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3 years ago

Two functions are given below: f(x) and h(x). State the axis of symmetry for each function and explain how to find it.

Mathematics

1 answer:

Luden [163]3 years ago

5 0

Answer:

Answers are below

Step-by-step explanation:

The axis of symmetry for f(x) is at x = 4 because it splits the function in half. I found it by looking at the vertex of the parabola and using the x-value as the axis of symmetry.

The axis of symmetry for h(x) is at x = -2 because it splits the function in half. I found it by looking at the vertex of the parabola and using the x-value as the axis of symmetry.

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So I’m not smart so like help meh pleasee<br><br> What is 49% of 200?

____ [38]

Answer:98

Step-by-step explanation:

We assume, that the number 200 is 100%.if 200 is 100%, so we can write it down as 200=100%.We also know, that x is 49% of the output value, so we can write it down as x=49%.Now we have two simple equations:

1) 200=100%

2) x=49%

we multiply both sides of the equation by x.then we divide both sides of the equation.

3 0

3 years ago

An ice cube is melting, and the lengths of its sides are decreasing at a rate of 0.8 millimeters per minute At what rate is the

julia-pushkina [17]

Answer:

The rate of decrease is: $43.2mm^3/min$

Step-by-step explanation:

Given

$l = 18mm$

$\frac{dl}{dt} = -0.8mm/min$ ---- We used minus because the rate is decreasing

Required

Rate of decrease when: $l = 18mm$

The volume of the cube is:

$V = l^3$

Differentiate

$\frac{dV}{dl} = 3l^2$

Make dV the subject

$dV = 3l^2 \cdot dl$

Divide both sides by dt

$\frac{dV}{dt} = 3l^2 \cdot \frac{dl}{dt}$

Given that: $l = 18mm$ and $\frac{dl}{dt} = -0.8mm/min$

$\frac{dV}{dt} = 3 * (18mm)^2 * (-0.8mm/min)$

$\frac{dV}{dt} = 3 * 18 *-0.8mm^3/min$

$\frac{dV}{dt} = -43.2mm^3/min$

<em>Hence, the rate of decrease is: 43.2mm^3/min</em>

8 0

2 years ago

A triangle ABC has its vertices at A(-2, -3), B(2, 1), and C(5,-1).

maksim [4K]

$~\hfill \stackrel{\textit{\large distance between 2 points}}{d = \sqrt{( x_2- x_1)^2 + ( y_2- y_1)^2}}~\hfill ~ \\\\[-0.35em] ~\dotfill\\\\ A(\stackrel{x_1}{-2}~,~\stackrel{y_1}{-3})\qquad B(\stackrel{x_2}{2}~,~\stackrel{y_2}{1}) ~\hfill AB=\sqrt{( 2- (-2))^2 + ( 1- (-3))^2} \\\\\\ AB=\sqrt{(2+2)^2+(1+3)^2}\implies AB=\sqrt{32}\implies \boxed{AB=4\sqrt{2}} \\\\[-0.35em] ~\dotfill$

$B(\stackrel{x_1}{2}~,~\stackrel{y_1}{1})\qquad C(\stackrel{x_2}{5}~,~\stackrel{y_2}{-1}) ~\hfill BC=\sqrt{( 5- 2)^2 + ( -1- 1)^2} \\\\\\ BC=\sqrt{3^2+(-2)^2}\implies BC=\sqrt{9+4}\implies \boxed{BC=\sqrt{13}} \\\\[-0.35em] ~\dotfill\\\\ C(\stackrel{x_1}{5}~,~\stackrel{y_1}{-1})\qquad A(\stackrel{x_2}{-2}~,~\stackrel{y_2}{-3}) ~\hfill CA=\sqrt{( -2- 5)^2 + ( -3- (-1))^2} \\\\\\ CA=\sqrt{(-7)^2+(-3+1)^2}\implies CA=\sqrt{49+(-2)^2}\implies \boxed{CA=\sqrt{53}}$