Which statement is true

Mathematics

1 answer:

AlladinOne [14]2 years ago

8 0

3) an $8 delivery fee and $1.50 per litre of water

This is because the $8 is a constant baseline, then adding $1.50 times the amount of litres purchased.

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A supermarket needs to display 64 boxes of detergent.

Amiraneli [1.4K]

All the listed combination of row and boxes are correct and can be chosen from.

From the question, we are asked to find the row arrangement of a stack of boxes. Let us take a look at each one of the aforelisted answers

x rows with 16 boxes in each row
8 rows with x boxes in each row
16 rows with x boxes in each row

Since we are told that the total number of boxes to be stacked is 64, then for the first arrangement, we can deduce that the "x" is going to be 4. This is because multiplying 4 by 16 gives 64. For the second row, using the same calculation, we find that "x" is 8. And that of the third gives our "x" to be 4 as well. We can then adjudge that;

4 rows with 16 boxes in each row
8 rows with 8 boxes in each row
16 rows with 8 boxes in each row

Bearing this in mind, we can see that each of the 3 answers fits into the requirements we are asked.

At least 3 rows, and at least 4 boxes in each row.

Hence, all of the options are correct.

To learn more about rows and columns, see here brainly.com/question/13196178

5 0

2 years ago

18x-10-6x=50 This is another question on my quiz I really need help

prohojiy [21]

The answer is got is X=5 :)

7 0

2 years ago

WILL MARK BRAINLIEST what is the domain of (f/g) (x)?

tatyana61 [14]

Given that $f(x) = \sqrt{7-x}$ and $g(x) = \sqrt{x + 2}$ , we can say the following:

$\Bigg(\dfrac{f}{g}\Bigg)(x) = \dfrac{f (x)}{g(x)} = \dfrac{\sqrt{7 - x}}{\sqrt{x+2}}$

Now, remember what happens if we have a negative square root: it becomes an imaginary number. We don't want this, so we want to make sure whatever is under a square root is greater than 0 (given we are talking about real numbers only).

Thus, let's set what is under both square roots to be greater than 0:

$\sqrt{7 - x} \Rightarrow 7 - x \geq 0 \Rightarrow x \leq 7$

$\sqrt{x + 2} \Rightarrow x + 2 \geq 0 \Rightarrow x \geq -2$

Since both of the square roots are in the same function, we want to take the union of the domains of the individual square roots to find the domain of the overall function.

$x \leq 7 \,\,\cup x \geq -2 = \boxed{-2 \leq x \leq 7}$

Now, let's look back at the function entirely, which is:

$\Bigg( \dfrac{f}{g} \Bigg)(x) = \dfrac{\sqrt{7 - x}}{\sqrt{x+2}}$

Since $\sqrt{x + 2}$ is on the bottom of the fraction, we must say that $\sqrt{x + 2} \neq 0$ , since the denominator can't equal 0. Thus, we must exclude $\sqrt{x + 2} = 0 \Rightarrow x + 2 = 0 \Rightarrow x = -2$ from the domain.