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shusha [124]
3 years ago
15

Hey there! posted picture of question help please

Mathematics
2 answers:
Amiraneli [1.4K]3 years ago
8 0
For this case we have the following function:
 f (x) = (1/6) ^ x
 We must evaluate the function for x = 3
 We have then:
 f (3) = (1/6) ^ 3
 Rewriting:
 f (3) = (1/216)
 Answer:
 
The function evaluated at x = 3 is:
 
f (3) = (1/216)
 
option C
Zanzabum3 years ago
8 0
The answer is C..

F(3) = (1/6)³
F(3) = 1/216

All you had to do is cube the 6 since the cube of 1 is the same.

Hope this helps :)
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Let m=x²-5. which equation is equivalent to (x²-5)²-3x²+15=-2 in term of m?
SpyIntel [72]
The answer would be m^2+3m+2=0 because the first “part” of the equation (x^2-5) is equal to m; in the equation it’s squared so it can be replaced by m^2. next, 3m is the same as 3(x^2+5) which is what you have when you see 3x^2+15. finally, you have to bring the -2 over from the other side so you have positive 2. hope this helps!
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3 years ago
Katalin drove 210 miles on her vacation. She drove an average of 1.4 times faster on the second 105 miles of her trip than she d
konstantin123 [22]

The answer is D. 180 divided by x. .


5 0
3 years ago
PLS HELP ME ON THIS QUESTION I WILL MARK YOU AS BRAINLIEST!!
aliya0001 [1]

Answer:

The answer is A: 120yd^2

Step-by-step explanation:

Volume = Width x length x height

Volume = 8 x 3 x 5

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3 0
3 years ago
Read 2 more answers
Suppose that the functions s and t are defined for all real numbers x as follows.
PolarNik [594]

Answer:

(s-t)(-1) = -1

(s+t)(-1) = -7

Step-by-step explanation:

Given the following set of functions

s(x)=2x-2

t(x)=3x

(s-t)(x) = s(t) - t(x)

(s-t)(x) = 2x - 2 - 3x

(s-t)(x)  = -x -2

(s-t)(-1) = -(-1) - 2

(s-t)(-1) = 1-2

(s-t)(-1) = -1

(s+t)(x) = s(t) + t(x)

(s+t)(x) = 2x - 2 + 3x

(s+t)(x)  = 5x -2

(s+t)(-1) = 5(-1) - 2

(s+t)(-1) = -5-2

(s+t)(-1) = -7

8 0
3 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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