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3 years ago

Please i need your help !

Mathematics

2 answers:

inna [77]3 years ago

7 0

Answer:

the last one

$\\ \huge\boxed{\ddot\smile}$ $\\ \huge\boxed{\ddot\smile}$ $\\ \huge\boxed{\ddot\smile}$ $\\ \huge\boxed{\ddot\smile}$ $\\ \huge\boxed{\ddot\smile}$ $\\ \huge\boxed{\ddot\smile}$

Butoxors [25]3 years ago

3 0

The answer is D the last one

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If the circumference is 3/2 pi what is the area

11111nata11111 [884]

Here is your answer

$\frac{9}{16} \pi \: {units}^{2}$

REASON:

Circumference(C)= 2πr

, where r is radius

Given

C= (3/2) π

So,

$2\pi \: r = \frac{3}{2}\pi$

$r = \frac{\frac{3}{2}\pi}{2\pi}$

$r = \frac{3}{4}$

Now,

$area = \pi {r}^{2}$

$= \pi \times { (\frac{3}{4}) }^{2}$

$= \frac{9}{16}\pi \: {unit}^{2}$

HOPE IT IS USEFUL

8 0

3 years ago

What is b1 and r of geometrical sequence if b3 −b1 = 16 and b5 −b3 = 144.

Naily [24]

Answer:

b1 = 2 ; r = 3

Step-by-step explanation:

Given that :

if b3 −b1 = 16 and b5 −b3 = 144.

For a geometric series :

Ist term = a

Second term = ar

3rd term = ar^2

4th term = ar^3

5th term = ar^4 ;...

If b3 - b1 = 16;

ar^2 - a = 16

a(r^2 - 1) = 16 - - - (1)

b5 - b3 = 144

ar^4 - ar^2 = 144

ar^2(r^2 - 1) = 144 - - - - (2)

Divide (1) by (2)

a(r^2 - 1) / ar^2(r^2 - 1) = 16 /144

a / ar^2 = 1 / 9

ar^2 = 9a

Substitute for a in ar^2 - a = 16

9a - a = 16

8a = 16

a = 2

From ar^2 - a = 16

2r^2 - 2 = 16

2r^2 = 16 + 2

2r^2 = 18

r^2 = 18 / 2

r^2 = 9

r = √9

r = 3

Hence ;

a = b1 = 2 ; r = 3

7 0

3 years ago

Jack uses a probability simulator to roll a six-sided number cube 100 times and to flip a coin 100 times. The results of the exp

Mandarinka [93]

The experimental probability is the number of specific outcomes divided by the sample size...

P(6)=27/100 (27%)

P(H)=41/100 (41%)

Not sure, but if you meant rolling a 6 AND getting a head then:

P(6 AND H)=(27/100)(41/100)=1107/10000 (11.07%)

5 0

3 years ago

Read 2 more answers

<img src="https://tex.z-dn.net/?f=%20%5Crm%20%5Cint_%7B0%7D%5E2%20%5Cleft%28%20%5Csqrt%5B3%5D%7B%20%7Bx%7D%5E%7B2%7D%20%20%2B%20

Verizon [17]

If f(x) has an inverse on [a, b], then integrating by parts (take u = f(x) and dv = dx), we can show

$\displaystyle \int_a^b f(x) \, dx = b f(b) - a f(a) - \int_{f(a)}^{f(b)} f^{-1}(x) \, dx$

Let $f(x) = \sqrt{1 + x^3}$ . Compute the inverse:

$f\left(f^{-1}(x)\right) = \sqrt{1 + f^{-1}(x)^3} = x \implies f^{-1}(x) = \sqrt[3]{x^2-1}$

and we immediately notice that $f^{-1}(x+1)=\sqrt[3]{x^2+2x}$ .

So, we can write the given integral as

$\displaystyle \int_0^2 f^{-1}(x+1) + f(x) \, dx$

Splitting up terms and replacing $x \to x-1$ in the first integral, we get

$\displaystyle \int_1^3 f^{-1}(x) \, dx + \int_0^2 f(x) \, dx = 2 f(2) - 0 f(0) = 2\times3+0\times1=\boxed{6}$

7 0

1 year ago

I need help fast in these questions.

cupoosta [38]

Answer:

all...

Step-by-step explanation:

3 0

3 years ago