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3 years ago

5

What is the number between 100,000 and 200,00

2 answers:

Lena [83]3 years ago

7 0

I believe 150,000. You're welcome ;-)

yawa3891 [41]3 years ago

6 0

150000
An easy way to work out the inbetween number or median, is to add the numbers together (100000 plus 200000 equals 3000000) and then divide by how many numbers their were (300000 divided by 2 equals 1500000)

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Tank walls must be at least 1 Foot tall so the turtle can't climb out. there must be 400 square inches of floor space for the tu

melamori03 [73]

To find the dimensions of the tank, you will use the volume of a rectangular prism formula.

V = BH, where B is the area of the base and H is the height.
400 in^2 x 12 in
V = 4800 in^3

This volume would be less than 5000 cubic inches, so to find the dimensions, you will need to find 2 factors that equal 400 (20 in x 20 in).

All the dimensions would be length 20 in, width, 20 in, and height 12 in.

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3 years ago

A spinner is spun and a cube is rolled. Find the probability.<br> P(A or B and a 2)

Semenov [28]

Probability

You will add both probabilities after finding them both.

The spinner is 1/2 because a and b are half of the full spinner.

The cube is 1/6 because 2 is only one unit of the cube.

1/2+1/6=2/3

Meaning, the probability of both these events happening is 66.6% approximately.

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3 years ago

Use the alternative form of the derivative to find the derivative at x = c (if it exists). (If the derivative does not exist at

KIM [24]

Answer:

The derivative of the function does not exist.

Step-by-step explanation:

The alternative form of a derivative is given by:

$f'(c)= \lim_{x \to c} \dfrac{f(x)-f(c)}{x-c}$

Our function is defined as:

h(x)=|x+8|

i.e. h(x)= -(x+8) when x+8<0

and x+8 when x+8≥0

i.e. h(x)= -x-8 when x<-8

and x+8 when x≥-8

Hence now we find the derivative of the function at c=-8

i.e. we need to find the Left hand derivative (L.H.D.) and Right hand derivative (R.H.D) of the function.

The L.H.D at a point 'a' is calculated as:

$\lim_{x \to a^-} \dfrac{f(x)-f(a)}{x-a}\\\\=\lim_{h \to0} \dfrac{f(a-h)-f(a)}{a-h-a}= \lim_{h\to 0} \dfrac{f(a-h)-f(a)}{-h}$

Similarly R.H.D is given by:

$\lim_{x \to a^+} \dfrac{f(x)-f(a)}{x+a}\\\\=\lim_{h \to 0} \dfrac{f(a+h)-f(a)}{a+h-a}= \lim_{h\to 0} \dfrac{f(a+h)-f(a)}{h}$

Now for L.H.D we have to use the function h(x) =-x-8

and for R.H.D. we have to use the function h(x)=x+8

L.H.D.

we have a=-8

$\lim_{x \to (-8)^-} \dfrac{h(x)-h(-8)}{x-(-8)}\\\\= \lim_{h \to0} \dfrac{h(-8-h)-h(-8)}{-8-h-(-8)}= \lim_{h\to 0} \dfrac{h(-8-h)-h(-8)}{-h}$

= $\lim_{h \to 0} \dfrac{8+h-8-0}{-h}= \lim_{h \to 0}\dfrac{h}{-h}=-1$

similarly for R.H.D.

$\lim_{x \to (-8)^+} \dfrac{h(x)-h(-8)}{x-(-8)}\\\\=\lim_{h \to 0} \dfrac{h(-8+h)-h(-8)}{-8+h-(-8)}= \lim_{h\to 0} \dfrac{h(-8+h)-h(-8)}{h}$

$\lim_{h \to 0} \dfrac{-8+h+8-0}{h}=\lim_{h \to 0}\dfrac{h}{h}=1$

Now as L.H.D≠R.H.D.

Hence, the function is not differentiable.

4 0

3 years ago

Marika solved the equation (6x + 15)2 + 24 = 0. Her work is below. 1. (6x + 15)2 = –24 2. = 3. 6x + 15 = 4. 6x = – 15 5. x = Ana

Licemer1 [7]

The answer would be B, there are no real solutions to the secretion because the Square root of a negative number is not a real number.

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Just in case some of you have this question

Anna007 [38]

This actually helped me

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