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3 years ago

Please help! If these two shapes are similar, what is the measure of missing length h?

Mathematics

2 answers:

Drupady [299]3 years ago

8 0

Answer:

Step-by-step explanation:

Katyanochek1 [597]3 years ago

6 0

Answer:

Step-by-step explanation:

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What is the image of (1, -6) for a 90° counterclockwise rotation about the origin?

maksim [4K]

Answer:

Step-by-step explanation:

As the counter clockwise rule is A(x,y) becomes A'(-y,x).

So (1, -6) becomes (6, 1)

Hope it helps you

5 0

3 years ago

Cody was 165cm tall on the first day of school this year, which was 10% taller than he was on the first day of school last year.

klemol [59]

Answer: 148.5

Step-by-step explanation:

well there is no real question here but cody is 165 which means you have to multiply that by 10% or 0.10 which gives you 16.5 and subtract that from the total

4 0

3 years ago

ASAP WILL MARK BRAINLEST

Brut [27]

Answer:

1296π m^3

Step-by-step explanation:

The volume of a cylinder is given by

V=πr^2h

We are given the diameter. To find the radius, divide the diameter in half.

r = d/2 = 24/2 = 12

Substituting in what we know

V = pi * (12)^2 * 9

V = pi *1296 m^3

7 0

4 years ago

Read 2 more answers

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

IF TAN 45 = x/20, solve for x.

amm1812

Answer:

x=20

Step-by-step explanation:

math

6 0

4 years ago

Read 2 more answers