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3 years ago

8

The equation of a line is y = 2x + 3. What is the equation of the line that is parallel to the first line and passes through (2,

–1)?

1 answer:

Masja [62]3 years ago

8 0

Answer:

y = 2x - 5

Step-by-step explanation:

The slope for a parallel line is the same as the slope in the given equation. Therefore the slope of the new line will be 2 m= 2

use that slope and (2, -1) to find b

-1 = 2(2) + b

-1 = 4 + b

-5 = b

Plug m and b into y = mx + b

y = 2x - 5

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Here's the question.

OleMash [197]

Answer:

The value of T₂₀ - T₁₅ is <u>-20</u>.

Step-by-step explanation:

<u>Given</u> :

>> If for an A.P, d = -4

<u>To</u><u> </u><u>Find</u> :

>> T₂₀ - T₁₅

<u>Using Formula</u> :

General term of an A.P.

$\star{\small{\underline{\boxed{\sf{\red{ T_n = a + (n - 1)d}}}}}}$

>> Tₙ = nᵗʰ term
>> a = first term
>> n = no. of terms
>> d = common difference

<u>Solution</u> :

Firstly finding the A.P of T₂₀ by substituting the values in the formula :

${\dashrightarrow{\pmb{\sf{ T_n = a + (n - 1)d}}}}$

${\dashrightarrow{\sf{ T_{20} = a + (20 - 1) d}}}$

${\dashrightarrow{\sf{ T_{20} = a + (19)d}}}$

${\dashrightarrow{\sf{ T_{20} = a + 19 \times d}}}$

${\dashrightarrow{\sf{ T_{20} = a + 19d}}}$

${\star \: {\underline{\boxed{\sf{\pink{ T_{20} = a + 19d}}}}}}$

Hence, the value of T₂₀ is a + 19d.

$\rule{190}1$

Secondly, finding the A.P of T₁₅ by substituting the values in the formula :

${\dashrightarrow{\pmb{\sf{ T_n = a + (n - 1)d}}}}$

${\dashrightarrow{\sf{ T_{15}= a + (15 - 1) d}}}$

${\dashrightarrow{\sf{ T_{15}= a + (14) d}}}$

${\dashrightarrow{\sf{ T_{15}= a + 14 \times d}}}$

${\dashrightarrow{\sf{ T_{15}= a + 14d}}}$

${\star{\underline{\boxed{\sf \pink{ T_{15}= a + 14d}}}}}$

Hence, the value of T₁₅ is a + 14d

$\rule{190}1$

Now, finding the difference between T₂₀ - T₁₅ :

${\dashrightarrow{\pmb{\sf{T_{20} - T_{15}}}}}$

${\dashrightarrow{\sf{(a + 19d) - (a + 14d)}}}$

${\dashrightarrow{\sf{a + 19d - a - 14d}}}$

${\dashrightarrow{\sf{a - a + 19d - 14d}}}$

${\dashrightarrow{\sf{0+ 19d - 14d}}}$

${\dashrightarrow{\sf{19d - 14d}}}$

${\dashrightarrow{\sf{5 \times - 4}}}$

${\dashrightarrow{\sf{ - 20}}}$

${\star \: \underline{\boxed{\sf{\pink{T_{20} - T_{15} = - 20}}}}}$

Hence, the value of T₂₀ - T₁₅ is -20.

$\underline{\rule{220pt}{3.5pt}}$

3 0

3 years ago

Let φ(x, y) = arctan (y/x) .

Alexandra [31]

Answer:

a) $\large F(x,y)=(-\frac{y}{x^2+y^2},\frac{x}{x^2+y^2})$

b) $\large \mathbb{R}^2-\{(0,0)\}$

c) the points of the form (x, -x) for x≠0

Step-by-step explanation:

a)

If φ(x, y) = arctan (y/x), the vector field F = ∇φ would be

$\large F(x,y)=(\frac{\partial \phi}{\partial x},\frac{\partial \phi}{\partial y})$

On one hand we have,

$\large \frac{\partial \phi}{\partial x}=\frac{\partial arctan(y/x)}{\partial x}=\frac{-y/x^2}{1+(y/x)^2}=-\frac{y/x^2}{1+y^2/x^2}=\\\\=-\frac{y/x^2}{(x^2+y^2)/x^2}=-\frac{y}{x^2+y^2}$

On the other hand,

$\large \frac{\partial \phi}{\partial y}=\frac{\partial arctan(y/x)}{\partial y}=\frac{1/x}{1+(y/x)^2}=\frac{1/x}{1+y^2/x^2}=\\\\=\frac{1/x}{(x^2+y^2)/x^2}=\frac{x}{x^2+y^2}$

So

$\large F(x,y)=(-\frac{y}{x^2+y^2},\frac{x}{x^2+y^2})$

b)

The domain of definition of F is

$\large \mathbb{R}^2-\{(0,0)\}$

i.e., all the plane X-Y except the (0,0)

c)

Here we want to find all the points such that

$\large (-\frac{y}{x^2+y^2},\frac{x}{x^2+y^2})=(k,k)$

where k is a real number other than 0.

But this means

$\large -\frac{y}{x^2+y^2}=\frac{x}{x^2+y^2}\Rightarrow y=-x$

So, all the points in the line y = -x except (0,0) are parallel to the vector field F, that is, the points (x, -x) with x≠ 0

8 0

3 years ago

(7+5)-2=(5+7)-2 What property is this??

SSSSS [86.1K]

Pretty sure the distributive property my friend

5 0

3 years ago

Read 2 more answers

Keith has 39 books in his library he brought several books at a yard sale over the weekend he now has 84 in his library how many

garri49 [273]

Subtract original amount from new amount:

84 - 39 = 45

He bought 45 books.

3 0

3 years ago

Zappos is an online retailer based in Nevada and employs 1,300 employees. One of their competitors, Amazon, would like to test t

Amanda [17]

Answer:

$t=\frac{33.9-36}{\frac{4.1}{\sqrt{22}}}=-2.402$

$df = n-1= 22-1=21$

$t_{\alpha/2}= -2.08$

Since the calculated values is lower than the critical value we have enough evidence to reject the null hypothesis at the significance level of 2.5% and we can say that the true mean is lower than 36 years old

Step-by-step explanation:

Data given

$\bar X=33.9$ represent the sample mean

$s=4.1$ represent the sample standard deviation

$n=22$ sample size

$\mu_o =26$ represent the value that we want to test

$\alpha=0.025$ represent the significance level for the hypothesis test.

t would represent the statistic (variable of interest)

$p_v$ represent the p value for the test (variable of interest)

System of hypothesis

We need to conduct a hypothesis in order to check if the true mean is less than 36 years old, the system of hypothesis would be:

Null hypothesis: $\mu \geq 36$

Alternative hypothesis: $\mu < 36$

The statistic is given by:

$t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}$ (1)

And replacing we got:

$t=\frac{33.9-36}{\frac{4.1}{\sqrt{22}}}=-2.402$

Now we can calculate the critical value but first we need to find the degreed of freedom:

$df = n-1= 22-1=21$

So we need to find a critical value in the t distribution with df =21 who accumulates 0.025 of the area in the left and we got:

$t_{\alpha/2}= -2.08$

Since the calculated values is lower than the critical value we have enough evidence to reject the null hypothesis at the significance level of 2.5% and we can say that the true mean is lower than 36 years old

3 0

3 years ago