Ask question

Ask question

All categories

3 years ago

9

Question 2 Unsaved Cat is 5 less than twice Dog's age. Cat is 11 years old. Find Dog's age.

2 answers:

madam [21]3 years ago

6 0

Let the dog's age be called x.

From the given statement, the following equation can be framed

Cat's age = 2*x -5

We are given to know that Cat's age is 11 years

So, we have:

11 = 2x -5

2x = 16

x = 16/2

x = 8

Dog's age = 8 years

PilotLPTM [1.2K]3 years ago

4 0

The dog is 8 years old because If you take 8 and multiply it by 2 you get 16 and then you subtract 5 and you get 11 years old or the cats age

You might be interested in

Pls help me on this

FinnZ [79.3K]

Answer:500

Step-by-step explanation:So first u mutiply by 100 then divide by 90

8 0

2 years ago

Julio bought 4 pounds of apples and one bag of oranges at the store. The apples cost $1.75 per pound and the bag of oranges cost

vaieri [72.5K]

Answer: $9.99

Step-by-step explanation:

There are 4 pounds of apples with each pound costing $1.75.

There is one bag of orange with a bag costing $2.99.

Total = (4 * 1.75) + 2.99

= 7 + 2.99

= $9.99

5 0

3 years ago

How do i turn 24% into a fraction

PolarNik [594]

You can remove the % sign if you divide the number by 100:

$\frac{24}{100}$

This is because % is defined as 1/100, and the name (per-cent) actually meant this originally!
we can also simplify it: let's divide it by 4:

$\frac{6}{25}$

7 0

3 years ago

How many pages will you have read after the 6 nights

Gnom [1K]

Answer:

168 pages

Step-by-step explanation:

each night you read 28 pages, so 6 nights you have to read for 28x6=168 pages.

6 0

2 years ago

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