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brilliants [131]

3 years ago

14

In the film, Zeke the Zombie Slayer Versus the Zombies from Space, Zeke slayed 140 zombies. In the sequel, Zeke the Zombie Slaye

r Versus the Zombies from Space, Part 2, he slayed 5% more zombies. How many zombies were slain by Zeke in the sequel?

1 answer:

ad-work [718]3 years ago

4 0

Answer:

147 zombies were slain by Zeke.

Step-by-step explanation:

In the part 1 of Zeke the Zombie Slayer Versus the Zombies from Space, Zeke slayed 140 zombies. Then in Zombie Slayer Versus the Zombies from Space part 2, Zeke slayed 5% more Zombies. That means he slayed 140 + 5% of 140 zombies more.

So by calculating the value of 140 + 5% of 140 we can tell the real numbers of zombies slayed.

Therefore numbers will be $140 + \frac{5}{100}(140) = 140 + 7 = 147$

The answer is 147 zombies were slain by Zeke in sequel.

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It's a question from real and complex numbers which I can't solve. so someone PLZ HeLp

Semmy [17]

Answer:

$\frac{1}{5}$

Step-by-step explanation:

Using the rules of exponents

$a^{m}$ × $a^{n}$ = $a^{(m+n)}$ , $\frac{a^{m} }{a^{n} }$ = $a^{(m-n)}$ , $(a^m)^{n}$ = $a^{mn}$

Simplifying the product of the first 2 terms

$\frac{a^{p^2+pq} }{a^{pq+q^2} }$ × $\frac{a^{q^2+qr} }{a^{qr+r^2} }$

= $a^{p^2-q^2}$ × $a^{q^2-r^2}$

= $a^{p^2-r^2}$

Simplifying the third term

5( $(a^p+r)^{p-r}$

= 5 $a^{(p+r)(p-r)}$ = 5 $a^{(p^2-r^2)}$

Performing the division, that is

$\frac{a^{(p^2-r^2)} }{5a^{(p^2-r^2)} }$ ← cancel $a^{(p^2-r^2)}$ on numerator/ denominator leaves

= $\frac{1}{5}$

4 0

3 years ago

A car valued at £18000 at the start of 2017, depreciated in value by 5% each year for 3 years. How much did it lose in value ove

Anna11 [10]

<u>Answer:</u>

The amount lost over the 3 years s 2567.25£

<u>Explanation:</u>

$\mathrm{F}=\mathrm{I} \times\left(1-\left(\frac{r}{100}\right)\right)^{\mathrm{n}}$

where F = final value after n years

I = initial value of the car in 2017 = £18000 (given)

Since the value is depreciated 5% every year for 3 years,

r = percentage rate of depreciation = 5% (given)

n = 3 years

Substituting these values in formula, we get

$\mathrm{F}=18000 \times\left(1-\frac{5}{100}\right)^{3}$

= $18000 \times\left(\frac{95}{100}\right)^{3}$

= 15432.75£ which is the value of the car after 3 years

Finally 18000-15432.75 = 2567.25£ is the amount lost over this period.

6 0

3 years ago

Which of the following statements are true of literal equations?

Elden [556K]

Answer:

1. They consist primarily of variables.

2. They are often called formulas.

4. They often describe real-world relationships.

4 0

4 years ago

What value is equivalent to 52 + 53 − 54? A) -475 B) -350 C) -125 D) 5

Dafna1 [17]

None of these answer choices are correct. The answer would be 51. 52+53=105

105-54=51

7 0

3 years ago

Read 2 more answers

What is the greatest common factor for 12 16 and 28

gayaneshka [121]

$12=\fbox2\times\fbox2\times3\\\\16=\fbox2\times\fbox2\times2\times2\\\\28=\fbox2\times\fbox2\times7\\\\GCF(12;\ 16;\ 28)=\fbox2\times\fbox2=\boxed{\boxed{4}}$

5 0

3 years ago