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3 years ago

Which of the following is NOT true about the given equation: x = 4? Question 10 options:

Mathematics

1 answer:

Ilya [14]3 years ago

7 0

Answer:

Step-by-step explanation:

the line with equation x = 4 is the equation of a vertical line parallel to the y- axis. Not a horizontal line

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What is the range of f(x)?

adoni [48]

Answer:

best ans is { f(x)| –∞ < f(x) ≤ 4}

Step-by-step explanation:

usually to find the range of a fn, u need to see what the value can be. the graph show only a small range of x values. but u can see f(x) is definitely <u><</u> 4. so the ans is { f(x)| –∞ < f(x) ≤ 4}

6 0

4 years ago

Read 2 more answers

How is a cylinder different from a prism?

Nikitich [7]

Answer:

third option

Step-by-step explanation:

good luck on the rest of it

3 0

3 years ago

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A model rocket is launched from the top of a building. The height (in meters) of the rocket above the grown is given by h(t)=-6t

Iteru [2.4K]

Answer:

47.5 Meters

Step-by-step explanation:

Find where the derivative changes from positive to negative (local maximum)

h(t)= $-6t^2+30t+10$

h'(t)= $-12t+30$

Find where the derivative equals zero

Theres no spot where the derivative wont exist because its a polynomial.

-12t=-30

t= $\frac{5}{2}$

Ill stop here because this is a parabola opening down and this is the only spot where there will be a horizontal tangent line so this is the x-coordinate of the vertex. AKA you don't need to test positive and negative on both sides.

Now plug into original equation

h(2.5)= $-6(2.5)^2+30(2.5)+10$

=-37.5+75+10

=47.5

5 0

3 years ago

I really stuck on trying to prove this

Alisiya [41]

$\displaystyle\sum_{r=1}^nr(r+1)\cdots(r+p-1)$

When $n=1$ ,

$\displaystyle\sum_{r=1}^1r(r+1)\cdots(r+p-1)=1(1+1)(1+2)\cdots(1+p-2)(1+p-1)=p!$

Meanwhile, you have on the right

$\dfrac{(1)(1+1)(1+2)\cdots(1+p-2)(1+p-1)(1+p)}{p+1}=(1)(1+1)(1+2)\cdots(p-1)(p)=p!$

so the equality holds for $n=1$ .

Assume it holds for $n=k$ , i.e. that

$\displaystyle\sum_{r=1}^kr(r+1)\cdots(r+p-1)=\frac{k(k+1)(k+2)\cdots(k+p-1)(k+p)}{p+1}$

Now for $n=k+1$ , you have

$\displaystyle\sum_{r=1}^{k+1}r(r+1)\cdots(r+p-1)=\sum_{r=1}^kr(r+1)\cdots(r+p-1)+(k+1)(k+2)\cdots(k+1+p-1)$
$=\displaystyle\frac{k(k+1)(k+2)\cdots(k+p-1)(k+p)}{p+1}+(k+1)(k+2)\cdots(k+1+p-2)(k+1+p-1)$
$=\displaystyle\frac{k(k+1)(k+2)\cdots(k+p-1)(k+p)}{p+1}+(k+1)(k+2)\cdots(k+p-1)(k+p)$
$=\left(\dfrac k{p+1}+1\right)(k+1)(k+2)\cdots(k+p-1)(k+p)$
$=\dfrac{k+p+1}{p+1}(k+1)(k+2)\cdots(k+p-1)(k+p)$
$=\dfrac{(k+1)(k+2)\cdots(k+p-1)(k+p)(k+p+1)}{p+1}$

as required.

3 0

3 years ago

Let f(X)=x^2+2 and g(X)=1-3x

balu736 [363]

Answer:

1. y = x2+2

2. g = (-3x+1)/x

Step-by-step explanation:

1. No explanation

Let's solve for g.

gx=1−3x

Step 1: Divide both sides by x

gx /x = −(3x+1 )/x

g = (-3x+1)/x

Hope it helped!

7 0

3 years ago