Ask question

Ask question

All categories

scoundrel [369]

3 years ago

6

Which has the greater circumference, circle A with radius = 6 inches, or circle B with diameter = 10 inches. Explain how you kno

w

2 answers:

Cerrena [4.2K]3 years ago

6 0

Answer:

The answer would be circle A

Step-by-step explanation:

circle A: 12 inches for diameter

circle B: 10 inches for diameter

Ksju [112]3 years ago

4 0

Answer:

skp

Step-by-step explanation:

You might be interested in

Simplify.<br><br> (12 + 8m) - 2m<br><br> 18m<br> 12 + 10m<br> 12 + 6m<br> 12 - 6m

Colt1911 [192]

Answer:

12 + 6m

Step-by-step explanation:

6 0

3 years ago

What is the value of k?<br> k = 28<br> k = 29<br> k = 31<br> k = 42

julia-pushkina [17]

Answer:

k=29

Step-by-step explanation:

we know that

-------> by supplementary angles

Solve for k

Combine like terms in the left side

Substract both sides

Divide by both sides

therefore

the answer is the option

B

6 0

3 years ago

Read 2 more answers

If triangle rst is an acute angle then m

gregori [183]

Less than 90 degrees

4 0

3 years ago

Point A is located at (3,6). The midpoint of line segment AB is point C(11,13). What are the coordinates of point B? Use the mid

lesya692 [45]

Answer: The coordinates of point B are (19 , 20).

Step-by-step explanation:

The mid point (x,y) of line joining (a,b) and (c,d) is given by :-

$(x,y)=(\dfrac{a+c}{2},\dfrac{b+d}{2})$ (1)

Given : Point A is located at (3,6). The midpoint of line segment AB is point C(11,13).

Let the coordinates of point B are (a,b) m, the according to (1) , we have

$(\dfrac{3+a}{2},\dfrac{6+b}{2})=(11,13)\\\\\Rightarrow\ \dfrac{3+a}{2}=11\ \ ,\ \ \dfrac{6+b}{2}=13\\\\\Rightarrow\ 3+a=22\ \ ,\ \ \ 6+b=26\\\\\Rightarrow\ a=19,\ \ b= 20$

Hence, the coordinates of point B are (19 , 20).

5 0

3 years ago

Let sin A = 12/13 with 90º≤A≤180º and tan B = -4/3 with 270º≤B≤360º. Find tan (A - B).

Juliette [100K]

$\tan(A-B)=\dfrac{\tan A-\tan B}{1+\tan A\tan B}=\dfrac{\tan A+\frac43}{1-\frac43\tan A}$

$\cos^2A=1-\sin^2A\implies\cos A=\pm\sqrt{1-\sin^2A}$

where the sign of the root is chosen depending on the value of $A$ . Since we know $90^\circ$ , it follows that $-1$ , i.e. $\cos A$ , which means

$\cos A=-\sqrt{1-\sin^2A}=-\dfrac5{13}$

$\implies\tan A=\dfrac{\sin A}{\cos A}=\dfrac{\frac{12}{13}}{-\frac5{13}}=-\dfrac{12}5$

$\implies\tan(A-B)=\dfrac{-\frac{12}5+\frac43}{1+\frac43\times\frac{12}5}=-\dfrac{16}{63}$

8 0

3 years ago

Read 2 more answers