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3 years ago

Which expression is equivalent to x² - 4x - 45?

Mathematics

2 answers:

dolphi86 [110]3 years ago

7 0

Answer:

(x - 9)(x+5)

Step-by-step explanation:

x² + 5x - 9x - 45

x · (x + 5) - 9 (x + 5)

(x - 9) (x + 5)

lisabon 2012 [21]3 years ago

6 0

Answer:

D. (x - 9)(x+5)

Step-by-step explanation:

$x^{2} - 4x - 45$ is your expression, so you have to factor it out. Since from the options available it's clear that the factors of -45 you use are some form of 9 and 5, we can use the process of elimination to figure out which are the correct ones.

First, through FOIL u know that in any factored form the last two terms in each bracket multiply together to make the last term in the expanded form.

So first, look at A. the last two terms are -9 and -5, which = 45 when multiplied together. 45 does not = -45, so it can't be A.

B has 9 and 5. When they're multiplied, they = 45, which does not = -45, so its not B

It can't be C either, because the first term of the first bracket in C is 20, and when multiplied with x it doesn't produce x^2.

Therefore it has to be D. u can check this by multiplying -9 and 5, which = -45, the final term in the expanded form. Also, -9 + 5 = -4, which is the middle term of the expanded form.

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26. Define a relation ∼ ∼ on R 2 R2 by stating that ( a , b ) ∼ ( c , d ) (a,b)∼(c,d) if and only if a 2 + b 2 ≤ c 2 + d 2 . a2+

Tresset [83]

Answer:

~ is reflexive.

~ is asymmetric.

~ is transitive.

Step-by-step explanation:

~ is reflexive:

i.e., to prove $\forall (a, b) \in \mathbb{R}^2$ , $(a, b) R(a, b)$ .

That is, every element in the domain is related to itself.

The given relation is $\sim: (a,b) \sim (c, d) \iff a^2 + b^2 \leq c^2 + d^2$

Reflexive:

$(a, b) \sim (a, b)$ since $a^2 + b^2 = a^2 + b^2$

This is true for any pair of numbers in $\mathbb{R}^2$ . So, $\sim$ is reflexive.

Symmetry:

$\sim$ is symmetry iff whenever $(a, b) \sim (c, d)$ then $(c, d) \sim (a, b)$ .

Consider the following counter - example.

Let (a, b) = (2, 3) and (c, d) = (6, 3)

$a^2 + b^2 = 2^2 + 3^2 = 4 + 9 = 13$

$c^2 + d^2 = 6^2 + 3^2 = 36 + 9 = 42$

Hence, $(a, b) \sim (c, d)$ since $a^2 + b^2 \leq c^2 + d^2$

Note that $c^2 + d^2 \nleq a^2 + b^2$

Hence, the given relation is not symmetric.

Transitive:

$\sim$ is transitive iff whenever $(a, b) \sim (c, d) \hspace{2mm} \& \hspace{2mm} (c, d) \sim (e, f)$ then $(a, b) \sim (e, f)$

To prove transitivity let us assume $(a, b) \sim (c, d)$ and $(c, d) \sim (e, f)$ .

We have to show $(a, b) \sim (e, f)$

Since $(a, b) \sim (c, d)$ we have: $a^2 + b^2 \leq c^2 + d^2$

Since $(c, d) \sim (e, f)$ we have: $c^2 + d^2 \leq e^2 + f^2$

Combining both the inequalities we get:

$a^2 + b^2 \leq c^2 + d^2 \leq e^2 + f^2$

Therefore, we get: $a^2 + b^2 \leq e^2 + f^2$

Therefore, $\sim$ is transitive.

Hence, proved.

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Step-by-step explanation:

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A family consisting of three persons—A, B, and C—goes to a medical clinic that always has a doctor at each of stations 1, 2, and

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Answer:

Step-by-step explanation:

When A, B,C are equally likely to be assigned to any one of the stations 1,2 or 3

we find that each one assigned to one station has probability 1/3

Also each person is independent of the other.

Probability that

a) All three family members are assigned to the same station

= P(ABC) to same station

= P(ABC) to 1+P(ABC) to 2 +P(ABC) to 3

=3*P(A)P(B)P(C) since independent

= $3*(\frac{1}{3} )^3\\=\frac{1}{9}$

b) This would be equivalent to 1- all 3 to the same station

= $1-\frac{1}{9} \\=\frac{8}{9}$

c) Every member to a different station

A has 3 choices while B has remaining 2 and C has 1

Hence prob = $\frac{3*2*1}{3*3*3} =\frac{2}{9}$

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