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3 years ago

6

Ruben bought six new books for his collection. This increased his collection by 12%. How many books did he have before his purch

ases?

1 answer:

NeTakaya3 years ago

4 0

<span> 20. Let x represent the amount of books he had:
x + 6 = 1.12x

Subtract x from both sides:
6 = 0.12x

Divide both sides by 0.12; x = 50
</span>

<span>28. $6 + 0.025*$6 = $6 + $0.15 = $6.15 </span>

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Y = (x+3)² – 64<br> how do you write this in standard form

gregori [183]

Answer:

i think its y = x² + 6x - 55

7 0

3 years ago

Use the table to answer the question.

Vesna [10]

Given:

The table of values of a linear relationship.

x y

-1 -1

0 1

1 3

2 5

To find:

The equation for the given table of values.

Solution:

If a linear function passes through the two points, then the equation of the linear relationship is

$y-y_1=\dfrac{y_2-y_1}{x_2-x_1}(x-x_1)$

Consider any two point from the given table. Let the two points are (-1,-1) and (0,1). So, the equation of the linear relationship is

$y-(-1)=\dfrac{1-(-1)}{0-(-1)}(x-(-1))$

$y+1=\dfrac{1+1}{0+1}(x+1)$

$y+1=\dfrac{2}{1}(x+1)$

$y+1=2(x+1)$

Using distributive property, we get

$y+1=2(x)+2(1)$

$y+1=2x+2$

Subtracting 1 from both sides, we get

$y+1-1=2x+2-1$

$y=2x+1$

Therefore, the required equation is $y=2x+1$ . Hence, the correct option is D.

5 0

3 years ago

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.

3 0

3 years ago

The value of x 3y , for x = -2 and y = 3, is -24. TrueFalse

LenaWriter [7]

False:))))))))))))))
-2x3x3=-18

8 0

3 years ago

Read 2 more answers

Factorise x2 + 5x + 6

Maurinko [17]

X² + 5x + 6.

When the x² is alone, you just think of two numbers that add up to give 5 and multiply to give 6.

those are 2 and 3.

<span>x² + 5x + 6
</span>
<span>x² + 2x + 3x + 6
</span>
x(x + 2) + 3(x + 2)

(x + 2)(x + 3)

6 0

3 years ago