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maria [59]
3 years ago
13

Write an odd cube number smaller than 100.

Mathematics
2 answers:
serious [3.7K]3 years ago
8 0

Answer:

(All numbers are 15, 22, 50, 114, 167, 175, 186, 212, 231, 238, 303, 364, 420, 428, 454.) >8, 27, 64 are the cube numbers < 100.

IRISSAK [1]3 years ago
4 0

Answer: >8, 27, 64 are the cube numbers < 100

Step-by-step explanation:

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what is the blank for

Step-by-step explanation:

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6 0
3 years ago
Find angle measures and arc measures
Vlad1618 [11]

Answer:

Step-by-step explanation:

a. We know that KH is a diameter of the circle and that along angles along it on either side adds up to 180° as it's a straight line.

-Therefore, we have that:

m\angle a+ \angle GaH=180\textdegree\\\\\\m\angle a=180-35\\\\=145\textdegree

Hence, the angle a is equal to 180°

b.We know that GJ is a diameter of the circle and that along angles along it on either side adds up to 180° as it's a straight line.

-Therefore, we have that:

m\angle b+\angle GaK=180\textdegree\\\\m\angle b=180\textdegree -145\textdegree\\\\\\\\=35\textdegree

Hence, the angle b is equal to 35°

3 0
3 years ago
If the temperature outside is 30°C, what is the temperature in F°​
emmasim [6.3K]

Answer:

<u>86°F</u>

Step-by-step explanation:

4 0
3 years ago
A Cepheid variable star is a star whose brightness alternately increases and decreases. For a certain star, the interval between
sattari [20]

Answer:

a)

B'(t) = \dfrac{0.9\pi}{4.4}\cos\bigg(\dfrac{2\pi t}{4.4}\bigg)

b) 0.09

Step-by-step explanation:

We are given the following in the question:

B(t) = 4.2 +0.45\sin\bigg(\dfrac{2\pi t}{4.4}\bigg)

where B(t) gives the brightness of the star at time t, where t is measured in days.

a) rate of change of the brightness after t days.

B(t) = 4.2 +0.45\sin\bigg(\dfrac{2\pi t}{4.4}\bigg)\\\\B'(t) = 0.45\cos\bigg(\dfrac{2\pi t}{4.4}\bigg)\times \dfrac{2\pi}{4.4}\\\\B'(t) = \dfrac{0.9\pi}{4.4}\cos\bigg(\dfrac{2\pi t}{4.4}\bigg)

b) rate of increase after one day.

We put t = 1

B'(t) = \dfrac{0.9\pi}{4.4}\cos\bigg(\dfrac{2\pi t}{4.4}\bigg)\\\\B'(1) = \dfrac{0.9\pi}{4.4}\bigg(\cos(\dfrac{2\pi (1)}{4.4}\bigg)\\\\B'(t) = 0.09145\\B'(t) \approx 0.09

The rate of increase after 1 day is 0.09

8 0
3 years ago
Solve for the area using heron´s formula
Vadim26 [7]

9514 1404 393

Answer:

  252.8 cm²

Step-by-step explanation:

The missing side of the right triangle can be found from the Pythagorean theorem:

  s² = 20² -16² = 400 -256 = 144

  s = 12 . . . . cm

The area of a right triangle is more easily found using the traditional area formula:

  A = 1/2bh

  A = 1/2(12 cm)(16 cm) = 96 cm² (left-side triangle)

The area of the triangle on the right can be found from Heron's formula. The semiperimeter is ...

  s = (16 +20 +23)/2 = 29.5

The area is ...

  A = √(29.5(29.5 -16)(29.5 -20)(29.5 -23)) = √(29.5·13.5·9.5·6.5)

  A = √24591.9375 ≈ 156.818 . . . . . cm² (right-side triangle)

Then the total area of the figure is ...

  A = 96 cm² +156.818 cm² = 252.818 cm² . . . . total area

4 0
3 years ago
Read 2 more answers
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