-x_1)$

Substitute the values then we get

The equation of tangent line is given by

$y-1=\frac{1}{\sqrt3}(x+3\sqrt3)$

$y-1=\frac{x}{\sqrt3}+3$

$y=\frac{x}{\sqrt3}+3+1$

$y=\frac{x}{\sqrt3}+4$

This is required equation of tangent line to the given curve at given point.

8 0

3 years ago

In a different plan for area codes, the first digit could be any number from 4 through 8, the second digit was either 3, 4, 5,

Snezhnost [94]

For the first digit, we have 5 options that are 4,5,6,7,8 . For the second digit, we have 4 options which are 3,4,5 or 6 and for the third digit, we have the options of all numbers except 2 or 5 that is 1,3,4,6,7,8,9,0 . SO we have 8 options for third digit . So to find the total number of options, we need to multiply all the possible options for each digit that is 5 times 4 times 8 = 160 . So the number of possible options are 160 .

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3 years ago

Can anyone PLEASE HELP ME with this question i really need help

choli [55]

The missing angle is 49

7 0

3 years ago

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4a²-8ab+4b=???????????? ​

4a²-8ab+4b=????????????