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3 years ago

Obtuse equilateral Which name correctly classifies this triangle?

Mathematics

1 answer:

professor190 [17]3 years ago

3 0

The answer is B.
Hope that this helps!

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How many meters in six feet? i need it to night

anyanavicka [17]

1.8288 meters = 6 feet

3 0

3 years ago

Given O A ‾ ⊥ O C ‾ OA ⊥ OC start overline, O, A, end overline, \perp, start overline, O, C, end overline m ∠ B O C = 6 x − 6 ∘

Sphinxa [80]

Answer:

x = 8

m∠BOC = 37°

m∠AOB = 53°

Step-by-step explanation:

If OA is perpendicular to OC, this means that <AOC = 90°

Given

m∠BOC=6x−6

m ∠AOB = 5 x + 8

The expression is true

m∠BOC+m∠AOB= m∠AOC

6x-6+5x+8 = 90

Find x:

11x + 2 = 90

11x = 90-2

11x = 88

x = 88/11

x =8

Get m∠BOC:

m∠BOC= 6x-6

m∠BOC= 6(8)-11

m∠BOC = 48-11

m∠BOC= 37°

Get m∠AOB;

m∠AOB = 90-m∠BOC

m∠AOB =90-37

m∠AOB = 53°

7 0

3 years ago

Read 2 more answers

If x is 6, What is X+4

hichkok12 [17]

Answer:

x = 10

Step-by-step explanation:

We know x = 6

asked: x + 4

Plus it in =>

6 + 4

=> 10

8 0

4 years ago

The area of the square, the

const2013 [10]

Answer:

Step-by-step explanation:

I think your question is missed of key information, allow me to add in and hope it will fit the original one.

Please have a look at the attached photo.

My answer:

As given in the question, we know that:

The ratio of the area of the circle to the area of the square is π/4

The formula to find the volume of the cone is:

V = 1/3*the height*the base area

<=> V1 = 1/3*h*π $r^{2}$

The formula to find the volume of the pyramid is:

V2 = 1/3*the height*the base area

<=> V = 1/3*h*4 $r^{2}$

=> the ratio of volume of the cone to the pyramid is:

= $\frac{V1}{V2}$

= (1/3*h*π $r^{2}$ ) / ( 1/3*h*4 $r^{2}$ )

= π/4

S we can conclude that the volume of the cone equals π/4 the volume of the pyramid

Hope it will find you well.

7 0

3 years ago

A Cepheid variable star is a star whose brightness alternately increases and decreases. Suppose that Cephei Joe is a star for wh

4vir4ik [10]

Answer:

$\dfrac{0.7\pi}{5.3}\cos(\dfrac{2\pi t}{5.3})$

0.16

Step-by-step explanation:

According to the question the mathematical model should be

$B(t)=2.9+0.35\sin(\dfrac{2\pi t}{5.3})$

Differentiating with respect to time we get

$B'(t)=0.35\cos(\dfrac{2\pi t}{5.3})\times(\dfrac{2\pi}{5.3})\\\Rightarrow B'(t)=\dfrac{0.7\pi}{5.3}\cos(\dfrac{2\pi t}{5.3})$

So, the rate of change of brightness after t days is $\dfrac{0.7\pi}{5.3}\cos(\dfrac{2\pi t}{5.3})$

After 1 day means $t=1$

$B'(1)=\dfrac{0.7\pi}{5.3}\cos(\dfrac{2\pi\times 1}{5.3})\\\Rightarrow B'(1)=0.16$

The rate of increase after one day is 0.16.

8 0

3 years ago