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4 years ago

In deductive reasoning, you start from a _____ set of rules and conditions to determine what must be true.

Mathematics

2 answers:

emmainna [20.7K]4 years ago

7 0

Answer: B. given

Step-by-step explanation:

The logical deduction or deductive reasoning is a process that is logical in which a conclusion is made on the bases of given premises that are commonly assumed to be true.

Thus, we start from a given set of rules and conditions which we call as premises to determine what must be true.

Therefore, In deductive reasoning, you start from a given set of rules and conditions to determine what must be true.

inna [77]4 years ago

5 0

The correct answer is:

B) given.

Explanation:

Deductive reasoning is called top-down reasoning. It starts with given rules, theorems, etc. and uses these to prove that other statements are true.

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The zoo downtown has 9 times as many monkeys as elephants. They have 60 animals all together. How many elephants are there?

dolphi86 [110]

Answer:

Step-by-step explanation:

e = number of elephants

9 times as many monkeys as elephants

m = 9e

------------------------

60 animals all together

e + m = 60

e + 9e = 60

Combine like terms

10e = 60

Divide both sides by 10

e = 6

3 0

3 years ago

Read 2 more answers

Find the length of the radius of the circle, which is inscribed into a right trapezoid with lengths of bases a and b.

egoroff_w [7]

Answer:

r = (ab)/(a+b)

Step-by-step explanation:

Consider the attached sketch. The diagram shows base b at the bottom and base a at the top. The height of the trapezoid must be twice the radius. The point where the slant side of the trapezoid is tangent to the inscribed circle divides that slant side into two parts: lengths (a-r) and (b-r). The sum of these lengths is the length of the slant side, which is the hypotenuse of a right triangle with one leg equal to 2r and the other leg equal to (b-a).

Using the Pythagorean theorem, we can write the relation ...

((a-r) +(b-r))^2 = (2r)^2 +(b -a)^2

a^2 +2ab +b^2 -4r(a+b) +4r^2 = 4r^2 +b^2 -2ab +a^2

-4r(a+b) = -4ab . . . . . . . . subtract common terms from both sides, also -2ab

r = ab/(a+b) . . . . . . . . . divide by the coefficient of r

The radius of the inscribed circle in a right trapezoid is r = ab/(a+b).

_____

The graph in the second attachment shows a trapezoid with the radius calculated as above.

6 0

3 years ago

Michelle made 26 more sales calls last month than Allen. Together, they made 120 calls. How many calls were made by Michelle?

algol13

Answer

Step-by-step explanation:

you need to subtract the 26 from 120 to get 94 divide that by 2 to get 47 so thats haw many calls they each made then add on the 26 to find her amount to get 73

6 0

3 years ago

If the sum of the zereos of the quadratic polynomial is 3x^2-(3k-2)x-(k-6) is equal to the product of the zereos, then find k?

lys-0071 [83]

Answer:

Step-by-step explanation:

So I'm going to use vieta's formula.

Let u and v the zeros of the given quadratic in ax^2+bx+c form.

By vieta's formula:

1) u+v=-b/a

2) uv=c/a

We are also given not by the formula but by this problem:

3) u+v=uv

If we plug 1) and 2) into 3) we get:

-b/a=c/a

Multiply both sides by a:

-b=c

Here we have:

a=3

b=-(3k-2)

c=-(k-6)

So we are solving

-b=c for k:

3k-2=-(k-6)

Distribute:

3k-2=-k+6

Add k on both sides:

4k-2=6

Add 2 on both side:

4k=8

Divide both sides by 4:

k=2

Let's check:

$3x^2-(3k-2)x-(k-6) \text{ with }k=2$ :

$3x^2-(3\cdot 2-2)x-(2-6)$

$3x^2-4x+4$

I'm going to solve $3x^2-4x+4=0$ for x using the quadratic formula:

$\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

$\frac{4\pm \sqrt{(-4)^2-4(3)(4)}}{2(3)}$

$\frac{4\pm \sqrt{16-16(3)}}{6}$

$\frac{4\pm \sqrt{16}\sqrt{1-(3)}}{6}$

$\frac{4\pm 4\sqrt{-2}}{6}$

$\frac{2\pm 2\sqrt{-2}}{3}$

$\frac{2\pm 2i\sqrt{2}}{3}$

Let's see if uv=u+v holds.

$uv=\frac{2+2i\sqrt{2}}{3} \cdot \frac{2-2i\sqrt{2}}{3}$

Keep in mind you are multiplying conjugates:

$uv=\frac{1}{9}(4-4i^2(2))$

$uv=\frac{1}{9}(4+4(2))$

$uv=\frac{12}{9}=\frac{4}{3}$

Let's see what u+v is now:

$u+v=\frac{2+2i\sqrt{2}}{3}+\frac{2-2i\sqrt{2}}{3}$

$u+v=\frac{2}{3}+\frac{2}{3}=\frac{4}{3}$

We have confirmed uv=u+v for k=2.

4 0

3 years ago

Two cards are drawn without replacement from a standard deck of 52 cards. What is the probability that one of the cards is red a

alukav5142 [94]

Total cards = 52
Total red cards = 26
Total black cards = 26

P(one red and one black) = P(red and then black)+ P(black and then red)
P(one red and one black) =(26/52)(26/51) + (26/52)(26/51)
P(one red and one black) = 26/51 (Answer A)

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Answer: 26/51 (Answer A)
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5 0

3 years ago