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devlian [24]
3 years ago
11

PLZ I NEED HELP!

Mathematics
1 answer:
lakkis [162]3 years ago
4 0

Answer:

a. (6x-5)^2

b. 4x(2x^2+1)

c. -4(x-12)(x+3)

d. (2x+1)(4x-3)

Step-by-step explanation:

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1. 8y = 9x - 14; y = 5
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3. 8(5) = 9x - 14
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7 0
3 years ago
Write the equation of a line that passes through the point (1,4) and has a slope of 9.
Kryger [21]

Answer:

y=9x-5

Step-by-step explanation:

You want to find the equation for a line that passes through the point (1,4) and has a slope of 9.

First of all, remember what the equation of a line is:

y = mx+b

Where:

m is the slope, and

b is the y-intercept

To start, you know what m is; it's just the slope, which you said was 9. So you can right away fill in the equation for a line somewhat to read:

y=9x+b.

Now, what about b, the y-intercept?

To find b, think about what your (x,y) point means:

(1,4). When x of the line is 1, y of the line must be 4.

Because you said the line passes through this point, right?

Now, look at our line's equation so far: b is what we want, the 9 is already set and x and y are just two "free variables" sitting there. We can plug anything we want in for x and y here, but we want the equation for the line that specifically passes through the point (1,4).

So, why not plug in for x the number 1 and y the number 4? This will allow us to solve for b for the particular line that passes through the point you gave!.

(1,4). y=mx+b or 4=9 × 1+b, or solving for b: b=4-(9)(1). b=-5.The equation of the line that passes through the point (1,4) with a slope of 9

is

y=9x-5

4 0
3 years ago
Find y' by implicit differentiation √x +√y =1
AleksandrR [38]
Differentiate both sides with respect to x

\frac{d}{dx}(x^{\frac{1}{2}} + y^{\frac{1}{2}}) = \frac{d}{dx}(1)\\\\ \frac{1}{2x^{\frac{1}{2}}} + \frac{1}{2y^{\frac{1}{2}}}\frac{dy}{dx} = 0\\\\\frac{1}{2y^{\frac{1}{2}}}\frac{dy}{dx} = -\frac{1}{2x^{\frac{1}{2}}}\\\\\frac{1}{2\sqrt{y}}\frac{dy}{dx} = -\frac{1}{2\sqrt{x}} \\\\\frac{dy}{dx} = -\frac{1}{2\sqrt{x}} \times 2\sqrt{y}\\\\\frac{dy}{dx} = -\frac{\sqrt{y}}{\sqrt{x}} \\\\\boxed{\bf{y'= -\frac{\sqrt{y}}{\sqrt{x}}}}
5 0
3 years ago
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