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Reptile [31]
3 years ago
15

The graph of an exponential function of the form y = f(x) = a^x passes through the points and . The graph lies the x-axis.

Mathematics
1 answer:
Georgia [21]3 years ago
7 0

y=f(x)=a^x,\ a\in\mathbb{R^+}\ \wedge\ a\neq1\\\\f(0)=a^0=1\to\ \text{BLANK A}\ (0,\ 1)\\\\f(1)=a^1=a\to\text{BLANK B}\ (1,\ a)\\\\\text{BLANK C}\ above\\\text{because for each positive number its power is a positive number}

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Which equation represents a line which is parallel to the line y = -6x – 1?
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A particle moves along the x-axis with velocity v(t) = t2 - 1, with t measured in seconds and v(t) measured in feet per second.
Kisachek [45]

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0.667 feet

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8 0
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Given:
diamong [38]

Answer:

10-5\sqrt{2}

Step-by-step explanation:

As per the attached figure, right angled \triangle MDL has an inscribed circle whose center is I.

We have joined the incenter I to the vertices of the \triangle MDL.

Sides MD and DL are equal because we are given that \angle M = \angle L = 45 ^\circ.

Formula for <em>area</em> of a \triangle = \dfrac{1}{2} \times base \times height

As per the figure attached, we are given that side <em>a = 10.</em>

Using pythagoras theorem, we can easily calculate that side ML = 10\sqrt{2}

Points P,Q and R are at 90 ^\circ on the sides ML, MD and DL respectively so IQ, IR and IP are heights of  \triangleMIL, \triangleMID and \triangleDIL.

Also,

\text {Area of } \triangle MDL = \text {Area of } \triangle MIL +\text {Area of } \triangle MID+ \text {Area of } \triangle DIL

\dfrac{1}{2} \times 10 \times 10 = \dfrac{1}{2} \times r \times 10 + \dfrac{1}{2} \times r \times 10 + \dfrac{1}{2} \times r \times 10\sqrt2\\\Rightarrow r = \dfrac {10}{2+\sqrt2} \\\Rightarrow r = \dfrac{5\sqrt2}{\sqrt2+1}\\\text{Multiplying and divinding by }(\sqrt2 +1)\\\Rightarrow r = 10-5\sqrt2

So, radius of circle = 10-5\sqrt2

8 0
2 years ago
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