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3 years ago

A rectangle has sides of length 8 cm and 3 cm. Find the length of it's diagonals.

Mathematics

2 answers:

True [87]3 years ago

6 0

Diagonal formula: a^2 + b^2 = c^2

Now, solve for the length of the diagonal.

8^2 + 3^2 = c^2

64 + 9 = c^2

73 = c^2

√73 = √c^2

8.5 ≈ c (This is the diagonal and its rounded).

Best of Luck!

slava [35]3 years ago

4 0

Using pythagoras a^2 + b^2 = c^2
8^2 + 3^2 = c^2
64 + 9 = 73
The square root of 73 is 8.54
So the length of the diagonals are 8.54cm

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Integrate with respect to x 1/√(1-2x)

Ilia_Sergeevich [38]

Answer:

-√(1 - 2x) + C

Step-by-step explanation:

1/√(1-2x)

We want to integrate it. Thus;

∫1/√(1 - 2x) dx

Let u = 1 - 2x

Thus;

du/dx = -2

Thus, dx = -½du

Thus,we now have;

-½∫1/√(u) du

By application of power rule, we will now have;

-½∫1/√(u) du = -√(u) + C

Plugging in the value of u, we will have;

-√(1 - 2x) + C

3 0

3 years ago

PLEASE ANSWER! Given the functions f(x) = x2 + 6x - 1, g(x) = -x2 + 2, and h(x) = 2x2 - 4x + 3, rank them from least to greatest

Ratling [72]

Answer:

he rank from least to great based on their axis of symmetry:

0, 1, -3 ⇒ g(x), h(x), f(x)

So, option C is correct.

Step-by-step explanation:

A quadratic equation is given by:

$ax^2+bx+c =0$

Here, a, b and c are termed as coefficients and x being the variable.

Axis of symmetry can be obtained using the formula

$x = \frac{-b}{2a}$

Identification of a, b and c in f(x), g(x) and h(x) can be obtained as follows:

$f(x) = x^2 + 6x - 1$

⇒ a = 1, b = 6 and c = -1

$g(x) = -x^2 + 2$

⇒ a = -1, b = 0 and c = 2

$h(x) = 2^2 - 4x + 3$

⇒ a = 2, b = -4 and c = 3

So, axis of symmetry in $f(x) = x^2 + 6x - 1$ will be:

$x = \frac{-b}{2a}$

x = -6/2(1) = -3

and axis of symmetry in $g(x) = -x^2 + 2$ will be:

$x = \frac{-b}{2a}$

x = -(0)/2(-1) = 0

and axis of symmetry in $h(x) = 2^2 - 4x + 3$ will be:

$x = \frac{-b}{2a}$

x = -(-4)/2(2) = 1

So, the rank from least to great based on their axis of symmetry:

0, 1, -3 ⇒ g(x), h(x), f(x)

So, option C is correct.

Keywords: axis of symmetry, functions

Learn more about axis of symmetry from brainly.com/question/11800108

#learnwithBrainly

7 0

3 years ago

Points M, N, and P are respectively the midpoints of sides AC , BC , and AB of △ABC. Prove that the area of △MNP is on fourth of

Hunter-Best [27]

Answer:

The area of △MNP is one fourth of the area of △ABC.

Step-by-step explanation:

It is given that the points M, N, and P are the midpoints of sides AC, BC and AB respectively. It means AC, BC and AB are median of the triangle ABC.

Median divides the area of a triangle in two equal parts.

Since the points M, N, and P are the midpoints of sides AC, BC and AB respectively, therefore MN, NP and MP are midsegments of the triangle.

Midsegments are the line segment which are connecting the midpoints of tro sides and parallel to third side. According to midpoint theorem the length of midsegment is half of length of third side.

Since MN, NP and MP are midsegments of the triangle, therefore the length of these sides are half of AB, AC and BC respectively. In triangle ABC and MNP corresponding side are proportional.

$\triangle ABC \sim \triangle NMP$