A driver-rating factor table and a 6-month basic rate schedule for an

uysha [10]

×imagine this is a flattened pyramid box. When you fold along the dotted lines you can see which sides are equivalent.

Add up the areas of each piece.

Two identical rectangular pieces 7 × 10
2(7)(10) = 140

One small rectangular piece 6 × 7
7(6) = 42

Two identical triangles
2(0.5)(6)(8) = 48

Add up all the pieces
140 + 42 + 48 = 230

3 0

3 years ago

A pack of 6 batteries of the same type cost £2.79 By calculating the price per battery,

maks197457 [2]

$\large{\rm{\underline{\underline{Answer:}}}}$

Given, 6 batteries for £2.79

(They are of same type, so each of them are of same cost. )

~Simply divide the total price of 6 batteries by 6 to get the price of each battery,

$= \dfrac{£2.79 }{6}$

$= £ \dfrac{279}{100 \times 6}$

$= £ \dfrac{46.5}{100}$

$= £0.465$

Thus, cost of each battery is £ 0.465 (ans)

8 0

2 years ago

A limited-edition poster increases in value each year. After 1 year, the poster is worth $20.70. After 2 years, it is worth $23.

kumpel [21]

Assuming that this is a linear relationship you must find the rate of change...$23.81-$20.70= $3.11. 3.11 is how much it changed that year. What that also gives you is how much you bought the post for at 0 years by subtracting $20.70 and the $3.11 = the price of the poster. This gives us the equation y=3.11(x)+17.59
Where Y is the value and X is the amount of years.

3 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.