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3 years ago

How many permutations in the word calculus

1 answer:

6 0

<span>The word CALCULUS contains

2 C's + 1 A + 2 L's + 2 U's + 1 S = 8 letters, in all.

The number of distinct permutations of all these 8 letters

taken together is

... ( 8! ) / [ (2!) (1!) (2!) (2!) (1!) ]

= [ 8 · (7!) ] / 8

= 7 · (6!)

= 7 ( 720 )

= 5040 ........................................... . </span>

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Tina ate 21 starbursts. If she ate 35% of the package of Starbursts, how many Starbursts are in the package?

mojhsa [17]

Answer:

60 starbursts

Step-by-step explanation:

We can set up an equation to find the answer. 21/x=0.35/1 because we're trying to find how many starbursts make up 100% of the package. So then to solve, we take 21x1 and then divide that by 0.35. From this we find that x=60.

5 0

2 years ago

Help me please m, it due soon

allsm [11]

Surface area= 2(6x15)+2(7x15)+2(6x7)
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3 years ago

Find the area of the shaded region. Round the the nearest tenth.

Brums [2.3K]

Answer:

226.084 sq.m.

Step-by-step explanation:

Radius of circle = 18.6 cm

Area of sector = $\frac{\theta}{360^{\circ} }\pi r^2$

Area of sector = $\frac{123}{360} \times 3.14 \times (18.6)^2$

Area of sector = $371.157 m^2$

Area of triangle = $\frac{1}{2} \times a \times b \times Sin x$

a = OA = radius = 18.6 cm

b = OB = radius = 18.6 cm

x =123 °

Substitute the values in the formula

Area of triangle = $\frac{1}{2} \times 18.6 \times 18.6 \times Sin 123$

Area of triangle = $145.073 m^2$

So, Area of shaded region = Area of sector - Area of triangle

= $371.157 m^2-145.073 m^2$

= $226.084 m^2$

Hence the area of the shaded region is 226.084 sq.m.

5 0

3 years ago

POSSIBLE

Levart [38]

I believe the answer is supplementary

7 0

2 years ago

Please help me figure out this problem

tiny-mole [99]

Answer:

$f(g(x))=f\left(-\dfrac{3x+21}{5}\right)=x\\\\g(f(x))=g\left(-\dfrac{5}{3}x-7\right)=x$

Step-by-step explanation:

Fill in the given expressions and simplify.

$f(g(x))=f\left(-\dfrac{3x+21}{5}\right)=-\dfrac{5}{3}\left(-\dfrac{3x+21}{5}\right)-7\\\\=\left(-\dfrac{5}{3}\right)\left(-\dfrac{3x}{5}\right)-\left(\dfrac{5}{3}\right)\left(-\dfrac{21}{5}\right)-7\\\\=\dfrac{15x}{15}+\dfrac{5\cdot 21}{3\cdot 5}-7=x+7-7=x$

Going the other way, we have ...

$g(f(x))=\displaystyle -\frac{3\left(-\frac{5}{3}x-7\right)+21}{5}=-\frac{-5x-21+21}{5}=x$

5 0

3 years ago