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11Alexandr11 [23.1K]
3 years ago
9

What is the common ratio for the sequence 5, -10, 20, -40, ...?

Mathematics
2 answers:
Stels [109]3 years ago
7 0

Common ratio is -2

 5 x (-2) = -10

-10 x (-2) = 20

20 x (-2) = -40

Anna [14]3 years ago
6 0

Answer:

the common ratio is -2


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SOMEONE PLEASE HELP.>>>>>>
alexira [117]

Answer:

The answer would be B



6 0
3 years ago
Forensic specialists can estimate the height of a deceased person from the lengths of the person's bones. These lengths are subs
Lera25 [3.4K]

Answer:

130.2845\leq h\leq 137.7245

Step-by-step explanation:

Given an inequality that relates the height h, in centimeters, of an adult female and the length f, in centimeters, of her femur by the equation

|h - (2.47f + 54.10)| \leq  3.72

If an adult female measures her femur as 32.25 centimeters, we can determine the possible range of her height by plugging f = 32.25cm into the modelled equation as shown:

|h - (2.47(32.25) + 54.10)| \leq  3.72\\|h - (79.9045 + 54.10)| \leq  3.72\\|h - (134.0045)| \leq  3.72\\

If the modulus function is positive then:

h - 134.0045 \leq  3.72\\h \leq 3.71+134.0045\\h\leq 137.7245

If the modulus function is negative then:

-(h - 134.0045) \leq  3.72\\-h+134.0045 \leq 3.72\\-h\leq 3.72-134.0045\\-h\leq -130.2845\\

multiply through by -1

-(-h)\geq  -(-130.2845)\\h\geq 130.2845\\130.2845\leq h

combining the resulting inequalities, the estimate of the possible range of heights will be 130.2845\leq h\leq 137.7245

8 0
3 years ago
What is the value of x?​
Kazeer [188]

Answer:

Step-by-step explanation:

The volume of a sphere is

V=\frac{4}{3}\pi r^3

We are told our volume is 500pi/3, so we fill that in and solve for r:

\frac{500\pi}{3}=\frac{4}{3}\pi r^3

Start by multiplying both sides by 3 over 4 pi:

(\frac{3}{4\pi})\frac{500\pi}{3}=(\frac{3}{4\pi})\frac{4\pi}{3}r^3

Simplifying gives us

125=r^3

Take the cubed root of both sides to get that

r = 5

6 0
3 years ago
Does there exist a di↵erentiable function g : [0, 1] R such that g'(x) = f(x) for all x 2 [0, 1]? Justify your answer
agasfer [191]

Answer:

No; Because g'(0) ≠ g'(1), i.e. 0≠2, then this function is not differentiable for g:[0,1]→R

Step-by-step explanation:

Assuming:  the function is f(x)=x^{2} in [0,1]

And rewriting it for the sake of clarity:

Does there exist a differentiable function g : [0, 1] →R such that g'(x) = f(x) for all g(x)=x² ∈ [0, 1]? Justify your answer

1) A function is considered to be differentiable if, and only if  both derivatives (right and left ones) do exist and have the same value. In this case, for the Domain [0,1]:

g'(0)=g'(1)

2) Examining it, the Domain for this set is smaller than the Real Set, since it is [0,1]

The limit to the left

g(x)=x^{2}\\g'(x)=2x\\ g'(0)=2(0) \Rightarrow g'(0)=0

g(x)=x^{2}\\g'(x)=2x\\ g'(1)=2(1) \Rightarrow g'(1)=2

g'(x)=f(x) then g'(0)=f(0) and g'(1)=f(1)

3) Since g'(0) ≠ g'(1), i.e. 0≠2, then this function is not differentiable for g:[0,1]→R

Because this is the same as to calculate the limit from the left and right side, of g(x).

f'(c)=\lim_{x\rightarrow c}\left [\frac{f(b)-f(a)}{b-a} \right ]\\\\g'(0)=\lim_{x\rightarrow 0}\left [\frac{g(b)-g(a)}{b-a} \right ]\\\\g'(1)=\lim_{x\rightarrow 1}\left [\frac{g(b)-g(a)}{b-a} \right ]

This is what the Bilateral Theorem says:

\lim_{x\rightarrow c^{-}}f(x)=L\Leftrightarrow \lim_{x\rightarrow c^{+}}f(x)=L\:and\:\lim_{x\rightarrow c^{-}}f(x)=L

4 0
3 years ago
Quadrilateral ABCD is translated 2 units left and 3 units up to form quadrilateral A'B'C'D'
TiliK225 [7]
The answer would be
<span>B) 
A' (-8, 5)
B' (-8, 8)
C' (-5, 7)
D' (-4.5, 5)</span>
3 0
3 years ago
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