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Aleksandr [31]
3 years ago
12

What is the area of this parallelogram PLEASE HELP QUICKLY!!!20 points!!!

Mathematics
2 answers:
olga nikolaevna [1]3 years ago
8 0

Answer:

72

Step-by-step explanation:

mario62 [17]3 years ago
4 0

Area of a parallelogram is Length X width.

Length = 9+5 = 14

Width = 8

Area = 14 * 8 = 112 m^2

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A water sprinkler has a range of 5 meters as shown. The jet of water from the sprinkler sweeps out at an angle of 85° as the noz
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I found the image that accompanied this problem.
We need to solve for the area of the sector.

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A = 85/360 * 3.14 * 5²
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Answer is C. 18.5 square meters.
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3 years ago
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If Triangle ABC is rotated 90 degrees clockwise about the origin, what will be the new coordinates of vertex B?
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ithe answer to your question is going to be c

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3 years ago
Determine which of the sets of vectors is linearly independent. A: The set where p1(t) = 1, p2(t) = t2, p3(t) = 3 + 3t B: The se
defon

Answer:

The set of vectors A and C are linearly independent.

Step-by-step explanation:

A set of vector is linearly independent if and only if the linear combination of these vector can only be equalised to zero only if all coefficients are zeroes. Let is evaluate each set algraically:

p_{1}(t) = 1, p_{2}(t)= t^{2} and p_{3}(t) = 3 + 3\cdot t:

\alpha_{1}\cdot p_{1}(t) + \alpha_{2}\cdot p_{2}(t) + \alpha_{3}\cdot p_{3}(t) = 0

\alpha_{1}\cdot 1 + \alpha_{2}\cdot t^{2} + \alpha_{3}\cdot (3 +3\cdot t) = 0

(\alpha_{1}+3\cdot \alpha_{3})\cdot 1 + \alpha_{2}\cdot t^{2} + \alpha_{3}\cdot t = 0

The following system of linear equations is obtained:

\alpha_{1} + 3\cdot \alpha_{3} = 0

\alpha_{2} = 0

\alpha_{3} = 0

Whose solution is \alpha_{1} = \alpha_{2} = \alpha_{3} = 0, which means that the set of vectors is linearly independent.

p_{1}(t) = t, p_{2}(t) = t^{2} and p_{3}(t) = 2\cdot t + 3\cdot t^{2}

\alpha_{1}\cdot p_{1}(t) + \alpha_{2}\cdot p_{2}(t) + \alpha_{3}\cdot p_{3}(t) = 0

\alpha_{1}\cdot t + \alpha_{2}\cdot t^{2} + \alpha_{3}\cdot (2\cdot t + 3\cdot t^{2})=0

(\alpha_{1}+2\cdot \alpha_{3})\cdot t + (\alpha_{2}+3\cdot \alpha_{3})\cdot t^{2} = 0

The following system of linear equations is obtained:

\alpha_{1}+2\cdot \alpha_{3} = 0

\alpha_{2}+3\cdot \alpha_{3} = 0

Since the number of variables is greater than the number of equations, let suppose that \alpha_{3} = k, where k\in\mathbb{R}. Then, the following relationships are consequently found:

\alpha_{1} = -2\cdot \alpha_{3}

\alpha_{1} = -2\cdot k

\alpha_{2}= -2\cdot \alpha_{3}

\alpha_{2} = -3\cdot k

It is evident that \alpha_{1} and \alpha_{2} are multiples of \alpha_{3}, which means that the set of vector are linearly dependent.

p_{1}(t) = 1, p_{2}(t)=t^{2} and p_{3}(t) = 3+3\cdot t +t^{2}

\alpha_{1}\cdot p_{1}(t) + \alpha_{2}\cdot p_{2}(t) + \alpha_{3}\cdot p_{3}(t) = 0

\alpha_{1}\cdot 1 + \alpha_{2}\cdot t^{2}+ \alpha_{3}\cdot (3+3\cdot t+t^{2}) = 0

(\alpha_{1}+3\cdot \alpha_{3})\cdot 1+(\alpha_{2}+\alpha_{3})\cdot t^{2}+3\cdot \alpha_{3}\cdot t = 0

The following system of linear equations is obtained:

\alpha_{1}+3\cdot \alpha_{3} = 0

\alpha_{2} + \alpha_{3} = 0

3\cdot \alpha_{3} = 0

Whose solution is \alpha_{1} = \alpha_{2} = \alpha_{3} = 0, which means that the set of vectors is linearly independent.

The set of vectors A and C are linearly independent.

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3 years ago
What is the simplified quotient?
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Answer:

Give me one sec i got the anser tho

Step-by-step explanation:

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3 years ago
Which expression is equivalent to RootIndex 3 StartRoot 32 x Superscript 8 Baseline y Superscript 10 Baseline EndRoot?
evablogger [386]

Answer:

The equivalent expression to the given expression is \sqrt[3]{32x^8y^{10}}=2x^2y^3\sqrt[3]{4x^2y}

Step-by-step explanation:

The given expression is \sqrt[3]{32x^8y^{10}}

To find the equivalent expression:

\sqrt[3]{32x^8y^{10}}=(32x^8y^{10})^{\frac{1}{3}}

We may write the above expression as below:

=(32^{\frac{1}{3}})((x^8)^{\frac{1}{3}})((y^{10})^{\frac{1}{3}})

=(2)((4)^{\frac{1}{3}})(x^6)\times (x^2)(x^{\frac{2}{3}})(y^3)(y^{\frac{1}{3}}) (using square root properties)

=(2\sqrt[3]{4})(x^2\sqrt[3]{x^2})(y^3\sqrt[3]{y}) (combining the like terms and doing multiplication )

=2x^2y^3\sqrt[3]{4x^2y}

Therefore \sqrt[3]{32x^8y^{10}}=2x^2y^3\sqrt[3]{4x^2y}

Therefore the equivalent expression to the given expression is \sqrt[3]{32x^8y^{10}}=2x^2y^3\sqrt[3]{4x^2y}

5 0
3 years ago
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