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2 years ago

What is the area of the shaded region?

Mathematics

1 answer:

ella [17]2 years ago

3 0

33cm

Because add all of them then subtract then do the parenthasees and you get 33

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I don't get it plz help

Salsk061 [2.6K]

Just use the order of operations. PEMDAS, parentheses, exponents, multiplication, division, addition and subtraction.

In the first problem 2 + 6*3^2, there are no parentheses, so next would be exponents.
There is an exponent so you do that first.
3^2 = 9

Now we have 2 + 6*9
Multiplication is next and it is in the problem, so we do 6*9 = 54
So now we have 2 + 54 which equals 56.

6 0

3 years ago

Read 2 more answers

Answer this volume based Question. I will make uh brainliest + 50 points

Harlamova29_29 [7]

Answer:

$\huge{\purple {r= 2\times\sqrt[3]3}}$

$\huge 2\times \sqrt [3]3 = 2.88$

Step-by-step explanation:

For solid iron sphere:
radius (r) = 2 cm (Given)

Formula for $V_{sphere}$ is given as:

$V_{sphere} =\frac{4}{3}\pi r^3$

$\implies V_{sphere} =\frac{4}{3}\pi (2)^3$

$\implies V_{sphere} =\frac{32}{3}\pi \:cm^3$

For cone:
r : h = 3 : 4 (Given)
Let r = 3x & h = 4x

Formula for $V_{cone}$ is given as:

$V_{cone} =\frac{1}{3}\pi r^2h$

$\implies V_{cone} =\frac{1}{3}\pi (3x)^2(4x)$

$\implies V_{cone} =\frac{1}{3}\pi (36x^3)$

$\implies V_{cone} =12\pi x^3\: cm^3$

It is given that: iron sphere is melted and recasted in a solid right circular cone of same volume
$\implies V_{cone} = V_{sphere}$

$\implies 12\cancel{\pi} x^3= \frac{32}{3}\cancel{\pi}$

$\implies 12x^3= \frac{32}{3}$

$\implies x^3= \frac{32}{36}$

$\implies x^3= \frac{8}{9}$

$\implies x= \sqrt[3]{\frac{8}{3^2}}$

$\implies x={\frac{2}{ \sqrt[3]{3^2}}}$

$\because r = 3x$

$\implies r=3\times {\frac{2}{ \sqrt[3]{3^2}}}$

$\implies r=3\times 2(3)^{-\frac{2}{3}}$

$\implies r= 2\times (3)^{1-\frac{2}{3}}$

$\implies r= 2\times (3)^{\frac{1}{3}}$

$\implies \huge{\purple {r= 2\times\sqrt[3]3}}$
Assuming log on both sides, we find:

$log r = log (2\times \sqrt [3]3)$

$log r = log (2\times 3^{\frac{1}{3}})$

$log r = log 2+ log 3^{\frac{1}{3}}$

$log r = log 2+ \frac{1}{3}log 3$

$log r = 0.4600704139$

Taking antilog on both sides, we find:

$antilog(log r )= antilog(0.4600704139)$

$\implies r = 2.8844991406$

$\implies \huge \red{r = 2.88\: cm}$

$\implies 2\times \sqrt [3]3 = 2.88$

8 0

2 years ago

Mr. Poyser uses 4 2/3 cups of flour to make 7 dozen oatmeal-raisin cookies.

Jobisdone [24]

Answer:X unknown cu. of flour/ 5 dozen cookies

3 0

1 year ago

Estimate the solution of the system x+y=4 2x-y=6 by sketching a graph of each linear function. Then solve the system algebraical

Juli2301 [7.4K]

Answer:

(10/3, 2/3)

Step-by-step explanation:

You may find it easier to write these two equations x+y=4 2x-y=6 in a column:

x+y=4

2x-y=6

Adding these together will eliminate y:

3x = 10. Then x = 10/3.

Substituting 10/3 for x in the first equation results in:

10/3 + y = 4

Clear fractions by multiplying all three terms by 3:

10 + 3y = 12

Then 3y = 2, and y = 2/3.

The solution is (10/3, 2/3)

8 0

3 years ago

Which of the following is an asymptote of y = sec(x)?

babunello [35]

Answer: option d. x = 3π/2

Solution:

function y = sec(x)

1) y = 1 / cos(x)

2) When cos(x) = 0, 1 / cos(x) is not defined

3) cos(x) = 0 when x = π/2, 3π/2, 5π/2, 7π/2, ...

4) limit of sec(x) = lim of 1 / cos(x).

When x approaches π/2, 3π/2, 5π/2, 7π/2, ... the limit →+/- ∞.

So, x = π/2, x = 3π/2, x = 5π/2, ... are vertical asymptotes of sec(x).

Answer: 3π/2

The figures attached will help you to understand the graph and the existence of multiple asymptotes for y = sec(x).

8 0

3 years ago

Read 2 more answers