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STatiana [176]
3 years ago
6

What is the formula for the volume of a trapezoidal prism?

Mathematics
1 answer:
larisa86 [58]3 years ago
6 0

Answer:

Ok so basically, the formula for Volume of a Trapezoidal Prism. If the prism length is L,trapezoid base width B, trapezoid top width A, and trapezoid height H, then the volume of the prism is given by the four-variable formula: V(L, B, A, H) = LH(A + B)/2. In other words, multiply together the length, height, and average of A and B.

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What's the negative reciprocal of 5? Question 12 options: A) 1∕5 B) –5 C) –1∕5 D) 5
levacccp [35]

Answer:

\huge\boxed{C)\ -\dfrac{1}{5}}

Step-by-step explanation:

The reciprocal of 5: \dfrac{1}{5}

The negative reciprocal of 5: -\dfrac{1}{5}

7 0
2 years ago
PLEAASE HELP
gladu [14]
Well you are given the equation so let's plug in for kaylib and see how many miles she can see
distance = sqrt [(3 * height) / 2]
d = sqrt [(3 *48) / 2]
d = sqrt (144 / 2)
d = sqrt (72)
d = sqrt (3 * 3 * 2 * 2 * 2)
d = 6 * sqrt (2)

You you did not list Addisons height but I will say she is at x feet above sea level. we plug in x for height:
d = sqrt [(3x) / 2]

It it says how much farther for Addison which means she can see farther. to find difference we just subtract kaylibs distance from Addison. so:
sqrt [(3x) / 2] - 6 * sqrt (2)

plug in your x and use a calculator to get a decimal approximation
8 0
3 years ago
Read 2 more answers
An experiment started with 100 bacteria. They double in number every hour. Find the number of bacteria after 8 hours.
zzz [600]
There will be 25,6000
8 0
3 years ago
Given that 'n' is any natural numbers greater than or equal 2. Prove the following Inequality with Mathematical Induction
Oliga [24]

The base case is the claim that

\dfrac11 + \dfrac12 > \dfrac{2\cdot2}{2+1}

which reduces to

\dfrac32 > \dfrac43 \implies \dfrac46 > \dfrac86

which is true.

Assume that the inequality holds for <em>n</em> = <em>k </em>; that

\dfrac11 + \dfrac12 + \dfrac13 + \cdots + \dfrac1k > \dfrac{2k}{k+1}

We want to show if this is true, then the equality also holds for <em>n</em> = <em>k</em> + 1 ; that

\dfrac11 + \dfrac12 + \dfrac13 + \cdots + \dfrac1k + \dfrac1{k+1} > \dfrac{2(k+1)}{k+2}

By the induction hypothesis,

\dfrac11 + \dfrac12 + \dfrac13 + \cdots + \dfrac1k + \dfrac1{k+1} > \dfrac{2k}{k+1} + \dfrac1{k+1} = \dfrac{2k+1}{k+1}

Now compare this to the upper bound we seek:

\dfrac{2k+1}{k+1}  > \dfrac{2k+2}{k+2}

because

(2k+1)(k+2) > (2k+2)(k+1)

in turn because

2k^2 + 5k + 2 > 2k^2 + 4k + 2 \iff k > 0

6 0
2 years ago
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Evaluate the expression 10^2+(3+5)^2 - 5=
vampirchik [111]

Answer:

159

Step-by-step explanation:

10^2+(3+5)^2-5

-> 100 + (3+5)^2 -5

-> 100 + 8^2 - 5

-> 100 + 64 - 5

-> 159

8 0
2 years ago
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