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3 years ago

Let a, b, and c be integers. Consider the following conditional statement:

Mathematics

1 answer:

Andrew [12]3 years ago

6 0

Step-by-step explanation:

Given that the logical statement is

"If a divides bc, then a divides b or a divides c"

we can see that a must divide one either b or c from the statement above

A) If a does not divide b or a does not divide c, then a does not divide bc.

This is False because a can divide b or c

B) If a does not divide b and a does not divide c, then a does not divide bc.

this is True for a to divide bc it must divide b or c (either b or c)

C) If a divides bc and a does not divide c, then a divides b.

This is True since a can divide bc and it cannot divide c, it must definitely divide b

D) If a divides bc or a does not divide b, then a divides c.

This is True since a can divide bc and it cannot divide b, it must definitely divide c

E) a divides bc, a does not divide b, and a does not divide c.

This is False for a to divide bc it must divide one of b or c

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Pls help !!! pleaseee.

ladessa [460]

I guess B. That statement is grammatically incorrect.

8 0

2 years ago

BRAINLIEST!!!

Liula [17]

Answer:

SA = 615.4 yd²

Step-by-step explanation:

The surface area (SA) of a sphere is calculated as

SA = 4πr² ← r is the radius

Here diameter = 14, thus r = 14 ÷ 2 = 7, then

SA = 4π × 7² = 4π × 49 = 196π = 196 × 3.14 ≈ 615.4 yd²

8 0

3 years ago

What is the quotient when 4x3 + 2x + 7 is divided by x + 3?

Arte-miy333 [17]

Answer:

The quotient of this division is $(4x^2 -12x + 38)$ . The remainder here would be $-26$ .

Step-by-step explanation:

The numerator $4x^3 + 2x + 7$ is a polynomial about $x$ with degree $3$ .

The divisor $x + 3$ is a polynomial, also about $x$ , but with degree $1$ .

By the division algorithm, the quotient should be of degree $3 - 1 = 2$ , while the remainder shall be of degree $1 - 1 = 0$ (i.e., the remainder would be a constant.) Let the quotient be $a\,x^2 + b\, x + c$ with coefficients $a$ , $b$ , and $c$ .

$4x^3 + 2x + 7 = \left(a\,x^2 + b\, x + c\right)(x + 3)$ .

Start by finding the first coefficient of the quotient.

The degree-three term on the left-hand side is $4 x^3$ . On the right-hand side, that would be $a\, x^3$ . Hence $a = 4$ .

Now, given that $a = 4$ , rewrite the right-hand side:

$\begin{aligned}&\left(4\,x^2 + b\, x + c\right)(x + 3) \cr =& \left(4x^2 + (b\, x + c)\right)(x + 3) \cr =& 4x^2(x + 3) + (bx + c)(x + 3) \cr =& 4x^3 + 12x^2 + (bx + c)(x + 3)\end{aligned}$ .

Hence:

$4x^3 + 2x + 7 = 4x^3 + 12x^2 + (b\,x + c)(x + 3)$

Subtract $\left(4x^3 + 12x^2\right$ from both sides of the equation:

$-12x^2 + 2x + 7 = (b\,x + c)(x + 3)$ .

The term with a degree of two on the left-hand side has coefficient $(-12)$ . Since the only term on the right hand side with degree two would have coefficient $b$ , $b = -12$ .

Again, rewrite the right-hand side:

$\begin{aligned}&\left(-12 x + c\right)(x + 3) \cr =& \left(-12 x+ c\right)(x + 3) \cr =& (-12x)(x + 3) + c(x + 3) \cr =& -12x^2 -36x + (bx + c)(x + 3)\end{aligned}$ .