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2 years ago

What is the value of f(4) in the function below? f(x) = -1/1·2x A. 16 OB. 8 C. 2 OD. 4

Mathematics

1 answer:

exis [7]2 years ago

3 0

Answer: 4

Step-by-step explanation:

$f(4)=\frac{1}{4} \cdot 2^{4}=\frac{1}{4} \cdot 16=4$

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2 years ago

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Answer:

132

Step-by-step explanation:

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3 years ago

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anzhelika [568]

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2 years ago

The ratio of Michael and Neil’s ages is 7:6. If Neil is really 42 years old, how old is Michael?

Anit [1.1K]

Answer:

49 years old.

Step-by-step explanation:

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6 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

What is the value of f(4) in the function below? f(x) = -1/1·2x A. 16 OB. 8 C. 2 OD. 4​

What is the value of f(4) in the function below? f(x) = -1/1·2x A. 16 OB. 8 C. 2 OD. 4