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3 years ago

How to write an equation in slope intercept form given a point and a slope?

Mathematics

1 answer:

pogonyaev3 years ago

4 0

Given that the slope = m
and Point (a, b)
The equation of the line is
y - b = m(x - a)

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Order these from least to greatest:<br><br> |–15|, |4|, –|15|, |–5|

Alenkinab [10]

-/15/, /4/, /-5/,/ -15/

6 0

3 years ago

Read 2 more answers

Prove: If C⊂A and D⊂B then D−A⊂B−C

nalin [4]

Let $x\in D\setminus A$ , so that $x\in D$ but $x\not\in A$ . Since $D\subset B$ , it follows that $x\in B$ , and since $C\subset A$ , it follows that $x\not\in C$ , which means $x\in B\setminus C$ .

3 0

3 years ago

The region bounded by y=(3x)^(1/2), y=3x-6, y=0

Ganezh [65]

Answer:

4.5 sq. units.

Step-by-step explanation:

The given curve is $y = (3x)^{\frac{1}{2} }$

⇒ $y^{2} = 3x$ ...... (1)

This curve passes through (0,0) point.

Now, the straight line is y = 3x - 6 ....... (2)

Now, solving (1) and (2) we get,

$y^{2} - y - 6 = 0$

⇒ (y - 3)(y + 2) = 0

⇒ y = 3 or y = -2

We will consider y = 3.

Now, y = 3x - 6 has zero at x = 2.

Therefor, the required are = $\int\limits^3_0 {(3x)^{\frac{1}{2} } } \, dx - \int\limits^3_2 {(3x - 6)} \, dx$

= $\sqrt{3} [{\frac{x^{\frac{3}{2} } }{\frac{3}{2} } }]^{3} _{0} - [\frac{3x^{2} }{2} - 6x ]^{3} _{2}$

= $[\frac{\sqrt{3}\times 2 \times 3^{\frac{3}{2} } }{3}] - [13.5 - 18 - 6 + 12]$

= 6 - 1.5

= 4.5 sq. units. (Answer)

7 0

3 years ago

3( − 3) − 5 >− 3 − 6<br> (show your work)

egoroff_w [7]

Answer:

x > 3

Step-by-step explanation:

3x-9-5x > -3x - 6

3x-5x+3x > -6+9

x > 3

3 0

3 years ago

Which zero pair could be added to the function f(x)=x2+12x+6 so that the function can be written in vertex form?

alex41 [277]

Find the vertex.

x=b/-2a

12/-2(1)
x=-6

then plug into equation

-6^2+12(-6)+6

36-72+6=-30

(-6,-30)

8 0

3 years ago