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3 years ago

Find the common difference of the sequence shown 1/6, 1/4, 1/3

Mathematics

2 answers:

Anvisha [2.4K]3 years ago

8 0

Answer:

Common difference of the sequence is = $\frac{1}{12}$

Step-by-step explanation:

Common difference of any sequence is defined by the difference between a term and its successive term.

Sequence is $\frac{1}{6}, \frac{1}{4},\frac{1}{3}.....$

Common difference = $T_{2}-T_{1}=\frac{1}{4}-\frac{1}{6}$

= $\frac{2}{24}$

= $\frac{1}{12}$

Now we will do the same for $T_{3}$ and $T_{2}$

$T_{3}-T_{2}$ = $\frac{1}{3}-\frac{1}{4}$

= $\frac{4-3}{12}$

= $\frac{1}{12}$

Therefore, answer is $\frac{1}{12}$

Damm [24]3 years ago

7 0

1/4 - 1/6 = (6-4)/24 = 2/24 = 1/12
1/3 - 1/4 = (4-3)/12 = 1/12

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What is the equation of the line passing through the points (–25, 50) and (25, 50) in slope-intercept form?

allochka39001 [22]

The equation of the line passing through the points (-25, 50) and (25, 50) in slope-intercept form is y = 50. Hence, 4th option is the right choice.

The slope-intercept form of a line is written as y = mx + b, where m is the slope of the line, and b is the y-intercept.

The slope of a line passing through the points (x₁, y₁) and (x₂, y₂) can be calculated using the formula, slope (m) = (y₂ - y₁)/(x₂ - x₁).

Therefore, slope of the line passing through the points (-25, 50) and (25, 50) can be calculated as m = (50 - 50)/(25 - (-25)) = 0/(-50) = 0.

We can find the equation of the line using the point-slope formula, according to which, a line having a slope m and passing through the point (x₁, y₁) can be written as y - y₁ = m(x - x₁).

Therefore, the equation of the given line can be written as:

y - 50 = 0(x - 25)

or, y - 50 = 0,

or, y = 50.

Therefore, the equation of the line passing through the points (-25, 50) and (25, 50) in slope-intercept form is y = 50. Hence, 4th option is the right choice.

Learn more about the equation of a line at

brainly.com/question/18831322

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2 years ago

Solve the inequality –8 < x – 14.

kodGreya [7K]

The answer is x=6. Hope this helped.

7 0

3 years ago

Using polar coordinates, evaluate the integral which gives the area which lies in the first quadrant below the line y=5 and betw

vfiekz [6]

First, complete the square in the equation for the second circle to determine its center and radius:

x ² - 10x + y ² = 0

x ² - 10x + 25 + y ² = 25

(x - 5)² + y ² = 5²

So the second circle is centered at (5, 0) with radius 5, while the first circle is centered at the origin with radius √100 = 10.

Now convert each equation into polar coordinates, using

x = r cos(θ)

y = r sin(θ)

Then

x ² + y ² = 100 → r ² = 100 → r = 10

x ² - 10x + y ² = 0 → r ² - 10 r cos(θ) = 0 → r = 10 cos(θ)

y = 5 → r sin(θ) = 5 → r = 5 csc(θ)

See the attached graphic for a plot of the circles and line as well as the bounded region between them. The second circle is tangent to the larger one at the point (10, 0), and is also tangent to y = 5 at the point (0, 5).

Split up the region at 3 angles θ₁, θ₂, and θ₃, which denote the angles θ at which the curves intersect. They are

θ₁ = 0 … … … by solving 10 = 10 cos(θ)

θ₂ = π/6 … … by solving 10 = 5 csc(θ)

θ₃ = 5π/6 … the second solution to 10 = 5 csc(θ)

Then the area of the region is given by a sum of integrals:

$\displaystyle \frac12\left(\left\{\int_0^{\frac\pi6}+\int_{\frac{5\pi}6}^{2\pi}\right\}\left(10^2-(10\cos(\theta))^2\right)\,\mathrm d\theta+\int_{\frac\pi6}^{\frac{5\pi}6}\left((5\csc(\theta))^2-(10\cos(\theta))^2\right)\,\mathrm d\theta\right)$

$=\displaystyle 50\left\{\int_0^{\frac\pi6}+\int_{\frac{5\pi}6}^{2\pi}\right\} \sin^2(\theta)\,\mathrm d\theta+\frac12\int_{\frac\pi6}^{\frac{5\pi}6}\left(25\csc^2(\theta) - 100\cos^2(\theta)\right)\,\mathrm d\theta$

To compute the integrals, use the following identities:

sin²(θ) = (1 - cos(2θ)) / 2

cos²(θ) = (1 + cos(2θ)) / 2

and recall that

d(cot(θ))/dθ = -csc²(θ)

You should end up with an area of

$=\displaystyle25\left(\left\{\int_0^{\frac\pi6}+\int_{\frac{5\pi}6}^{2\pi}\right\}(1-\cos(2\theta))\,\mathrm d\theta-\int_{\frac\pi6}^{\frac{5\pi}6}(1+\cos(2\theta))\,\mathrm d\theta\right)+\frac{25}2\int_{\frac\pi6}^{\frac{5\pi}6}\csc^2(\theta)\,\mathrm d\theta$

$=\boxed{25\sqrt3+\dfrac{125\pi}3}$

We can verify this geometrically:

• the area of the larger circle is 100π

• the area of the smaller circle is 25π

• the area of the circular segment, i.e. the part of the larger circle that is bounded below by the line y = 5, has area 100π/3 - 25√3

Hence the area of the region of interest is

100π - 25π - (100π/3 - 25√3) = 125π/3 + 25√3

as expected.

3 0

2 years ago

Please help answer and explanation im very confused.

Veseljchak [2.6K]

Answer:

1) 18m²

2) 25cm²

Step-by-step explanation:

a = ½ × b × h
a = ½ × 9 × 4
a = ½ × 36
a = 18m²

a = ½ × b × h
a = ½ × 10 × 5
a = 5 × 5
a = 25 cm²

6 0

2 years ago

Can someone please help me on these questions!!!

34kurt

Answer:

#1 is 28

Step-by-step explanation:

Set the perimeters equal, as 6x - 2 = 8x - 12, solve for x = 5, and plug in.

3 0

3 years ago