MATHEMATICS HIGH SCHOOL

A sector of a circle is shown. Find its area, rounded to the nearest tenth. A)
9.8 cm2

B)
10.5 cm2

C)
11.2 cm2

D)
12.3 cm2

I would like to know how you did it too

Answers

Answer 1

Answer:

The area of the sector of the circle, to the nearest tenth is: B. 10.5 cm²

What is the Area of the Sector of a Circle?

Area of the sector of a circle = ∅/360 × πr².

r = radius.

Given:

∅ = 68°

Radius (r) = 4.2 cm

Substitute

Area of the sector = 68/360 × π × 4.2²

Area of the sector = 10.5 cm²

Learn more about area of sector on:

brainly.com/question/22972014

Answer 2

Answer: Area of a circle = pi * radius^2
But we're not talking about a whole circle, but a portion of the circle.
The proportion of the area of the whole circle occupied by this sector is equal to the ratio of the created angle to the angle of a whole circle;
that may seem a bit confusing but basically:
Area of the sector = pi * radius^2 * 68/360
so...
Area = pi * (4.2)^2 * 68/360
= 10.467...
= 10.5 cm^2

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Write an algebraic expression for two fifths of the square of a number

Answers

The algebraic expression for two-fifths of the square of a number will be (2/5)x².

What is Algebra?

The analysis of mathematical representations is algebra, and the handling of those symbols is logic.

The square of the number means the number is multiplied by itself.

The algebraic expression for two-fifths of the square of a number will be

Let the number be x.

Then the square of the number will be

⇒ x²

And the two-fifth number is expressed as,

⇒ 2/5

The algebraic expression for two-fifths of the square of a number will be

⇒ (2/5)x²

Thus, the algebraic expression for two-fifths of the square of a number will be (2/5)x².

More about the Algebra link is given below.

brainly.com/question/953809

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"of" means you are multiplying and just use "x" to represent "a number". Therefore:

(2/5)x^2

Which expression is equivalent to x^4-y^2 ?

Answers

$x^(4)$ - y²
x² · x² - y ·y
(x² - y)(x + y) = $x^(4)$ - y²

I need help ion understand this

Answers

Answer:

-1 7/8

Step-by-step explanation:

What's the answer do this question??

Answers

It is A) skewed left because the long tail in the box plot is on the left hand side. The mean is also on the left hand side of the peak so it is skewed left.

Hope this helps :)

If the x- intercept is 2 and the y- intercept is -5 and the slope is 1/3. Write in equation in slope intercept form and standard form.

Answers

Answer:

The equation in slope-intercept form is

$y=(5)/(2)x-5$

The equation in the standard form will be:

$(5)/(2)x-y=5$

Step-by-step explanation:

The x-intercept is obtained when we set the value y=0

As the x-intercept is 2, therefore the point representing

the x-intercept will be: (2, 0)

The y-intercept is obtained when we set the value x=0

As the y-intercept is -5, therefore the point representing

the y-intercept will be: (0, -5)

So we get the two points

(2, 0)

(0, -5)

Finding the slope between (2, 0) and (0, -5)

$\left(x_1,\:y_1\right)=\left(2,\:0\right),\:\left(x_2,\:y_2\right)=\left(0,\:-5\right)$

$\mathrm{Slope}=(y_2-y_1)/(x_2-x_1)$

$m=(-5-0)/(0-2)$

$m=(5)/(2)$

Using the point-slope form of the line equation

$y-y_1=m\left(x-x_1\right)$

Here m is the slope

substituting the values m = 5/2 and the point (2, 0)

$y-0=(5)/(2)\left(x-2\right)$

so writing the equation in slope-intercept form

As we know that the slope-intercept form is

$y=mx+b$

here

m = gradient or slop

b = y-intercept

$y=(5)/(2)\left(x-2\right)$

$y=(5)/(2)x-5$

Hence, the equation in slope-intercept form is

$y=(5)/(2)x-5$

Writing the equation in the standard form form

As we know that the equation in the standard form is

$Ax+By=C$

where x and y are variables and A, B and C are constants

As we already know the equation in slope-intercept form

$y=(5)/(2)x-5$

so the equation in the standard form will be:

$(5)/(2)x-y=5$

For each quadratic function, find the domain range vertex and axis of symmetry y=(x-1) squared +1

Answers

$y=(x-1)^2 +1 \n \n For \ the \ quadratic \ function \n \n f(x) =a(x-h)^2+k \n \n where \ a, \ h \ and \ k \ are \ real \ numbers \ with \ a \neq 0 , \n \n the \ vertex \ is \ (h,k) =(1,1) \n \n a \ symmetry \ about \ the \ vertical \ line \ x = h \n \n x= 1$