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3 years ago

Can you find the perimeter and area of the figure?

Mathematics

2 answers:

s344n2d4d5 [400]3 years ago

7 0

Answer:

perimeter of triangle=48 units.

area of triangle=96cm²

stellarik [79]3 years ago

3 0

Answer:

perimeter = 46

Area = 96 units^2

Step-by-step explanation:

The given is a right triangle and in right triangle the square length of hypotenuse is equal to sum of other two sides's square length

12^2 + 16^2 = hypotenuse^2

144 + 256 = hypotenuse^2

400 = hypotenuse^2

20 = hypotenuse

The perimeter is equal to the sum of all side lengths

12 + 16 + 18 = 46

The area of a triangle is calculated by multiplying height to the base and that divided by two

16 × 12 ÷ 2 = 96 units^2

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Find the sum of the reciprocals of two consecutive even integers if the larger integer is x. please answer;-;

Harman [31]

Step-by-step explanation:

Let the larger integer be x

Hence, the smaller integer = x-2

Hence,

Reciprocal of the x= 1/x

Reciprocal of x-2= 1/ x-2

Hence. their sum=

$\frac{1}{x} + \frac{1}{x - 2} \\ = \frac{x - 2 + x}{x(x - 2)} \\ = \frac{2x - 2}{ {x}^{2} - 2x}$

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3 years ago

In the figure above line m is parallel to line n.

enot [183]

Given:

Line m is parallel to line n.

The measure of ∠1 is (4x + 15)°

The measure of ∠2 is (9x + 35)°

We need to determine the measure of ∠1

Value of x:

From the figure, it is obvious that ∠1 and ∠2 are linear pairs.

Thus, we have;

$\angle 1+\angle 2=180^{\circ}$

Substituting the measures of ∠1 and ∠2, we get;

$4x+15+9x+35=180$

$13x+50=180$

$13x=130$

$x=10$

Thus, the value of x is 10.

Measure of ∠1:

The measure of ∠1 can be determined by substituting x = 10 in the measure of ∠1

Thus, we have;

$\angle 1 =4(10)+15$

$=40+15$

$\angle 1=55^{\circ}$

Thus, the measure of ∠1 is 55°

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3 years ago

If the null hypothesis is true in a chi-square test, discrepancies between observed and expected frequencies will tend to be

babymother [125]

Answer:

If the null hypothesis is true in a chi-square test, discrepancies between observed and expected frequencies will tend to be small enough to qualify as a common outcome.

Step-by-step explanation:

Here in this question, we want to state what will happen if the null hypothesis is true in a chi-square test.

If the null hypothesis is true in a chi-square test, discrepancies between observed and expected frequencies will tend to be small enough to qualify as a common outcome.

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In the question however, since the null is true, the discrepancies we will be expecting will thus be small and common.

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3 years ago

What is the equation of the graphed line written in standard form?

Dmitry_Shevchenko [17]

Answer:

Step-by-step explanation:

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3 years ago

Identify the diameter of the circular base created by folding the figure into a right cone. HELP ASAP PLEASE!!

Akimi4 [234]

let's notice something, we have a circle with a radius of 12 and one 90° sector is cut off, so only three 90° sectors of the circle are left shaded, so namely the cone will be using 3/4 of that circle.

think of it as, this shaded area is some piece of paper, and you need to pull it upwards and have the cutoff edges meet, and when that happens, you'll end up with a cone-shaped paper cup, and pour in some punch.

now, once we have pulled up the center of the circle to make our paper cup, there will be a circular base, its diameter not going to be 24, it'll be less, but whatever that base is, we know that is going to have the same circumference as those in the shaded area. Well, what is the circumference of that shaded area?

$\bf \textit{circumference of a circle}\\\\ C=2\pi r~~ \begin{cases} r=radius\\[-0.5em] \hrulefill\\ r=12 \end{cases}\implies C=2\pi 12\implies C=24\pi \implies \stackrel{\textit{three quarters of it}}{24\pi \cdot \cfrac{3}{4}} \\\\\\ 6\pi \cdot 3\implies 18\pi$

well then, the circumference of that circle at the bottom will be 18π, so, what is the diameter of a circle with a circumferenc of 18π?

$\bf \textit{circumference of a circle}\\\\ C=2\pi r~~ \begin{cases} r=radius\\[-0.5em] \hrulefill\\ C=18\pi \end{cases}\implies 18\pi =2\pi r\implies \cfrac{18\pi }{2\pi }=r\implies 9=r \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ ~\hfill \stackrel{\textit{diameter is twice the radius}}{d=18}~\hfill$

3 0

3 years ago