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3 years ago

What is the circumference area?

Mathematics

1 answer:

tia_tia [17]3 years ago

4 0

I believe the circumference is 52.15.

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The product of three different positive integers is 8 what is the sum of these integers

lys-0071 [83]

Answer:

(4)(2)(1)=

4+2+1=

explanation:

6 0

2 years ago

What is 10000 more than 46952 and what is the missing<br> number

Morgarella [4.7K]

10000 more than 46952 is 56952

3 0

3 years ago

Read 2 more answers

Find exact values for sin θ, cos θ and tan θ if csc θ = 3/2 and cos θ < 0.

kumpel [21]

Answer:

Part 1) $sin(\theta)=\frac{2}{3}$

Par 2) $cos(\theta)=-\frac{\sqrt{5}}{3}$

Part 3) $tan(\theta)=-\frac{2\sqrt{5}}{5}$

Step-by-step explanation:

step 1

Find the $sin(\theta)$

we have

$csc(\theta)=\frac{3}{2}$

Remember that

$csc(\theta)=\frac{1}{sin(\theta)}$

therefore

$sin(\theta)=\frac{2}{3}$

step 2

Find the $cos(\theta)$

we know that

$sin^{2}(\theta) +cos^{2}(\theta)=1$

we have

$sin(\theta)=\frac{2}{3}$

substitute

$(\frac{2}{3})^{2} +cos^{2}(\theta)=1$

$\frac{4}{9} +cos^{2}(\theta)=1$

$cos^{2}(\theta)=1-\frac{4}{9}$

$cos^{2}(\theta)=\frac{5}{9}$

square root both sides

$cos(\theta)=\pm\frac{\sqrt{5}}{3}$

we have that

$cos(\theta) < 0$ ---> given problem

$cos(\theta)=-\frac{\sqrt{5}}{3}$

step 3

Find the $tan(\theta)$

we know that

$tan(\theta)=\frac{sin(\theta)}{cos(\theta)}$

we have

$sin(\theta)=\frac{2}{3}$

$cos(\theta)=-\frac{\sqrt{5}}{3}$

substitute

$tan(\theta)=\frac{2}{3}:-\frac{\sqrt{5}}{3}=-\frac{2}{\sqrt{5}}$

Simplify

$tan(\theta)=-\frac{2\sqrt{5}}{5}$

6 0

3 years ago

(05.08A)Triangle ABC is transformed to similar triangle A′B′C′ below:

Romashka-Z-Leto [24]

Answer:

1st option is correct i.e., 1 over 2

Step-by-step explanation:

Given that Triangle ABC transformed to Triangle A'B'C'.

Coordinated of Triangle ABC are A(2,6) , B(2,4) and C(4,4)

Coordinated of Triangle A'B'C' are A'(1,3), B'(1,2) and C'(2,2)

Scale factor of Dilation is the no. of times coordinates of vertices of a figure increase or decrease.i.e.,

if A(a,b) ⇒ A'(c,d)

where, a.k=c & b.k=d then k is scale factor of dilation.

Value of k is same for all vertices.

For coordinates of A & A' ,we have $2.k=1\implies k=\frac{1}{2} \:\: and \:\: 6.k=3\implies k=\frac{1}{2}$

For coordinates of B & B' ,we have $2.k=1\implies k=\frac{1}{2} \:\: and \:\: 4.k=2\implies k=\frac{1}{2}$

For coordinates of C & C' ,we have $4.k=2\implies k=\frac{1}{2} \:\: and \:\: 4.k=2\implies k=\frac{1}{2}$

Therefore, the scale factor of dilation is $\:\frac{1}{2}\:$ since in all cases $k=\frac{1}{2}$

1st option is correct i.e., 1 over 2

3 0

3 years ago

20 points!! Can you help me with this math problem??

umka2103 [35]

D. 663 square meters is the anwser

4 0

3 years ago