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2 years ago

18

Mathematics

1 answer:

asambeis [7]2 years ago

7 0

Answer:

Step-by-step explanation:

When we have a horizontal translation on the x-axis, it means the translation in question would be affecting only the x component of our function

With respect to the question, what we have here is that we are going to take out some values from x (or add some values) to it

Thus;

f(x) = x^2 would be;

g(x) = (x-4)^2

Corresponding to a shift to the right of upto 4 units on the x axis

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Help asap please and please explain so I could try the rest on my own

kiruha [24]

Answer:

Step-by-step explanation:

It has a 45 45 90 ratio, so if the hypotenuse is 7 root 2, then the two sides have to be 7.

4 0

2 years ago

Read 2 more answers

Find the area of the triangle with vertices: q(3,-4,-5), r(4,-1,-4), s(3,-5,-6).

maw [93]

Answer:

(√6)/2 square units

Step-by-step explanation:

The area of a triangle is half the magnitude of the cross product of the vectors representing adjacent sides.

QR = (4-3, -1-(-4), -4-(-5)) = (1, 3, 1)

QS = (3 -3, -5-(-4), -6-(-5)) = (0, -1, -1)

The cross product is the determinant ...

$\text{det}\left|\begin{array}{ccc}i&j&k\\1&3&1\\0&-1&-1\end{array}\right|=-2i+j-k$

The magnitude of this is ...

|QR × QS| = √((-2)² +1² +(-1)²) = √6

The area of the triangle is half this value:

Area = (1/2)√6 . . . . square units

3 0

3 years ago

Describe how to transform the graph of f(x) = x² to obtain

lord [1]

Answer:

The functions given are:

f(x) = x²

g(x) = f(-4x-3) + 1

First, find f(-4x-3):

f(x) = x²

f(-4x-3) = (-4x-3)²

Find g(x):

g(x) = f(-4x-3) + 1

g(x) = (-4x-3)² + 1

g(x) = (-1)² (4x+3)² + 1

g(x) = (4x+3)² + 1

First take

y = (x)²

Compress the graph along x axis by multiplying x with 4

y = (4x)²

Shift the graph left by 0.75 units, by adding 3 to x term.

y = (4x+3)²

Shift the graph up by 1 unit by adding 1 to the total terms.

y = (4x+3)² +1

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3 years ago

The height h (in feet) of an object dropped from a ledge after x seconds can be modeled by h(x)=−16x2+36 . The object is dropped

kakasveta [241]

Check the picture below.

$\bf ~~~~~~\textit{initial velocity in feet} \\\\ h(t) = -16t^2+v_ot+h_o \quad \begin{cases} v_o=\textit{initial velocity}&\\ \qquad \textit{of the object}\\ h_o=\textit{initial height}&\\ \qquad \textit{of the object}\\ h=\textit{object's height}&\\ \qquad \textit{at "t" seconds} \end{cases}$

so the object hits the ground when h(x) = 0, hmmm how long did it take to hit the ground the first time anyway?

$\bf h(x)=-16x^2+36\implies \stackrel{h(x)}{0}=-16x^2+36\implies 16x^2=36 \\\\\\ x^2=\cfrac{36}{16}\implies x^2 = \cfrac{9}{4}\implies x=\sqrt{\cfrac{9}{4}}\implies x=\cfrac{\sqrt{9}}{\sqrt{4}}\implies x = \cfrac{3}{2}~~\textit{seconds}$

now, we know the 2nd time around it hit the ground, h(x) = 0, but it took less time, it took 0.5 or 1/2 second less, well, the first time it took 3/2, if we subtract 1/2 from it, we get 3/2 - 1/2 = 2/2 = 1, so it took only 1 second this time then, meaning x = 1.

$\bf ~~~~~~\textit{initial velocity in feet} \\\\ h(x) = -16x^2+v_ox+h_o \quad \begin{cases} v_o=\textit{initial velocity}&0\\ \qquad \textit{of the object}\\ h_o=\textit{initial height}&\\ \qquad \textit{of the object}\\ h=\textit{object's height}&0\\ \qquad \textit{at "t" seconds}\\ x=\textit{seconds}&1 \end{cases} \\\\\\ 0=-16(1)^2+0x+h_o\implies 0=-16+h_o\implies 16=h_o \\\\[-0.35em] ~\dotfill\\\\ ~\hfill h(x) = -16x^2+16~\hfill$