Ask question

Ask question

All categories

3 years ago

13

Marisol buys 3 notebooks that each cost the same amount and 1 calculator that costs $9. The total cost is $21. How much does eac

h notebook cost?

2 answers:

adoni [48]3 years ago

4 0

Answer: 4 dollars for each notebook

Step-by-step explanation:

21-9= 12

12/3=4

each notebooks costs 4 dollars

kondaur [170]3 years ago

4 0

Answer:

$4

Step-by-step explanation:

You might be interested in

Solve for x.<br> x^2 + 6 = 10

Ostrovityanka [42]

Steps To Solve:

x² + 6 = 10

~Subtract 6 to both sides

x² + 6 - 6 = 10 - 6

~Simplify

x² = 4

~Take square root of 4

x² = ±√4

~Simplify

x = -2 or x = 2

Best of Luck!

5 0

3 years ago

6x^2-5x+1<br><br> Factor polynomial

Tamiku [17]

6x²-5x+1
=6x²-3x-2x+1
=3x(2x-1)-1(2x-1)
=(3x-1)(2x-1)

8 0

3 years ago

I don’t know how to work i this out

snow_tiger [21]

Equation of the line:

y = -0.4x + 1.6

or
y =
- 2 x
5
+ 8
5
When x=0, y = 1.6
When y=0, x = 4

Mark brainliest please

Hope this helps you

8 0

3 years ago

Let x be 3 <br> 4 + x + x + x = ?

IRINA_888 [86]

Answer:

13

Step-by-step explanation:

We replace each X with a 3, so

4 + 3 + 3 + 3

which equals to 13

7 0

3 years ago

a person invest in 9000 in a bank. the bank pays 5% interest compounded semi annually. to the nearest tenth of a year, how long

QveST [7]

Answer:

10.1 years.

Step-by-step explanation:

It is given that,

Principal = 9000

Rate of interest = 5%

No. of times interest compounded = 2 times in an year

Amount after certain time = 14800

The formula for amount:

$A=P(1+\frac{r}{n})^{nt}$

where, P is principal, r is rate of interest, n is no. of times interest compounded in an year and t is time in years.

Substitute the given values in the above formula.

$14800=9000(1+\frac{0.05}{2})^{2t}$

$\frac{14800}{9000}=(1+0.025)^{2t}$

$1.644=(1.025)^{2t}$

Taking log both sides.

$\log(1.644)=\log(1.025)^{2t}$

$\log(1.644)=2t\log(1.025)$ $[\because \log a^b=b\log a]$

$\frac{\log(1.644)}{2\log(1.025)}=t$

$t=10.066$

$t=10.1$

Therefore, the required time is 10.1 years.

3 0

3 years ago