All categories

3 years ago

If the square root of x is greater than x then x could be

Mathematics

1 answer:

Shtirlitz [24]3 years ago

8 0

Any decimal under 1, i believe.

You might be interested in

What is 15% of 82? <br><br> PLEASE HELP ME IT I DON'T KNOW IT

Rufina [12.5K]

Answer: the answer is 12.3

Step-by-step explanation:

Just do 15/100(82)

5 0

2 years ago

Read 2 more answers

What is being constructed based on the markings in the following diagram?

Liula [17]

Answer:

perpendicular line through a point on a line

Step-by-step explanation:

The circle centered at C seems intended to produce point D at the same distance as point B. That is, C is the midpoint of BD.

The circles centered at B and D with radius greater than BC seems intended to produce intersection points G and H. (It appears accidental that those points are also on circle C. As a rule, that would be difficult to do in one pass.)

So. points G and H are both equidistant from points B and D. A line between them will intersect point C at right angles to AB.

Segment GH is perpendicular to AB through point C (on AB).

6 0

3 years ago

Read 2 more answers

Solve the following quadratic for the values x

mamaluj [8]

Answer:

x² + 6x - 27 = 0

x² + 9x - 3x - 27 = 0

x ( x + 9 ) - 3 ( x + 9 ) = 0

( x + 9 ) ( x - 3 ) = 0

Thus ,

x = -9 or x = 3

I hope it helps you

#Indian : )

8 0

3 years ago

Read 2 more answers

Find cotθ if tanθ = 2 and 180° < θ < 270° <br> Plz help

Musya8 [376]

Answer:

1/2

Step-by-step explanation:

cotangent is the reciprocal of a tangent

tanθ = 2

cotθ = 1/2

6 0

2 years ago

Find the solution of the problem (1 3. (2 cos x - y sin x)dx + (cos x + sin y)dy=0.

lakkis [162]

Answer:

$2*sin(x)+y*cos(x)-cos(y)=C_1$

Step-by-step explanation:

Let:

$P(x,y)=2*cos(x)-y*sin(x)$

$Q(x,y)=cos(x)+sin(y)$

This is an exact differential equation because:

$\frac{\partial P(x,y)}{\partial y} =-sin(x)$

$\frac{\partial Q(x,y)}{\partial x}=-sin(x)$

With this in mind let's define f(x,y) such that:

$\frac{\partial f(x,y)}{\partial x}=P(x,y)$

and

$\frac{\partial f(x,y)}{\partial y}=Q(x,y)$

So, the solution will be given by f(x,y)=C1, C1=arbitrary constant

Now, integrate $\frac{\partial f(x,y)}{\partial x}$ with respect to x in order to find f(x,y)

$f(x,y)=\int\ 2*cos(x)-y*sin(x)\, dx =2*sin(x)+y*cos(x)+g(y)$

where g(y) is an arbitrary function of y

Let's differentiate f(x,y) with respect to y in order to find g(y):

$\frac{\partial f(x,y)}{\partial y}=\frac{\partial }{\partial y} (2*sin(x)+y*cos(x)+g(y))=cos(x)+\frac{dg(y)}{dy}$

Now, let's replace the previous result into $\frac{\partial f(x,y)}{\partial y}=Q(x,y)$ :

$cos(x)+\frac{dg(y)}{dy}=cos(x)+sin(y)$

Solving for $\frac{dg(y)}{dy}$

$\frac{dg(y)}{dy}=sin(y)$

Integrating both sides with respect to y:

$g(y)=\int\ sin(y) \, dy =-cos(y)$

Replacing this result into f(x,y)

$f(x,y)=2*sin(x)+y*cos(x)-cos(y)$

Finally the solution is f(x,y)=C1 :

$2*sin(x)+y*cos(x)-cos(y)=C_1$

7 0

3 years ago