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3 years ago

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.

1 answer:

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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Answer:

6x+y=-8 because it will be y=-6x-8 having same slope with the line given

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3 years ago

12. A man planted trees on both sides of

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2 years ago

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]

Answer:

0.667 feet

Step-by-step explanation:

To get the total distance travelled, we would integrate the velocity v(t) equation with respect to time t.

Given that the particle moves from t = 0 to t = 2 seconds

Integrating v(t) = t2 - 1

we get

= t^3/3 - t

substituting the boundaries from t = 0 to t = 2 seconds, we get

= 2^3/3 - 2 - (0^3/3 - 0)

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8 0

2 years ago

Read 2 more answers

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 area of a $\triangle = \dfrac{1}{2} \times base \times height$

As per the figure attached, we are given that side a = 10.

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 $\triangle$ MIL, $\triangle$ MID and $\triangle$ DIL.

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