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Mama L [17]
4 years ago
5

What is the rule/output?

Mathematics
2 answers:
tankabanditka [31]4 years ago
7 0

Answer:

The rule is y = 4x - 5.

Step-by-step explanation:

Notice that if we start with x = 1 and increase x by 1, we get 2.  Simultaneously, y starts with -1 and becomes 3.  Thus, the slope is m = rise / run = 4/1, or 4.

The rule is y = 4x - 5.

Check:  Suppose we pick input 4 from the table. Does this rule produce output 11?  Is 11 = 4(4) - 5 true?  YES.

labwork [276]4 years ago
6 0

Answer:

y = 4x - 5

Step-by-step explanation:

Have you been taught to set up 2 equations and 2 unknowns?

That is actually the only way I could do this.

y = mx + b

x = 2

y = 3

3 = 2m + b

x = 1

y = -1

-1 = m + b        Multiply by 2. That means that the m term will cancel.

================

-2 =2m +2b    

<u>3 = 2m + b </u>        Subtract

-5 = b

==================

3 = 2m + b       Substitute - 5 for b

3 = 2m - 5        Add 5 to both sides.

3+5= 2m-5+5   Combine

8 = 2m               Divide by 2

8/2=2m/2

m = 4

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Condense the following logs into a single log:
mamaluj [8]

QUESTION 1

The given logarithm is

8\log_g(x)+5\log_g(y)

We apply the power rule of logarithms; n\log_a(m)=\log_(m^n)

=\log_g(x^8)+\log_g(y^5)

We now apply the product rule of logarithm;

\log_a(m)+\log_a(n)=\log_a(mn)

=\log_g(x^8y^5)

QUESTION 2

The given logarithm is

8\log_5(x)+\frac{3}{4}\log_5(y)-5\log_5(z)

We apply the power rule of logarithm to get;

=\log_5(x^8)+\log_5(y^{\frac{3}{4}})-\log_5(z^5)

We apply the product to obtain;

=\log_5(x^8\times y^{\frac{3}{4}})-\log_5(z^5)

We apply the quotient rule; \log_a(m)-\log_a(n)=\log_a(\frac{m}{n} )

=\log_5(\frac{x^8\times y^{\frac{3}{4}}}{z^5})

=\log_5(\frac{x^8 \sqrt[4]{y^3} }{z^5})

7 0
3 years ago
5) The Barnett triplets went to the school carnival. Henry rode on 5 rides, bought 2 drinks, 1 bag of popcorn and
dusya [7]

Answer:

D

Step-by-step explanation:

5 0
3 years ago
Prove that if the set of vectors {u1, u2, u3} is linearly dependent and u4 is any vector, then the set {u1, u2, u3, u4} is linea
Lesechka [4]

Answer with Step-by-step explanation:

We are given that the set of vectors{u_1,u_2,u_3} is lineraly dependent set .

We have to prove that the set {u_1,u_2,u_3,u_4} is linearly dependent .

Linearly dependent vectors : If the vectors u_1,u_2.u_3,u_4

are linearly  dependent therefore the linear combination

a_1u_1+a_2u_2+a_3u_3+a_4u_4=0

Then ,there exit a scalar which is not equal to zero .

Let a_1\neq0 then the vector u_1 will be zero and remaining  other vectors are not zero.

Proof:

When u_1,u_2,u_3 are linearly dependent vectors therefore, linear combination of vectors of given set

a_1u_1+a_2u_2+a_3u_3=0

By definition of linearly dependent vector

There exist a scalar which is not equal to zero.

Suppose a_1\neq 0 then  u_1=0

The linear combination of the set {u_1,u_2,u_3,u_4}

a_1u_1+a_2u_2+a_3u_3+a_4u_4=0

When a_1\neq0\; and\; u_1=0

Therefore,the set {u_1,u_2,u_3,u_4} is linearly dependent because it contain a vector which is zero.

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Answer:

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Step-by-step explanation:

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The length of the arc is calculated as:

2 × π × 3 × 90/360

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Approximately to the nearest hundredth = 4.71 units

8 0
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Lyrx [107]

Answer:

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Step-by-step explanation:

4x = 7.20

x = 1.80

7 0
3 years ago
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