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3 years ago

How do you factor this in the form (x+b)(x+c)

Mathematics

1 answer:

faust18 [17]3 years ago

7 0

So you have your polynomial. To factor this out, you need to find the factors of -45 that will add up to 3. These two factors would be -5 and 8. So, you have factored your polynomial! (x-5)(x+8). If you are still unsure, simply use the FOIL method to check and see if you get the same polynomial back. Hope this helps!

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8. Is the following statement true or false? Justify your conclusion. If a and b are nonnegative real numbers and a + b =0, then

atroni [7]

Answer:

If $a + b = 0$ then $a = 0$ and $b =0$

a | b | a + b (answer)

0 | 0 | 0

0 | 1 | 1

0 | 2 | 2

1 | 0 | 1

2 | 0 | 2

1 | 1 | 2

2 | 1 | 3

Step-by-step explanation:

Considering the following conditions for the real numbers:

$a\geq 0\\b\geq 0$

Following the rules of these in-equations, it is possible to deduce:

$a + b \geq 0$

Then, if the proposed statement is:

$a + b = 0$

The conditions above shall comply the requirements established, but first, analyzing the statement:

If $a \geq 0$ and $b \geq 0$ then $a + b \geq a$ , $a + b \geq b$ and $a + b \geq 0$ .

If $a = 0$ and b a non negative real number, then $a + b \geq b$ , but because to $a = 0$ , then $a + b = b$ . Due to the commutative property of sums, the same behavior will be presented if $b = 0$ and a a non negative real number.

According to that, if $a + b = 0$ , then $a = 0$ and $b = 0$ .

7 0

2 years ago

Please help me solve this please

nlexa [21]

Answer:

2. (-3f)

3. (4f)

4. x - 11 (not sure if this one is correct)

Step-by-step explanation:

Move -3 to the left of f = -3f

Move 4 to the left of f = 4f

f(x)=x+11 f ( x ) = x + 11

8 0

2 years ago

Required information Skip to question A die (six faces) has the number 1 painted on two of its faces, the number 2 painted on th

grigory [225]

Answer:

The change to the face 3 affects the value of P(Odd Number)

Step-by-step explanation:

Analysing the question one statement at a time.

Before the face with 3 is loaded to be twice likely to come up.

The sample space is:

$S = \{1,1,2,2,2,3\}$

And the probability of each is:

$P(1) = \frac{n(1)}{n(s)}$

$P(1) = \frac{2}{6}$

$P(1) = \frac{1}{3}$

$P(2) = \frac{n(2)}{n(s)}$

$P(2) = \frac{3}{6}$

$P(2) = \frac{1}{2}$

$P(3) = \frac{n(3)}{n(s)}$

$P(3) = \frac{1}{6}$

P(Odd Number) is then calculated as:

$P(Odd\ Number) = P(1) + P(3)$

$P(Odd\ Number) = \frac{1}{3} + \frac{1}{6}$

Take LCM

$P(Odd\ Number) = \frac{2+1}{6}$

$P(Odd\ Number) = \frac{3}{6}$

$P(Odd\ Number) = \frac{1}{2}$

After the face with 3 is loaded to be twice likely to come up.

The sample space becomes:

$S = \{1,1,2,2,2,3,3\}$

The probability of each is:

$P(1) = \frac{n(1)}{n(s)}$

$P(1) = \frac{2}{7}$

$P(2) = \frac{n(2)}{n(s)}$

$P(2) = \frac{3}{7}$

$P(3) = \frac{n(3)}{n(s)}$

$P(3) = \frac{1}{7}$

$P(Odd\ Number) = P(1) + P(3)$

$P(Odd\ Number) = \frac{2}{7} + \frac{1}{7}$

Take LCM

$P(Odd\ Number) = \frac{2+1}{7}$

$P(Odd\ Number) = \frac{3}{7}$

Comparing P(Odd Number) before and after

$P(Odd\ Number) = \frac{1}{2}$ --- Before

$P(Odd\ Number) = \frac{3}{7}$ --- After

<em>We can conclude that the change to the face 3 affects the value of P(Odd Number)</em>

7 0

2 years ago

(d-4)(2d+3)<br><img src="https://tex.z-dn.net/?f=%20%5Cgeqslant%20" id="TexFormula1" title=" \geqslant " alt=" \geqslant " align

valina [46]

Answer:

5d×(d-4)

-20d answer

answer -20d

3 0

3 years ago

Solve the system using substitution x+5y=0 3y+2x=-21

Crazy boy [7]

Answer:

$\left \{ {{x = -15} \atop {y=3}} \right.$

Step-by-step explanation:

$\left \{ {{x + 5y = 0} (1) \atop {3y + 2x = -21}(2)} \right. \left \{ {{x = -5y} \atop {3y + 2x = -21}} \right. \left \{ {{x = -5y} \atop {3y + 2*(-5y) = -21}} \right. \left \{ {{x = -5y} \atop {3y - 10y = -21}} \right. \\ \left \{ {{x = -5y } \atop {-7y = -21}} \right. \left \{ {{x = -5y} \atop {y= 3 }} \right. \left \{ {{x = -5 * 3} \atop {y=3}} \right.=> \left \{ {{x=-15} \atop {y=3}} \right.$

5 0

2 years ago

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