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HACTEHA [7]
2 years ago
9

Do the write a relation as a set of five ordered pairs then express the relation using a:

Mathematics
1 answer:
PSYCHO15rus [73]2 years ago
4 0

Answer:

Step-by-step explanation:

Table

<h3>x I y</h3><h3>-------</h3><h3>1  I  2</h3><h3>2  I  4</h3><h3>3  I  6</h3><h3>4  I  8</h3><h3>5  I  10</h3>
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Student Produced Response - Calculator
faltersainse [42]

Answer:

\dfrac{a}{b}=0.75.

Step-by-step explanation:

If two equations a_1x+b_1y+c_1=0 and a_2x+b_2y+c_2=0 has infinitely many solutions, then

\dfrac{a_1}{a_2}=\dfrac{b_1}{b_2}=\dfrac{c_1}{c_2}

The given equations are

ax+by=10

3x+4y=20

It is given that the above system of equations has infinitely many solutions. So,

\dfrac{a}{3}=\dfrac{b}{4}=\dfrac{10}{20}

\dfrac{a}{3}=\dfrac{b}{4}=\dfrac{1}{2}

Now,

\dfrac{a}{3}=\dfrac{1}{2} and \dfrac{b}{4}=\dfrac{1}{2}

a=\dfrac{3}{2} and b=\dfrac{4}{2}

a=1.5 and b=2

So, a=1.5 and b=2.

Now,

\dfrac{a}{b}=\dfrac{1.5}{2}=0.75

Therefore, \dfrac{a}{b}=0.75.

5 0
3 years ago
How to Learn Math
Scorpion4ik [409]
What? This isn’t really a question!
6 0
3 years ago
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Please help me with my question<br> Thank you <br> :))BRAINLIEST AVAILABLE
lara [203]
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4 0
3 years ago
A polyhedron has 6 faces and 7 vertices, how many edges does it have
iris [78.8K]

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Step-by-step explanation:

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3 years ago
Choose the correct simplification of the expression (-9a^3b^2c^10)^0
Flauer [41]

Option C: 1 is the simplification of the expression \left(-9 a^{3} b^{2} c^{10}\right)^{0}

Explanation:

The given expression is \left(-9 a^{3} b^{2} c^{10}\right)^{0}

We need to simplify the given expression.

<u>Simplification:</u>

To simplify the given expression, let us apply the exponent rule, a^{0}=1 where a \neq 0

Thus, the the exponent rule a^{0}=1 means that any value (except zero) to the power of zero is equal to one.

Thus, the expression \left(-9 a^{3} b^{2} c^{10}\right)^{0} becomes,

\left(-9 a^{3} b^{2} c^{10}\right)^{0}=1

Therefore, the simplified expression is 1

Hence, Option C is the correct answer.

8 0
3 years ago
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