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LiRa [457]
2 years ago
9

In a certain isosceles triangle, the base is $1\frac 12$ times as long as each leg and the perimeter is $63. $ how long is the b

ase?
SAT
1 answer:
V125BC [204]2 years ago
7 0

The length of the base of the isosceles triangle is: 27 units long.

<h3>What is an isosceles Triangle?</h3>

A triangle with two equal legs and a base is called an isosceles triangle.

Thus:

Let x represent the length of each of the equal legs of the isosceles triangle.

Therefore:

The length of the base = 1 1/2(x) = 1.5x units

Perimeter = 63

Therefore:

x + x + 1.5x = 63

3.5x = 63

x = 63/3.5

x = 18

The length of the base = 1.5x units

Plug in the value of x

The length of the base = 1.5(18) = 27 units

Therefore, the length of the base of the isosceles triangle is: 27 units long.

Learn more about isosceles triangle on:

brainly.com/question/11884412

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Answer:

c. A doctoral degree.

Explanation:

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Most tertiary institutions has made it a prerequisite for all applicants who are interested in being a lecturer (teacher) or research inclined positions, to possess a minimum of a doctorate degree.

Hence, a doctoral degree is necessary for individuals interested in researching and teaching positions.

3 0
2 years ago
Consider the following line integral. C xy dx + x2y3 dy, c is counterclockwise around the triangle with vertices (0, 0), (1, 0),
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It looks like the integral is

\displaystyle \int_C xy \, dx + x^2y^3 \, dy

Let's close the loop by adding a line integral over the line segment joining (1, 4) to (0, 0). Then the closed loop is the triangular region

T = {(x, y) : 0 ≤ x ≤ 1 and 0 ≤ y ≤ 4x}

Since the integrand has no singularities over or on the boundary of T, we have by Green's theorem

\displaystyle \int_C xy \, dx + x^2y^3 \, dy = \iint_T \left(\frac{\partial(x^2y^3)}{\partial x} - \frac{\partial(xy)}{\partial y}\right) \, dx \, dy = \int_0^1 \int_0^{4x} (2xy^3 - x) \, dy \, dx

Compute the double integral:

\displaystyle \int_0^1 \int_0^{4x} (2xy^3 - x) \, dy \, dx = \int_0^1 \int_0^{4x} \left(128x^5 - 4x^2\right) \, dx = \frac{60}3

From this result, we subtract the line integral over the extra line segment we added. Parameterize this path by

C' : {(1 - t, 4 - 4t) : 0 ≤ t ≤ 1}

The line integral over C' is

\displaystyle \int_{C'} xy \, dx + x^2y^3 \, dy = \int_0^1 (1-t)(4-4t)(-dt) + (1-t)^2(4-4t)^3(-4\,dt)

\displaystyle \int_{C'} xy \, dx + x^2y^3 \, dy = \int_0^1 \left(256 t^5-1280 t^4+2560 t^3-2564 t^2+1288 t-260 \right) \, dt = -44

so that the line integral over C alone is 60/3 - (-44) = 64.

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2 years ago
Which of the statements is true of the energies of the bonds that break and form in an endothermic reaction?.
Vedmedyk [2.9K]

Answer:

Hi

Explanation:

The energy needed to break the required bonds is greater than the energy released when the new bonds form.

Bonds that have been formed during specific reactions require much energy to be broken down again because of the closed atomic bond that have formed

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Answer:

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So, in the question above, the end points of the segment are 0 inch and 12 inch while the length of the segment is 12inch.

This is calculated by 12 inch - 0 inch = 12 inch.

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