All categories

1 year ago

Please answer the question below

Mathematics

1 answer:

alexandr402 [8]1 year ago

4 0

The value of x in the given sector is determined as 7.1.

<h3>Area of the sector</h3>

Total area of the sector = area of the shaded + area of the unshaded

Since shaded and unshaded areas are equal, the value of x is calculated as follows;

πr²θ/360 = πx²60/360 + πx²60/360

πr²θ/360 = 2(πx²θ/360)

πr² = 2(πx²)

r² = 2x²

x² = r²/2

x² = (10²)/2

x² = 50

x = √50

x = 7.1

Thus, the value of x in the given sector is determined as 7.1.

Learn more about area of sector here: brainly.com/question/22972014

#SPJ1

You might be interested in

Whats the missing varible for the following questions

marta [7]

Answers + Explanation:

1. x + 7 =10

if you know some number + 7 is 10, you know that number must be 3.

so, x = 3

the other way you can solve this would be:

x + 7 = 10

subtract 7 from both sides of the equation

x + 7 - 7 = 10 - 7

x = 3

2. Use the same strategy:

x - 8 = -5

add 8 to both sides of the equation

x - 8 + 8 = -5 + 8

x = 3

3. x + 3 = 6

subtract 3 from both sides of the equation

x + 3 - 3 = 6 - 3

x = 3

4. x + 6 = 3

subtract 6 from both sides of the equation

x + 6 - 6 = 3 - 6

x = -3

7 0

2 years ago

Find the value of x

Vilka [71]

<span>An exterior angle of a triangle is equal to the sum of the opposite interior
angles </span>⇒ x = 30 + 90 = 120°

6 0

3 years ago

(1/16)^x+3=(1/4)^x+1

Anika [276]

Answer:

x=1

....................

6 0

3 years ago

Determine if S could lie on the perpendicular bisector of QR with the given coordinates

ehidna [41]

Refer to picture for answers.

4 0

3 years ago

On a linear X temperature scale, water freezes at −115.0°X and boils at 325.0°X. On a linear Y temperature scale, water freezes

belka [17]

Answer:

The current temperature on the X scale is 1150 °X.

Step-by-step explanation:

Let is determine first the ratio of change in X linear temperature scale to change in Y linear temperature scale:

$r = \frac{\Delta T_{X}}{\Delta T_{Y}}$

$r = \frac{325\,^{\circ}X-(-115\,^{\circ}X)}{-25\,^{\circ}Y - (-65.00\,^{\circ}Y)}$

$r = 11\,\frac{^{\circ}X}{^{\circ}Y}$

The difference between current temperature in Y linear scale with respect to freezing point is:

$\Delta T_{Y} = 50\,^{\circ}Y - (-65\,^{\circ}Y)$

$\Delta T_{Y} = 115\,^{\circ}Y$

The change in X linear scale is:

$\Delta T_{X} = r\cdot \Delta T_{Y}$

$\Delta T_{X} = \left(11\,\frac{^{\circ}X}{^{\circ}Y} \right)\cdot (115\,^{\circ}Y)$

$\Delta T_{X} = 1265\,^{\circ}X$

Lastly, the current temperature on the X scale is:

$T_{X} = -115\,^{\circ}X + 1265\,^{\circ}X$

$T_{X} = 1150\,^{\circ}X$

The current temperature on the X scale is 1150 °X.

5 0

3 years ago