All categories

2 years ago

2 x 5 x 0 please help

Mathematics

2 answers:

igor_vitrenko [27]2 years ago

8 0

Answer: 0

Step-by-step explanation: hope it helps

Kazeer [188]2 years ago

8 0

0 multiplied by anything is going to always be 0

You might be interested in

Which is traveling faster, a car whose velocity vector is 201 + 25), or a car whose velocity vector is 30i, assuming that the un

ioda

Answer with explanation:

For any object having the velocity vector as

$\overrightarrow{v}=v_x\widehat{i}+v_y\widehat{j}+v_z\widehat{k}$

the magnitude of velocity is given by

$|v|=\sqrt{v_x^2+v_y^2+v_z^2}$

For car 1 the velocity vector is

$\overrightarrow{v}_1=20\widehat{i}+25\widehat{j}$

Therefore

$|v_1|=\sqrt{20^2+25^2}\\\\\therefore v_1=32.0156units$

Similarly for car 2 we have

$\overrightarrow{v}_2=30\widehat{i}$

Therefore

$|v_2|=\sqrt{30^2}\\\\\therefore v_2=30.0units$

Comparing both the values we find that car 1 has the greater speed.

7 0

3 years ago

Yx divide 2 use x=7 and y=2

miskamm [114]

Answer:

finish the sentence

Step-by-step explanation:

8 0

3 years ago

Which measure of central tendency is MOST EASILY affected by outliers

Serjik [45]

The average can be easily affected by outliers because an outlier can bring the average much higher or lower depending on the value. For example, if this was the data set

0,50,51,52
the average would be 38.25

But without the 0, the average would be 51.<span />

5 0

3 years ago

Find all solutions of the equation: 2cos^2x-cosx=1

Art [367]

If you're using the app, try seeing this answer through your browser: brainly.com/question/3166243

——————————

Solve the trigonometric equation:

$\mathsf{2\,cos^2\,x-cos\,x=1}\\\\ \mathsf{2\,cos^2\,x-cos\,x-1=0}$

Make a substitution:

$\mathsf{cos\,x=t\qquad (-1\le t\le 1)}$

and the equation becomes

$\mathsf{2t^2-t-1=0}$

Rewrite conveniently – t as + t – 2t, and then factor the left-hand side by grouping:

$\mathsf{2t^2+t-2t-1=0}\\\\ \mathsf{t\cdot (2t+1)-1\cdot (2t+1)=0}$

Factor out 2t + 1:

$\mathsf{(2t+1)\cdot (t-1)=0}\\\\ \begin{array}{rcl} \mathsf{2t+1=0}&~\textsf{ or }~&\mathsf{t-1=0}\\\\ \mathsf{2t=1}&~\textsf{ or }~&\mathsf{t=1}\\\\ \mathsf{t=\dfrac{\,1\,}{2}}&~\textsf{ or }~&\mathsf{t=1} \end{array}$

Substitute back for t = cos x:

$\begin{array}{rcl}\mathsf{cos\,x=\dfrac{\,1\,}{2}}&~\textsf{ or }~&\mathsf{cos\,x=1}\\\\ \mathsf{cos\,x=cos\,60^\circ}&~\textsf{ or }~&\mathsf{cos\,x=cos\,0} \end{array}$

Therefore,

$\begin{array}{rcl} \mathsf{x=\pm\,60^\circ+k\cdot 360^\circ}&~\textsf{ or }~&\mathsf{cos\,x=0+k\cdot 360^\circ} \end{array}$

where k is an integer.

Solution set:

$\mathsf{S=\left\{x\in\mathbb{R}:~~x=-\,60^\circ+k\cdot 360^\circ~~or~~x=60^\circ+k\cdot 360^\circ~~or~~x=k\cdot 360^\circ,~~k\in\mathbb{Z}\right\}}$

I hope this helps. =)

3 0

3 years ago

How do you find the variable when it’s on both sides of the equation

ASHA 777 [7]

If, for example, you had x on one side and x on the other, then subtracting x from both sides "disappears" x. If no other power of x shows up on either side, then your equation either has no solution or is always true.

3 0

3 years ago