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4 years ago

Please help! I'd appreciate it !

Mathematics

1 answer:

Alla [95]4 years ago

4 0

Get a tutor at the library

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Ayden wrote 54.077 as the standard form for fifty-four and seventy-seven hundredths. Is he correct? Explain.

cestrela7 [59]

Answer:

No, Ayden is incorrect

Step-by-step explanation:

Because the last 7 is in the thousandth place so therefore he is wrong, but this would be the correct answer 54.77 since the last 7 is in the hundredth place.

7 0

2 years ago

For the graphed exponential equation, calculate the average rate of change from x = −3 to x = 0.

PolarNik [594]

Answer:

Step-by-step explanation:

f(-3) = 2

f(0) = -5

[f(-3) - f(0)] ÷ -3-0

[2-(-5)] / -3

7/-3

8 0

3 years ago

Find cos θ given that cos 2θ = 5/6 and 0 ≤ θ < π/2. Give an exact answer

trasher [3.6K]

$\bf \textit{Double Angle Identities}
\\ \quad \\
sin(2\theta)=2sin(\theta)cos(\theta)
\\ \quad \\
cos(2\theta)=
\begin{cases}
cos^2(\theta)-sin^2(\theta)\\
1-2sin^2(\theta)\\
\boxed{2cos^2(\theta)-1}
\end{cases}
\\ \quad \\
tan(2\theta)=\cfrac{2tan(\theta)}{1-tan^2(\theta)}\\\\
-------------------------------\\\\
$

$\bf cos(2\theta)=\cfrac{5}{6}\implies 2cos^2(\theta)-1=\cfrac{5}{6}\implies 2cos^2(\theta)=\cfrac{5}{6}+1
\\\\\\
2cos^2(\theta)=\cfrac{11}{6}\implies cos^2(\theta)=\cfrac{11}{12}\implies cos(\theta)=\pm\sqrt{\cfrac{11}{12}}$

now, bear in mind, the square root gives us +/- versions, so, which is it? well, we know the angle is in the range of "0 ≤ θ < π/2", that simply means the 1st quadrant, so, we'll use the positive one then

 $\bf cos(\theta)=\cfrac{\sqrt{11}}{\sqrt{12}}\implies cos(\theta)=\cfrac{\sqrt{11}}{2\sqrt{3}}
\\\\\\
\textit{now, let's rationalize the denominator}
\\\\\\
\cfrac{\sqrt{11}}{2\sqrt{3}}\cdot \cfrac{\sqrt{3}}{\sqrt{3}}\implies \cfrac{\sqrt{11}\cdot \sqrt{3}}{2\sqrt{3^2}}\implies \cfrac{\sqrt{11\cdot 33}}{2\cdot 3}\implies \boxed{\cfrac{\sqrt{33}}{6}}$

3 0

4 years ago

Javier bought a painting for $150. Each year the painting's value increases by a factor of 1.15. What's the expression that give

lubasha [3.4K]

Multiply 1.15 times 7 to find the painting value

8 0

3 years ago

Read 2 more answers

Find the point on the plane ax + by + cz = d at minimum distance from the origin using the method of lagrange multipliers.

kupik [55]

The distance between some point $(x,y,z)$ and the origin is given by

$f(x,y,z)=\sqrt{x^2+y^2+z^2}$

so this is the function we're trying to minimize. But notice that $f(x,y,z)$ and $f(x,y,z)^2$ attain their critical points at the same $(x,y,z)$ , so we can solve the same problem by minimizing $x^2+y^2+z^2$ instead.

So let's take the Lagrangian to be

$L(x,y,z,\lambda)=x^2+y^2+z^2+\lambda(ax+by+cz-d)$

with partial derivatives (set equal to 0)

$L_x=2x+a\lambda=0$
$L_y=2y+b\lambda=0$
$L_z=2z+c\lambda=0$
$L_\lambda=ax+by+cz-d=0$

Now, notice that

$aL_x+bL_y+cL_z=2(ax+by+cz)+(a^2+b^c+c^2)\lambda=0$
$\implies\lambda=-\dfrac{2d}{a^2+b^2+c^2}$

and we can use this to solve for $x,y,z$ . We get

$x=\dfrac{ad}{a^2+b^2+c^2}$
$y=\dfrac{bd}{a^2+b^2+c^2}$
$z=\dfrac{cd}{a^2+b^2+c^2}$

At this critical point, we get a minimum distance of

$\sqrt{\left(\dfrac{ad}{a^2+b^2+c^2}\right)^2+\left(\dfrac{bd}{a^2+b^2+c^2}\right)^2+\left(\dfrac{cd}{a^2+b^2+c^2}\right)^2}=\sqrt{\dfrac{d^2}{a^2+b^2+c^2}}$

8 0

3 years ago