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leva [86]
2 years ago
13

4^4 x 4^7 How do I work this problem?​

Mathematics
1 answer:
mote1985 [20]2 years ago
6 0
4^4 is simply 4x4x4x4 and 4^7 is 4x4x4x4x4x4x4
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The answer is 4/3
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Solve -np-80&lt;60 for n. Show your work.<br><br> Solve 2a-5d=30 for d. Show your work.
xz_007 [3.2K]

-np-80 < 60\qquad\text{change the signs}\\\\np+80

------------------------------------------------------

2a-5d=30\qquad\text{subtract 2a from both sides}\\\\-5d=-2a+30\qquad\text{change the signs}\\\\5d=2a-30\qquad\text{divide both sides by 5}\\\\d=0.4a-6

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The sum of the measures of two angles in a triangle is 98°. What is the measure of the third angle?
ch4aika [34]

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A

Step-by-step explanation:

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  • A.

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<u>Here is how I found it:</u>

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5 0
2 years ago
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
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