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3 years ago

K(x) = X +-6 k(3) =

Mathematics

2 answers:

mina [271]3 years ago

7 0

Answer:

-3

Step-by-step explanation:

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vampirchik [111]3 years ago

3 0

Answer:

$x=-\frac{18k}{k-1};\quad \:k\ne \:1$

Step-by-step explanation:

$\mathrm{Divide\:both\:sides\:by\:}k-1;\quad \:k\ne \:1$

$\frac{x\left(k-1\right)}{k-1}=\frac{-18k}{k-1};\quad \:k\ne \:1$

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12 boys collected 16 aluminum cans each 15 girls collected 14 aluminum cans how meany cans did the girls collected than the boys

ANEK [815]

Answer:

Step-by-step explanation:

Since the boys collected 16 cans each and there are 12 of them it means they collected 16*12= 192 cans in total

there are 15 girls and each collected 14 cans. Therefore they collected 15*14=210 cans in total

210-192=18 cans the girls collected more than the boys

5 0

3 years ago

A ball is thrown in the air from a platform that is 96 feet above ground level with an initial vertical velocity of 32 feet per

pishuonlain [190]

Answer:

y = -16 (x - 1)^2 + 112

The object lands on the ground in approximately 3.6s

Explanation:

The equation given is that of a parabola.

Now the maximum (local) point of a parabola is the vertex. Therefore, if we want to rewrite our function in the form that would be used to find the maximum height, then that form must be the vertex form of a parabola.

The vertex form of a parabola is

$y=a(t-h)^2+k$

where (h, k) is the vertex.

The only question is, what is the vertex for our function h(t)?

Remember that if we have an equation of the form

$y=ax^2+bx+c$

then the x-coordinate of the vertex is

$h=-\frac{b}{2a}$

Now in our case b = 32 and a = -16; therefore,

$h=\frac{-32}{2(16)}=1$

We've found the value of the x-coordinate of the vertex. What about the y-coordinate? To get the y-coordinate, we put x = 1 into h(t) and get

$k=-16(1)+32(1)+96=112$

Hence, the y-coordindate is k = 112.

Therefore, the vertex of the parabola is (1, 112).

With the coordinates of the vertex in hand, we now write the equation of the parabola in vertex form.

$h(t)=a(t-1)^2+112$

The only problem is that we don't know what the value of a is. How do we find a?

Note that the point (0, 96) lies on the parabola. In other words,

$h(0)=-16(0)^2+32(0)+96=96$

Therefore, the vertex form of the parabola must also contain the point (0, 96).

Putting in t = 0, h = 96 into the vertex form gives

$96=a(0-1)^2+112$ $96=a+112$

subtracting 112 from both sides gives

$a=-16$

With the value of a in hand, we can finally write the equation of the parabola on vertex form.

$\boxed{h\mleft(t\mright)=-16\left(t-1\right)^2+112.}$

Now when does the object hit the ground? In other words, for what value of t is h(t) = 0? To find out we just have to solve the following for t.

$h(t)=0.$

We could either use h(t) = -16t^2 + 32t + 96 or the h(t) = -16(t - 1)^2 + 112 for the above equation. But it turns out, the vertex form is more convenient.

Thus we solve,

$-16\left(t-1\right)^2+112=0$

Now subtracting 112 from both sides gives

$-16(t-1)^2=-112$

Dividing both sides by -16 gives

$(t-1)^2=\frac{-112}{-16}$ $(t-1)^2=7$

taking the square root of both sides gives

$t-1=\pm\sqrt{7}$

adding 1 to both sides gives

$t=\pm\sqrt{7}+1$

Hence, the two solutions we get are

$t=\sqrt{7}+1=3.6$ $t=-\sqrt{7}+1=-1.6$

Now since time cannot take a negative value, we discard the second solution and say that t = 3.6 is our valid solution.

Therefore, it takes about 3.6 seconds for the object to hit the ground.

3 0

1 year ago

The graph of the parent function y = x cubed is horizontally stretched by a factor of One-fifth and reflected over the y-axis. W

sasho [114]

Answer: OPTION D.

Step-by-step explanation:

Below are some transformations for a function f(x):

If $-f(x)$ , then the function is reflected over the x-axis.

If $f(-x)$ , then the function is reflected over the y-axis.

If $f(bx)$ and $b>1$ , then the function is horizontally compressed.

If $f(bx)$ and $0 < b$ , then the function is horizontally stretched.

In this case you know that the parent function is:

$y=x^3$

According to the information given in the exercise, the parent function IS horizontally stretched by a factor of $\frac{1}{5}$ and it is also reflected over the y-axis.