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2 years ago

The formula for the volume of a prism is v=bh and the formula for the volume of a pyramid is v=1/3 bh. Explain b in formulas

Mathematics

1 answer:

inysia [295]2 years ago

4 0

The variable b represents the area of the base of the prism or pyramid. Hope that helps!

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6 1 2 3 + + 2 3 4 Situation: Find the value of this fraction computation: 6(1/2+3/4+2/3)

MArishka [77]

Answer:

11.5 gotchu 100% correct

Step-by-step explanation:

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2 years ago

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4 red balls 4 white balls what is the probability of drawing at least 1 red ball if two balls are drawn

vodomira [7]

The probability of drawing exactly one red ball is given by:
$P(1\ red\ ball)=\frac{4C1\times4C1}{8C2}=\frac{16}{28}$
the probability of drawing two red balls is given by:
$P(2\ red\ balls)=\frac{4C2\times4C0}{8C2}=\frac{6}{28}$
The probability of drawing at least one red ball is:
P(1 red ball) + P(2 red balls) = 16/28 + 6/28 = 11/14.
The answer is $\ \frac{11}{14}$

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3 years ago

this graph shows the relationship between the pound of cheese brought at a deli and the total cost in dollars for the cheese

Savatey [412]

Easy! The best answer is B!

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2 years ago

For each of the solutions of the equations find two consecutive integers between which the solution is located.

adoni [48]

Using squares of integers numbers, it is found that the solution of the equation is located between the integers x = 1 and x = 2.

The equation given is:

$x^2 = 3$

The solution of the equation given is:

$x = \sqrt{3}$

The squares of the integers numbers until the square root of 3 are:

$1^2 = 1$

$2^2 = 4$

Since $1 < 3 < 4$ , the square root of 3, which is the solution to the equation, is located between the integers x = 1 and x = 2.

A similar problem is given at brainly.com/question/3729492

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2 years ago

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1. Consider the following hypotheses:

Andrej [43]

Answer:

See deductions below

Step-by-step explanation:

a) p(y)∧q(y) for some y (Existencial instantiation to H1)

b) q(y) for some y (Simplification of a))

c) q(y) → r(y) for all y (Universal instatiation to H2)

d) r(y) for some y (Modus Ponens using b and c)

e) p(y) for some y (Simplification of a)

f) p(y)∧r(y) for some y (Conjunction of d) and e))

g) ∃x (p(x) ∧ r(x)) (Existencial generalization of f)

a) ¬C(x) → ¬A(x) for all x (Universal instatiation of H1)

b) A(x) for some x (Existencial instatiation of H3)

c) ¬(¬C(x)) for some x (Modus Tollens using a and b)

d) C(x) for some x (Double negation of c)

e) A(x) → ∀y B(y) for all x (Universal instantiation of H2)

f) ∀y B(y) (Modus ponens using b and e)

g) B(y) for all y (Universal instantiation of f)

h) B(x)∧C(x) for some x (Conjunction of g and d, selecting y=x on g)

i) ∃x (B(x) ∧ C(x)) (Existencial generalization of h)

3) We will prove that this formula leads to a contradiction.

a) ∀y (P (x, y) ↔ ¬P (y, y)) for some x (Existencial instatiation of hypothesis)

b) P (x, y) ↔ ¬P (y, y) for some x, and for all y (Universal instantiation of a)

c) P (x, x) ↔ ¬P (x, x) (Take y=x in b)

But c) is a contradiction (for example, using truth tables). Hence the formula is not satisfiable.

7 0

2 years ago