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2 years ago

2^4 - 2^3 + (2^2) simplify

Mathematics

2 answers:

Wittaler [7]2 years ago

6 0

The answer is 12

Hoped I help :)

Ipatiy [6.2K]2 years ago

6 0

This might help you answer your question

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Find solution to the system of linear equations. 5x1 + x2 = 0 , 25x1 + 5x2 = 0

Zigmanuir [339]

Answer:

the system of equation has infinite solution

Step-by-step explanation:

$5x_1 + x_2 = 0$ , $25x_1 + 5x_2 = 0$

Solve the first equation for x_2

Subtract 5x1 on both sides

$5x_1 + x_2 = 0$

$x_2 =-5x_1$

Now substitute -5x1 on second equation

$25x_1 + 5x_2 = 0$

$25x_1 + 5(-5x_1)= 0$

$25x_1-25x_1= 0$

0=0

So the system of equation has infinite solution

5 0

3 years ago

Read 2 more answers

Calculate the unit rate: write your answer as a decimal 1/2 mile in ever 1/3 hour

Alexandra [31]

Answer:

$unit \: rate = \frac{distance}{time}$

unit rate is similar to speed / velocity:

$unit \: rate = \frac{ (\frac{1}{2} )}{ (\frac{1}{3})} \\ \\ = \frac{1}{2} \div \frac{1}{3} \\ \\ = \frac{1}{2} \times \frac{3}{1} \\ \\ = > { \boxed{ \boxed{unit \: rate = \frac{3}{2} \: miles \: per \: hour}}}$

[unit rate must have units including per <time> ]

7 0

2 years ago

Write first 10 pirme numbers and find median

likoan [24]

Answer:

2 3 5 7 11 13 17 19 23 27

Step-by-step explanation:

12 I think

5 0

3 years ago

Calculate the max shear of cantilever balcony 2m wide 0.2m deep and 1.5m length.

Vikki [24]

Answer:

14.4 kN

Step-by-step explanation:

cantilever balcony 2m wide 0.2m deep and 1.5m length.

= 2 x 0.2 x 1.5

= 0.6 m³

weight of the balcony

= 0.6 x 24

= 14.4 kN.

therefore the max shear 14.4 kN and is acting at support.

8 0

3 years ago

Mason walked on Monday, Wensday, and Friday. Use these clues to find how far he walked each day. These distances were six-eights

krok68 [10]

Answer:

Monday he walked $\frac{1}{4}miles$

Wednesday he walked $\frac{6}{8}miles$

Friday he walked $\frac{1}{6}miles$

Step-by-step explanation:

Given Mason walked on Monday, Wednesday and Friday. These distances were six-eights mile, one-fourth mile, and one-sixth mile. He did not walk the farthest on Monday. He walked less on Friday than Monday. we have to find how far he walked each day.

distances are $\frac{6}{8}, \frac{1}{4}, \frac{1}{6}$ that are 0.75, 0.25 and 0.17 respectively.

Now, he didn't walk farthest on Monday and also walked less on Friday than Monday.

∴ Less distance travelled is $\frac{1}{6}$ which is on friday and then $\frac{1}{4}$ on monday.

Rest distance which is $\frac{6}{8}$ on wednesday.

Hence, Monday he walked $\frac{1}{4}miles$

Wednesday he walked $\frac{6}{8}miles$

Friday he walked $\frac{1}{6}miles$

4 0

3 years ago

Read 2 more answers