All categories

3 years ago

10 squared plus four times nine minus two cubed

2 answers:

7 0

Lets write this out:-

10² + 4 × 9 - 2³

Solve:-

10² = 10 × 10 = 100
100 + 4 × 9 - 2³
2³ = 2 × 2 × 2 = 8
100 + 4 × 9 - 8
4 × 9 = 36
100 + 36 - 8
100 + 36 = 136
136 - 8 = 128

10² + 4 × 9 - 2³ = 128

Good luck! :P

Anton [14]3 years ago

4 0

We need to follow PEMDAS.
Parenthesis
Exponents
Multiplication
Division
Addition
Subtraction

$10^2 + 4\times 9 -2^3$

First take care of all the exponents.

$100 + 4 \times 9 - 8$

Now do the multiplication.

$100 + 36 - 8$

Now add and subtract from left to right.

$136 - 8$

$128$

Your final answer is 128.

You might be interested in

Item 8 Solve for x. Use the quadratic formula. 2x2−5x−9=0 Enter the solutions, in simplified radical form, in the boxes.

Lubov Fominskaja [6]

Answer:

5+√97/4

Also 5-√97/4

Step-by-step explanation:

The Quadratic formula is x=-b+-√b^2-4ac/2a

This means that we should plug the values for A B AND C into the formula

We can work out that

A = 2

B=-5

C=-9

Once we have put these into the formula we get

5+√97/4 (all over 4) aka 3.71

Also 5-√97/4 (all over 4) aka -1.21

7 0

3 years ago

Read 2 more answers

4(7/8)+7(3/4+1/8)-10 what’s the answer to this? It’s pretty tough ;-;

NISA [10]

Hopes this helps:

Answer: -3/8

Have a great day.

6 0

3 years ago

PLEASE HELP ME THIS.. I REALLY NEED HELP

Colt1911 [192]

9514 1404 393

Answer:

5 hours

Step-by-step explanation:

A quick way to look at this is to compare the difference in hourly charge to the difference in 0-hour charge.

The first day, the charge is $3 more than $12 per hour.

The second day, the charge is $12 less than $15 per hour.

The difference in 0-hour charges is 3 -(-12) = 15. The difference in per-hour charges is 15 -12 = 3. The ratio of these is ...

$15/($3/h) = 5 h

The charges are the same after 5 hours.

If you write equations for the charges, they will look like ...

y1 = 15 + 12(x -1)

y2 = 3 + 15(x -1)

Equating these charges, we have ...

15 +12(x -1) = 3 + 15(x -1)

12x +3 = 15x -12 . . . . . . . . eliminate parentheses

15 = 3x . . . . . . . . . . add 12-12x

x = 15/3 = 5 . . . . . . divide by 3

You might notice that the math here is very similar to that described in words, above.

The charges are the same after 5 hours.

6 0

3 years ago

On the first day of travel, a driver was going at a speed of 40 mph. The next day, he increased the speed to 60 mph. If he drove

gogolik [260]

Velocity, distance and time:

This question is solved using the following formula:

$v = \frac{d}{t}$

In which v is the velocity, d is the distance, and t is the time.

On the first day of travel, a driver was going at a speed of 40 mph.

Time $t_1$ , distance of $d_1$ , v = 40. So

$v = \frac{d}{t}$

$40 = \frac{d_1}{t_1}$

The next day, he increased the speed to 60 mph. If he drove 2 more hours on the first day and traveled 20 more miles

On the second day, the velocity is $v = 60$ .

On the first day, he drove 2 more hours, which means that for the second day, the time is: $t_1 - 2$

On the first day, he traveled 20 more miles, which means that for the second day, the distance is: $d_1 - 20$

Thus

$v = \frac{d}{t}$

$60 = \frac{d_1 - 20}{t_1 - 2}$

System of equations:

Now, from the two equations, a system of equations can be built. So

$40 = \frac{d_1}{t_1}$

$60 = \frac{d_1 - 20}{t_1 - 2}$

Find the total distance traveled in the two days:

We solve the system of equation for $d_1$ , which gets the distance on the first day. The distance on the second day is $d_2 = d_1 - 20$ , and the total distance is: