All categories

3 years ago

∛(8+3√21) + ∛(8-3√21) solve without calculator

Mathematics

2 answers:

kykrilka [37]3 years ago

5 0

Answer:

Step-by-step explanation:

I was thinking about it and I prepared my paper and pen and started typing the problem but then I realized this...

8+ and 8-

Then I was like

OK....

Anna [14]3 years ago

3 0

Answer: 16

Step-by-step explanation:

8+3√21 + 8-3√21

= 8 +8 +3√21 -3√21

= 16

You might be interested in

Write the ratio as a fraction in simplest form 35 girls: 15 boys

nika2105 [10]

Answer:

7:3

Step-by-step explanation:

think of it as a fraction

6 0

2 years ago

If the cost of those white tiles are $ 3.50 what is the unit cost per white

ryzh [129]

Answer:

how many tiles are "those" bc you cant figure it out without that info sorry.

Step-by-step explanation: y

7 0

3 years ago

Help me please thanks

SVEN [57.7K]

Answer:

5/6

Step-by-step explanation:

(-6 - 4) / (3 * -4) [substitute a and b]

-10 / -12 [multiply the values in parenthesis]

5 / 6 [simplify by dividing by the common factor 2]

6 0

2 years ago

Read 2 more answers

Serengetectorupid 2013

Paha777 [63]

Answer:

Step-by-step explanation:

4 0

2 years ago

A particle moves with velocity vector

asambeis [7]

Answer:

$\vec{s(t)}=\frac{1}{2}t^2\hat{i}-(2t+1)\hat{j}+(\frac{1}{3}t^3+1)\hat{k}$

Step-by-step explanation:

We are given that velocity vector of a particle

$\vec{v(t)}=t\hat{i}-2\hat{j}+t^2\hat{k}$

When t=0 then the particle is at the point (0,-1,1).

We have to find the position of particle at time t.

We know that

Velocity = $\frac{Displacement }{time}=\frac{ds}{dt}$

Therefore, $\vec{v}=\frac{\vec{ds}}{dt}$

$\int{ds}=\int (t\hat{i}-2\hat{j}+t^2\hat{k})dt$

Integrate on both sides then we get

$\vec{s(t)}=\frac{1}{2}t^2\hat{i}-2t\hat{j}+\frac{1}{3}t^3\hat{k}+C$

$\int x^n dx=\frac{x^{n+1}}{n+1}+C$

Substitute the value of point at time t=0 then we get

C= $-\hat{j}+\hat{k}$

Substitute the value of C then we get

$\vec{s(t)}=\frac{1}{2}t^2\hat{i}-2t\hat{j}+\frac{1}{3}t^3\hat{k}-\hat{j}+\hat{k}$

Therefore, the position of particle at time t

$\vec{s(t)}=\frac{1}{2}t^2\hat{i}-(2t+1)\hat{j}+(\frac{1}{3}t^3+1)\hat{k}$

8 0

2 years ago