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Ostrovityanka [42]

2 years ago

14

Are the triangles congrouent? if so which postulate are the congrouent by?

1 answer:

ankoles [38]2 years ago

3 0

Answer:

ASA

Step-by-step explanation:

They give us one pair of angles and one pair of congruent sides for it to be ASA we need an additional angle but on the other side from where the last angle was. By vertical angles proof we can see that the angles at the intersection will be congruent so you have one angle then a side then another angle

Hope This Helps :)

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The perimeter of a certain rectangle is 16 times the width. The length is 12 cm more than the width. Find the length and width o

Blizzard [7]

Check the picture below

surely you know how much that is.

8 0

3 years ago

Which of the following polynomials has x = − 2/7 and x = 0 as solutions?

weeeeeb [17]

Answer:

what are the polynomials?

Step-by-step explanation:

8 0

2 years ago

Solve this and show work

just olya [345]

Answer:

1.71

Step-by-step explanation:

x/4 = 3/7

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12/7 = 1.714285714285714

Rounded to the nearest hundreths, the answer is 1.71

6 0

2 years ago

Two circles are concentric if they have the same center.

Tpy6a [65]

9514 1404 393

Answer:

(b) (x – 4)^2 + (y – 6)^2 = 16

Step-by-step explanation:

The equation for a circle with center (h, k) and radius r is ...

(x -h)^2 +(y -k)^2 = r^2

You want the equation for a circle with center (4, 6) and radius 4. That equation is ...

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3 0

2 years ago

Water is added to a cylindrical tank of radius 5 m and height of 10 m at a rate of 100 L/min. Find the rate of change of the wat

nirvana33 [79]

Answer:

$V = \pi r^2 h$

For this case we know that $r=5m$ represent the radius, $h = 10m$ the height and the rate given is:

$\frac{dV}{dt}= \frac{100 L}{min}$

$Q = 100 \frac{L}{min} *\frac{1m^3}{1000L}= 0.1 \frac{m^3}{min}$

And replacing we got:

$\frac{dh}{dt}=\frac{0.1 m^3/min}{\pi (5m)^2}= 0.0012732 \frac{m}{min}$

And that represent $0.127 \frac{cm}{min}$

Step-by-step explanation:

For a tank similar to a cylinder the volume is given by:

$V = \pi r^2 h$

For this case we know that $r=5m$ represent the radius, $h = 10m$ the height and the rate given is:

$\frac{dV}{dt}= \frac{100 L}{min}$

For this case we want to find the rate of change of the water level when h =6m so then we can derivate the formula for the volume and we got:

$\frac{dV}{dt}= \pi r^2 \frac{dh}{dt}$

And solving for $\frac{dh}{dt}$ we got:

$\frac{dh}{dt}= \frac{\frac{dV}{dt}}{\pi r^2}$

We need to convert the rate given into m^3/min and we got:

$Q = 100 \frac{L}{min} *\frac{1m^3}{1000L}= 0.1 \frac{m^3}{min}$

And replacing we got:

$\frac{dh}{dt}=\frac{0.1 m^3/min}{\pi (5m)^2}= 0.0012732 \frac{m}{min}$

And that represent $0.127 \frac{cm}{min}$

5 0

3 years ago