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2 years ago

M(p + q); use m = 2, p = 6, and q = 1

Mathematics

2 answers:

bija089 [108]2 years ago

6 0

I’m pretty sure it’s 14
m(p+q)
2(6+1)
2(7)
14

irina1246 [14]2 years ago

4 0

Answer: 14

Step-by-step explanation:

If they say that m(p+q) = 2(6+1) then we take what is in the parenthasis first (6+1) then we get (7). After that take m=2 times 7= 14

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What are the roots of the equatio x² - x -42=0

Alex17521 [72]

$x^2-x-42=0\\x^2+6x-7x-42=0\\x(x+6)-7(x+6)=0\\(x-7)(x+6)=0\\x=7 \vee x=-6$

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2 years ago

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PLEASE HELP QUICK, I NEED IT

just olya [345]

We take the equation d = -16t^2+12t and subtract d from both sides to get
 0 = -16t^2+12t - d
We apply the quadratic formula to solve for t. With a = -16, b = 12, c = -d, we have
 t = [ -(12) ± √( 12^2 - 4(-16)(-d) ) ] / [2 * -16]
 = [- 12 ± √(144-64d) ] / (-32)
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 = [- 12 ± 4√(9-4d)] / (-32)
 = 3/8 ±√(9-4d) / 8
The answer to your question is t = 3/8 ±√(9-4d) / 8

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3 years ago

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F⃗ (x,y)=−yi⃗ +xj⃗ f→(x,y)=−yi→+xj→ and cc is the line segment from point p=(5,0)p=(5,0) to q=(0,2)q=(0,2). (a) find a vector pa

DerKrebs [107]

a. Parameterize $C$ by

$\vec r(t)=(1-t)(5\,\vec\imath)+t(2\,\vec\jmath)=(5-5t)\,\vec\imath+2t\,\vec\jmath$

with $0\le t\le1$ .

b/c. The line integral of $\vec F(x,y)=-y\,\vec\imath+x\,\vec\jmath$ over $C$ is

$\displaystyle\int_C\vec F(x,y)\cdot\mathrm d\vec r=\int_0^1\vec F(x(t),y(t))\cdot\frac{\mathrm d\vec r(t)}{\mathrm dt}\,\mathrm dt$

$=\displaystyle\int_0^1(-2t\,\vec\imath+(5-5t)\,\vec\jmath)\cdot(-5\,\vec\imath+2\,\vec\jmath)\,\mathrm dt$

$=\displaystyle\int_0^1(10t+(10-10t))\,\mathrm dt$

$=\displaystyle10\int_0^1\mathrm dt=\boxed{10}$

d. Notice that we can write the line integral as

$\displaystyle\int_C\vecF\cdot\mathrm d\vec r=\int_C(-y\,\mathrm dx+x\,\mathrm dy)$

By Green's theorem, the line integral is equivalent to

$\displaystyle\iint_D\left(\frac{\partial x}{\partial x}-\frac{\partial(-y)}{\partial y}\right)\,\mathrm dx\,\mathrm dy=2\iint_D\mathrm dx\,\mathrm dy$

where $D$ is the triangle bounded by $C$ , and this integral is simply twice the area of $D$ . $D$ is a right triangle with legs 2 and 5, so its area is 5 and the integral's value is 10.

4 0

2 years ago

Write an equation of a line that passes through (-12, -14) with slope 6.

damaskus [11]

The slope intercept form of a line is y = mx + b

m = slope
b = y-intercept

Plug in the slope, 6, into m.

Rewrite the equation;

y = 6x + b
We need to find b, your y-intercept, to finish this equation.

Plug in your point coordinate, (x, y) ⇒ (-12, -14) into the equation.

-14 = 6(-12) + b

Solve for b to find the y-intercept.

-14 = -72 + b
58 = b

Your new equation (your answer) is y = 6x + 58.

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bazaltina [42]

Answer:

Use google scooby it's easier than this way.

Step-by-step explanation:

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