All categories

1 year ago

Which is more, 70 inches or 6 feet?

1 answer:

5 0

Correct Answer:
6 feet

Explanation:
6 feet = 12 x 6 = 72 inches
72 inches > 70 inches
Hence, 6 feet is greater than 70 inches.

I would appreciate it if you mark my answer as brainliest.

You might be interested in

5/7 + 1/7<br> Put in simplest form please!

Rasek [7]

Answer:

6/7

Step-by-step explanation:

5/7 + 1/7

Since the denominators are the same, we just add the numerators

5+1 = 6

Then we put the numerator over the denominator

6/7

5 0

4 years ago

-3+15mx=q <br><br> Solve for m

rjkz [21]

Answer:

m = q+3/15x

step-by-step explanations:

3 0

3 years ago

I don't understand this class and my teacher is stupid and won't go over it again to help me better understand

allochka39001 [22]

Answer:

Well i believe the answer is obtuse

Step-by-step explanation:

4 0

3 years ago

Read 2 more answers

prove that ( sin theta cos theta = cot theta ) is not a trigonometric identity by producing a counterexample

user100 [1]

Answer:

To prove that ( sin θ cos θ = cot θ ) is not a trigonometric identity.

Assume θ with a value and substitute with it.

Let θ = 45°

So, L.H.S = sin θ cos θ = sin 45° cos 45° = (1/√2) * (1/√2) = 1/2

R.H.S = cot θ = cot 45 = 1

So, L.H.S ≠ R.H.S

So, sin θ cos θ = cot θ is not a trigonometric identity.

8 0

3 years ago

Solve the equation using the substitution u = y/x. When u = y/x is substituted into the equation, the equation becomes separable

bekas [8.4K]

Answer:

$\frac{u^2+3}{-u^3+u^2-2u}u'=\frac{1}{x}$

Step-by-step explanation:

First step: I'm going to solve our substitution for y:

$u=\frac{y}{x}$

Multiply both sides by x:

$ux=y$

Second step: Differentiate the substitution:

$u'x+u=y'$

Third step: Plug in first and second step into the given equation dy/dx=f(x,y):

$u'x+u=\frac{x(ux)+(ux)^2}{3x^2+(ux)^2}$

$u'x+u=\frac{ux^2+u^2x^2}{3x^2+u^2x^2}$

We are going to simplify what we can.

Every term in the fraction on the right hand side of equation contains a factor of $x^2$ so I'm going to divide top and bottom by $x^2$ :

$u'x+u=\frac{u+u^2}{3+u^2}$

Now I have no idea what your left hand side is suppose to look like but I'm going to keep going here:

Subtract u on both sides:

$u'x=\frac{u+u^2}{3+u^2}-u$

Find a common denominator: Multiply second term on right hand side by $\frac{3+u^2}{3+u^2}$ :

$u'x=\frac{u+u^2}{3+u^2}-\frac{u(3+u^2)}{3+u^2}$

Combine fractions while also distributing u to terms in ( ):

$u'x=\frac{u+u^2-3u-u^3}{3+u^2}$

$u'x=\frac{-u^3+u^2-2u}{3+u^2}$

Third step: I'm going to separate the variables:

Multiply both sides by the reciprocal of the right hand side fraction.

$u' \frac{3+u^2}{-u^3+u^2-2u}x=1$

Divide both sides by x:

$\frac{3+u^2}{-u^3+u^2-2u}u'=\frac{1}{x}$

Reorder the top a little of left hand side using the commutative property for addition:

$\frac{u^2+3}{-u^3+u^2-2u}u'=\frac{1}{x}$

The expression on left hand side almost matches your expression but not quite so something seems a little off.

5 0

3 years ago