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3 years ago

Which are vertical angles?

Mathematics

2 answers:

Oksanka [162]3 years ago

4 0

Hello.

Vertical angles are two sets of angles congruent to each other, both on opposite sides, but measure the same.

In the diagram shown, both angles ∠BFC and ∠EFA are identical, it seems mirrored, thus they are vertical angles.

The correct answer is B.

guapka [62]3 years ago

3 0

Step-by-step explanation:

AFE-BFD<---------------

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How would I set up the equation to find mRQS on this circle?

Leni [432]

MRQS=360-mSR. or mRQS=31x+13...To solve the measure of the minor arc RQS you would use this... 31x+13+(6+18x/2. )=360.

3 0

3 years ago

Am I correct? (Perfect Square Form)

natka813 [3]

Answer:

$(x -3 )^2 = 10$

Step-by-step explanation:

Given

$x^2 - 6x - 1 = 0$

Required: Express as a perfect square

$x^2 - 6x - 1 = 0$

Add 10 to both sides

$x^2 - 6x - 1 +10= 0+10$

$x^2 - 6x +9= 10$

Expand

$x^2 - 3x - 3x+9= 10$

Factorize:

$x(x - 3) - 3(x - 3)= 10$

Factorize:

$(x -3 )(x - 3) = 10$

$(x -3 )^2 = 10$

8 0

3 years ago

Between 0 degrees Celsius and 30 degrees Celsius, the volume V (in cubic centimeters) of 1 kg of water at a temperature T is giv

Ira Lisetskai [31]

Answer:

T = 3.967 C

Step-by-step explanation:

Density = mass / volume

Use the mass = 1kg and volume as the equation given V, we will come up with the following equation

D = 1 / 999.87−0.06426T+0.0085043T^2−0.0000679T^3

= (999.87−0.06426T+0.0085043T^2−0.0000679T^3)^-1

Find the first derivative of D with respect to temperature T

dD/dT = $\dfrac{70000000\left(291T^2-24298T+91800\right)}{\left(679T^3-85043T^2+642600T-9998700000\right)^2}$

Let dD/dT = 0 to find the critical value we will get

$\dfrac{70000000\left(291T^2-24298T+91800\right)}$ = 0

Using formula of quadratic, we get the roots:

T = 79.53 and T = 3.967

Since the temperature is only between 0 and 30, pick T = 3.967

Find 2nd derivative to check whether the equation will have maximum value:

$-\dfrac{140000000\left(395178T^4-65993368T^3+3286558821T^2+2886200857800T-121415215620000\right)}{\left(679T^3-85043T^2+642600T-9998700000\right)^3}$

Substituting the value with T=3.967,

d2D/dT2 = -1.54 x 10^(-8) a negative value. Hence It is a maximum value

Substitute T =3.967 into equation V, we get V = 0.001 i.e. the volume when the the density is the highest is at 0.001 m3 with density of

D = 1/0.001 = 1000 kg/m3

Therefore T = 3.967 C

3 0

4 years ago

The width x of your iris decreases from 4 millimeters to 2 millimeters when you enter a dark room. How many times greater is the

inysia [295]

Answer:

The area of the Iris is 4 times greater.

Step-by-step explanation:

Firstly, we have to realize that we cannot solve this problem without knowing the radius of the iris. because from the information given, the radius of the iris could well be the size of the galaxy and contract from width of 4mm to 2mm and we wouldn't know!

The average radius of the iris is 6mm, so we take this value.

Now, initially the width of the iris is 4mm, that means the radius of the pupil is:

$6mm-4mm=2mm$