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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

<em>"If a divides bc, then a divides b or a divides c"</em>

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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OlgaM077 [116]

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3 years ago

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2 years ago

1+-w2+9w and I need help cuz I’m on 76 and I’m sooo close help

Gnesinka [82]

$\huge \boxed{\mathfrak{Question} \downarrow}$

Simplify :- 1 + - w² + 9w.

$\large \boxed{\mathfrak{Answer \: with \: Explanation} \downarrow}$

$\large \sf1 + - w ^ { 2 } + 9 w$

Quadratic polynomial can be factored using the transformation $\sf \: ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right)$ , where $\sf x_{1} and x_{2}$ are the solutions of the quadratic equation $\sf \: ax^{2}+bx+c=0$ .

$\large \sf-w^{2}+9w+1=0$

All equations of the form $\sf\:ax^{2}+bx+c=0$ can be solved using the quadratic formula: $\sf\frac{-b±\sqrt{b^{2}-4ac}}{2a}$ . The quadratic formula gives two solutions, one when ± is addition and one when it is subtraction.

$\large \sf \: w=\frac{-9±\sqrt{9^{2}-4\left(-1\right)}}{2\left(-1\right)} \\$

Square 9.

$\large \sf \: w=\frac{-9±\sqrt{81-4\left(-1\right)}}{2\left(-1\right)} \\$

Multiply -4 times -1.

$\large \sf \: w=\frac{-9±\sqrt{81+4}}{2\left(-1\right)} \\$

Add 81 to 4.

$\large \sf \: w=\frac{-9±\sqrt{85}}{2\left(-1\right)} \\$

Multiply 2 times -1.

$\large \sf \: w=\frac{-9±\sqrt{85}}{-2} \\$

Now solve the equation $\sf\:w=\frac{-9±\sqrt{85}}{-2}$ when ± is plus. Add -9 to $\sf\sqrt{85}$ .

$\large \sf \: w=\frac{\sqrt{85}-9}{-2} \\$

Divide -9+ $\sf\sqrt{85}$ by -2.

$\large \boxed{ \sf \: w=\frac{9-\sqrt{85}}{2}} \\$

Now solve the equation $\sf\:w=\frac{-9±\sqrt{85}}{-2}$ when ± is minus. Subtract $\sf\sqrt{85}$ from -9.

$\large \sf \: w=\frac{-\sqrt{85}-9}{-2} \\$

Divide $\sf-9-\sqrt{85}$ by -2.

$\large \boxed{ \sf \: w=\frac{\sqrt{85}+9}{2}} \\$

Factor the original expression using $\sf\:ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right)$ . Substitute $\sf\frac{9-\sqrt{85}}{2}$ for $\sf\:x_{1}$ and $\sf\frac{9+\sqrt{85}}{2}$ for $\sf\:x_{2}$ .

$\large \boxed{ \boxed {\mathfrak{-w^{2}+9w+1=-\left(w-\frac{9-\sqrt{85}}{2}\right)\left(w-\frac{\sqrt{85}+9}{2}\right) }}}$

Well, in the picture you inserted it says that it's 8th grade mathematics. So, I'm not sure if you have learned simplification with the help of biquadratic formula. So, if you want the answer simplified only according to like terms then your answer will be ⇨

$\large \sf \: 1 + - w {}^{2} + 9w \\ =\large \boxed{\bf \: 1 - {w}^{2} + 9w}$

This cannot be further simplified as there are no more like terms (you can use the biquadratic formula if you've learned it.)

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2 years ago

What is this functie<br> f(x) = 2x + 5

Aloiza [94]

Slope function
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3 years ago

Read 2 more answers

PLEASE HELP URGENTLY!!!!¡!!!!!!¡

DaniilM [7]

<h3>There are 3 answers: Choice A, Choice D, Choice F</h3>

Explanation:

2 & 1/2 = 2.5

10 + (-2.5) can be represented by starting with 10 and then dropping by 2.5, which is what choice A is saying

or we can start with -2.5 and add on 10. This works because we can add numbers in any order, example: 2+3 = 3+2 = 5. So starting with -2.5 and adding on 10 is represented by choice D's and choice F's examples

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3 years ago