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Effectus [21]
3 years ago
8

Find the area of a circle whose radius is 52 km. Round to nearest tenth

Mathematics
2 answers:
kozerog [31]3 years ago
8 0

Step-by-step explanation:

3.142×52^2=8494.866535

8494.9 nearest tenth

goblinko [34]3 years ago
7 0

Answer:

8494.87 is the actual answer but the rounded version is 8494.9

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Please help I need workings
Aleonysh [2.5K]
So first thing is find the number of small boxes. Find 1% of 120, 1.2. Multiply 1.2 by 35, it’s 42.
Next find what the percent that 2/5 is. Divide two by five, it’s .4 or 40%. We already know one percent is 1.2 so multiply 1.2 by 40, 48.
Add this two.
42+48=90.
120-90=30

Small boxes:42
Medium boxes: 48
Large:30
42*6=252
48*10=480
15*30=450
1182 eggs total.
8 0
3 years ago
PLEASE HELP, 50 POINTS! This is a triangle investigation.
mylen [45]
A) they fornite a right angle and 2 acute angles(not 100% sure of what the Questions is asking)


B) they appear to form a triangle

C) the angles add up to 180 as they always do in all triangles





















B)
5 0
3 years ago
Read 2 more answers
Find the area of the regular polygon. Round to the nearest tenth.
ladessa [460]

Answer:

374.123 m^2

rounded it would be 374.1 m^2

4 0
3 years ago
Read 2 more answers
Click an item in the list or group of pictures at the bottom of the problem and, holding the button down, drag it into the corre
LiRa [457]

Answer:

x = 7, -7

Step-by-step explanation:

(2x-1)(3x+2)+x+292

3x - 2x +x = 2 - 1 + 292

(6)x - 294 = 6(x - 49)

Use communicative property of multiplication

Example: (A+B)(A-B)

Ax^{2} -AB+BA-Bx^{2} =Ax^{2} -AB+AB-Bx^{2} =\\Ax^{2} -Bx^{2}

49 is the square of 7

x square is the square of x to the power of 1

(x+7)(x-7)

5 0
3 years ago
GCSE 995381
nikdorinn [45]

Answer:

2.9 cm^2

Step-by-step explanation:

Given,

Angle of sector = 30°

Radius of circle, r = 8 cm

Area of the shaded region, A =?

The diagram is attached below.

Now,

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

Area of triangle = \frac{1}{2}\times base \times height

We know that,

\sin \theta = \dfrac{P}{H}

\sin 30^\circ= \dfrac{P}{8}

P = \dfrac{8}{2}

P = 4\ cm

And

\cos \theta =\dfrac{B}{H}

\cos 30^\circ = \dfrac{B}{8}

B = \dfrac{8\times \sqrt{3}}{2}

B = 4\sqrt 3

Area of sector = \dfrac{\theta}{360^\circ}\times \pi r^2

Area of sector =  \dfrac{30^\circ}{360^\circ}\times \pi \times 8^2

                        = 16.75 cm^2

Area of triangle = \dfrac{1}{2} \times 4\sqrt 3 \times 4

                          = 13.85 cm^2

Area of shaded region = 16.75 - 13.85

                                      = 2.9 cm^2

7 0
2 years ago
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