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MissTica
3 years ago
5

Help me please again

Mathematics
1 answer:
sveticcg [70]3 years ago
7 0

Answer:

area = 282.74 yd^2

volume = 314.16 yd^2

Step-by-step explanation:

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1.5.3
SIZIF [17.4K]

Answer:

A. x^2 = (4x - 1)^2 + (6x + 7)^2

Step-by-step explanation:

The question seems to be incomplete. However, we'll make the following assumption.

Opposite = 6x + 7

Adjacent = 4x - 1

Hypothenuse = x

Pythagoras theorem states that:

Hypothenuse^2 = Opposite^2 + Adjacent^2

Substitute values for Hypothenuse, Opposite and Adjacent.

This gives

x^2 = (6x + 7)^2 + (4x - 1)^2

or

x^2 = (4x - 1)^2 + (6x + 7)^2

From the list of given options.

A answers the question

5 0
3 years ago
The dotted lines show how to cut the pie. Identify m∠LPM.
allsm [11]
Due to the cause it would end up being the second option
4 0
3 years ago
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Problem 1: Joseph is going to buy a cell phone plan on his own. The most important thing for him is being able to text. So, he h
andriy [413]

Step-by-step explanation: lol like a dozen

4 0
3 years ago
39 times 49 times 81
d1i1m1o1n [39]

Answer:

154,791

Step-by-step explanation:

7 0
3 years ago
Read 2 more answers
Solve 3x² + 4x = 2. (2 points)
uysha [10]

Answer:

Step-by-step explanation:

As it is a second order equation, it means that it has two possible answers and they are x_1 and x_2.

The famous quadratic formula for solving any second order equation is the following:

x_{1} = \frac{-b + \sqrt{b^2 - 4 ac}}{2a}\\x_{2}=\frac{-b - \sqrt{b^2 - 4 ac}}{2a}

Where a is the coefficient of x^2, b is the coefficient of x, and c is the free term. In other words,

a = 3\\b=4\\c=-2

as the equation should be in the following form:

a x^2 + bx+c = 0

Therefore the possible answer should be the following,

\frac{-4 + \sqrt{4^2 - 4*3*(-2)}}{2*3}=\frac{-4 +\sqrt{16 + 24}}{6} =\frac{-4 + \sqrt{40}}{6}=\\ \frac{-4 + \sqrt{4*10}}{6} = \frac{-4 + 2\sqrt{10}}{6}\frac{2*(-2 + \sqrt{10})}{2*3} = \frac{-2 + \sqrt{10}}{3}

by dividing the numerator and denominator by 2, we can deduce the following,

6 0
2 years ago
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