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3 years ago

1. Find the circumference and area of circle with radius 5 cm.

Mathematics

2 answers:

Natali5045456 [20]3 years ago

6 0

Answer:

Hey There!

Let's solve...

First let's find out circumference

We know the formula is

$2 \pi r \\ \\ = 2 \times 5 \times 3.14cm \\ \\ = 10 \times 3.14cm \\ \\ = 31.4cm$

Now let's find out area...

The formula is

$\pi r^{2} \\ = 3.14 \times {5}^{2} \\ = 3.14 \times 25 \\ = 78.50 {cm}^{2}$

<h2>I hope it is helpful to you...</h2><h3>Cheers!_____________</h3>

SOVA2 [1]3 years ago

4 0

Answer:

3l.42cm

Step-by-step explanation:

C =2(3.14) × 5cm

C= 31.42cm

Area = pie r2

area =3.14 × 5cm ×5cm

area= 78.5cm

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Quadrilateral abcd has vertices A(4,0),B(3,3),C(-3,1) and D(-2,-2)

bonufazy [111]

Answer:

very easy

Step-by-step explanation:

First of all

made a distance between

and diagnals

If the all are equal means it is a square

if diagnols are different means it is rectangle

if diagnals are not equal means it is parallelogram

thats it try your best

you can do it

best of luck

6 0

2 years ago

A small plane and a large plane are 6.8km from each other, at the same altitude (height). From an observation tower, the two air

Lemur [1.5K]

Answer:

7.9km

Step-by-step explanation:

(a)See attached for the diagram representing this situation.

(b)

In Triangle ABC

$\text{Using Law of Sines}\\\dfrac{\sin A}{a}=\dfrac{\sin C}{c} \\\dfrac{\sin A}{5.2}=\dfrac{\sin 58^\circ}{6.8} \\\sin A=5.2 \times \dfrac{\sin 58^\circ}{6.8}\\A=\arcsin (5.2 \times \dfrac{\sin 58^\circ}{6.8})\\A=40.43^\circ$

Next, we determine the value of Angle B.

$\angle A+\angle B+\angle C=180^\circ\\40.43+58+\angle B=180^\circ\\\angle B=180^\circ-(40.43+58)\\\angle B=81.57^\circ$

Finally, we find b.

$\text{Using Law of SInes}\\\dfrac{b}{\sin B}=\dfrac{c}{\sin C} \\\dfrac{b}{\sin 81.57^\circ}=\dfrac{6.8}{\sin 58^\circ} \\b=\dfrac{6.8}{\sin 58^\circ} \times \sin 81.57^\circ\\b=7.9km $ (to the nearest tenth of a kilometer)$

The distance between the small plane and the observation tower is 7.9km.

8 0

4 years ago

Pls help ill give brainliest

Ymorist [56]

Answer:

$\boxed {\boxed {\sf (5g^2+2)(2g-5)}}$

Step-by-step explanation:

We are given the expression:

$10g^3-25g^2+4g-10$

There are no common factors between the four numbers, however the first two have a common factor and the last two do too. Therefore, we can factor by grouping.

Group the first two terms and the last two.

$(10g^3-25g^2)+(4g-10)$

Find the greatest common factor (GCF) of the first group. It is 5g². Factor it out of the first group. You can do this by dividing both terms by the GCF.

10g³/5g²= 2g
-25g²/5g² = -5

$5g^2 (2g-5) + (4g-10)$

Repeat with the second group. The GCF is 2.

4g/2 = 2g
-10/2= -5

$5g^2 (2g-5) + 2(2g-5)$

There is another GCF: 2g-5. We can factor this out of both terms.

5g²(2g-5)/2g-5= 5g²
2(2g-5)/ 2g-5=2

$(5g^2+2)(2g-5)$

This cannot be factored further, so it is the answer.

4 0

3 years ago

Help! The picture is with it to see the question

bearhunter [10]

Answer:

Step-by-step explanation:

sqrt8*sqrt8 = sqrt8^2 = 8

4 0

3 years ago

Extra Credits:

Effectus [21]

Answer:

Missing Number) 55

Property being used) Distributive

Step-by-step explanation:

To find the missing number, just plug in reasonable numbers from the other equation. It is distributive because the 55 is being distributed into everything that is inside the parentheses. If you need more help, comment and I'll be more than glad to assist you! :P

8 0

3 years ago