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larisa86 [58]
3 years ago
10

Who was the first prince of Russia to be crowned as a czar

Mathematics
1 answer:
liraira [26]3 years ago
4 0
It should be Ivan IV
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Find the volume of a right circular cone that has a height of 3.9 ft and a base with a circumference of 16.3 ft. Round your answ
MissTica

Answer:

The answer to your question is Volume = 27.5 ft³

Step-by-step explanation:

Data

Volume = ?

height = 3.9 ft

Circumference = 16.3 ft

Process

1.- Calculate the radius of the Cone

   Circumference = 2πr = 16.3 ft

-Solve for r

                                    r = 16.3/2π

                                    r = 16.3/6.28

                                    r = 2.5955 ft

2.- Calculate the volume

Volume = 1/3πr²h

-Substitution

Volume = 1/3(3.14)(2.5955)²(3.9)

-Simplification

Volume = 1/3(82.4966)

-Result

Volume = 27.5 ft³

8 0
3 years ago
Read 2 more answers
Simplify log(16x²) ± 2㏒(1÷×)
sp2606 [1]

Answer: just simplify condense it then your answer will be log(16)

8 0
2 years ago
Which of the following is a proportion?
poizon [28]

Answer:

Option C, 14/21=9/21 is the proportion.

Step-by-step explanation:

Given:

a.5/7 = 10/12

b.9/15 = 12/18

c.4/6 = 8/12

d.14/21 = 9/12

We have to find the proportions.

In proportion two ratios are equal, as it has an equality sign in between so both side must be of same ratio in its simplest form.

Let's work with option C

⇒ \frac{4}{6}=\frac{8}{12}

To find the simplest form we have to divide the numerator and denominator with same digit (or its factor).

Simplest form of  \frac{8/2}{12/2} = \frac{4}{6} , \frac{4/2}{6/2}= \frac{2}{3}

Simplest form of \frac{4}{6} = \frac{2}{3}

Both sides in option C have equal ratios of 2/3.

So 4/6=8/12 is in proportion.

6 0
3 years ago
Correct/best answer will get brainliest!
vodka [1.7K]
The answer would be b.100 because its clearly not a right triangle
6 0
3 years ago
Simplify the expression given below.<br><br> x+2/ 4x^{2} +5x+1 *4x+1/2x-4[/tex]
VashaNatasha [74]

Answer:

pls write understandably

7 0
3 years ago
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