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3 years ago

Equation for (6,17) (14,17)

Mathematics

1 answer:

bixtya [17]3 years ago

8 0

Answer:

y=17

Step-by-step explanation:

y=mx+b

m=(17-17)/(14-6) = 0

y=b

y=17

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May someone please help me on which to graph?

olga_2 [115]

Answer:

Option (4)

Step-by-step explanation:

Proportional relationship means,

y ∝ x

y = kx

$k=\frac{y}{x}$

Here, k = proportionality constant

Therefore, if the graph of a line passes through the origin (0, 0) table will represent the proportional relationship.

From table 1,

For a point (1, 2)

$k=\frac{2}{1}=2$

For another point (3, 2)

$k=\frac{3}{2}=1.5$

In both the cases 'k' is not same of constant.

Therefore, table (1) is not proportional.

For table (2),

Line passes through (2, 0).

That means there is a x-intercept → (2, 0)

Therefore, table doesn't represent a proportional relationship.

For table (3),

Line passes through a point (0, 1)

It means given line has a y-intercept → y = 1

Therefore, table doesn't represent a proportional relationship.

For table (4),

Line of this table passes through two points (1, 3) and (2, 6)

$k=\frac{3}{1}=3$

$k=\frac{6}{2}=3$

Therefore, proportionality constant for the given table is 3.

Now we can graph table (4).

6 0

3 years ago

Vera needs repairs on her car. The mechanic’s estimate says that he will provide 7.3 hours of labor. His rate is $55 per hour. T

Ksivusya [100]

Answer:

total cost = given hours × rate per hour

3 0

2 years ago

What times 3 equals something that is divisible by 2 that equals 8?

Delicious77 [7]

Let the unknown be n, then: n × 3 = 8 × 2 n × 3 = 16 n = 16 ÷ 3 n = 5.333

8 0

3 years ago

For this exercise assume that all matrices are ntimesn. Each part of this exercise is an implication of the form "If "statement

inna [77]

Answer:

C. True; by the Invertible Matrix Theorem if the equation Ax=0 has only the trivial solution, then the matrix is invertible. Thus, A must also be row equivalent to the n x n identity matrix.

Step-by-step explanation:

The Invertible matrix Theorem is a Theorem which gives a list of equivalent conditions for an n X n matrix to have an inverse. For the sake of this question, we would look at only the conditions needed to answer the question.

There is an n×n matrix C such that CA= $I_n$ .
There is an n×n matrix D such that AD= $I_n$ .
The equation Ax=0 has only the trivial solution x=0.
A is row-equivalent to the n×n identity matrix $I_n$ .
For each column vector b in $R^n$ , the equation Ax=b has a unique solution.
The columns of A span $R^n$ .

Therefore the statement:

If there is an n X n matrix D such that AD=I, then there is also an n X n matrix C such that CA = I is true by the conditions for invertibility of matrix:

The equation Ax=0 has only the trivial solution x=0.

A is row-equivalent to the n×n identity matrix $I_n$ .

The correct option is C.

5 0

4 years ago

<img src="https://tex.z-dn.net/?f=y%20%3D%20%20%5Cfrac%7B%28190%29%5E%7B3%7D%20%7B68%5E%7B2%7D%20%7D%20%20%5Ctimes%20%28%20%5Cfr

mrs_skeptik [129]

Answer:

$\dfrac{\sqrt[3]{95^2}}{17\cdot95^4}=\dfrac{\sqrt[3]{9\,025}}{1\,384\,660\,625}$

Step-by-step explanation:

The applicable rules of exponents are ...

(ab)^c = (a^c)(b^c)

(a^b)/(a^c) = a^(b-c)

$\dfrac{190^3}{68^2}\times\dfrac{34}{95^{\frac{19}{3}}}=\dfrac{(2\cdot 95)^3}{(2\cdot 34)^2}\cdot\dfrac{34}{95^6\cdot 95^{\frac{1}{3}}}=2^{3-2}95^{3-6-\frac{1}{3}}34^{1-2}\\\\=2\cdot 95^{-3\frac{1}{3}}\cdot 34^{-1}=2\cdot 95^{-4+\frac{2}{3}}\cdot 34^{-1}\\\\=\dfrac{2\sqrt[3]{95^2}}{95^4\cdot 34}=\dfrac{\sqrt[3]{95^2}}{17\cdot95^4}\\\\=\dfrac{\sqrt[3]{9\,025}}{1\,384\,660\,625}$

6 0

3 years ago