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2 years ago

Which statements about the line that passes through (−2, 0) and (2, −4) are true? Select all that apply.

Mathematics

1 answer:

Gennadij [26K]2 years ago

8 0

The statement about the line that passes through (-2, 0) and (2, -4) that are true are:

Option B:The line intersects the y-axis at (0, −2).

Option C: The equation of the line is y = −x − 2.

Option D: The line intersects the x-axis at (−2, 0).

Solution:

Given, two points are (-2, 0) and (2, -4)

We have to select the options that states true about line that passes through given two points.

Now, let us find the line equation that passes through given two points using point slope form.

$\mathrm{y}-\mathrm{y}_{1}=\mathrm{m}\left(\mathrm{x}-\mathrm{x}_{1}\right) \text { where } \mathrm{m} \text { is slope and }\left(\mathrm{x}_{1}, \mathrm{y}_{1}\right) \text { is a point on the line. }$

Now slope "m" is given as:

$\mathrm{m}=\frac{y_{2}-y_{1}}{x_{2}-x_{1}}=\frac{0-(-4)}{-2-2}=\frac{4}{-4}=-1$

Then, line equation = y – 0 = -1(x – (-2))

y = -1(x + 2)

y = -x – 2 --- equation 1

Now let us check options.

Option a) slope of line is 1

We know that slope of our line is -1 so this option is wrong.

Option b) line intersects y – axis at (0, -2)

When line meets y – axis x becomes 0 ⇒ 0 + y + 2 = 0 ⇒ y = -2 so point is (0, -2).

This option is right.

Option c) equation of line is y = -x – 2

The given equation is x + y + 2 = 0 ⇒ y = -x – 2 . so this option is correct.

Option d) line intersects at (-2, 0)

When line meets x – axis y becomes 0 ⇒ x + 0 + 2 = 0 ⇒ x = -2. So point is (-2, 0).

This option is right.

Hence, options b, c, d are correct.

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T= 300 (d−15) 2 +20space, T, equals, start fraction, left parenthesis, d, minus, 15, right parenthesis, start superscript, 2,

stealth61 [152]

As can be read from your statement written "T, equals, start fraction, left parenthesis, d, minus, 15, right parenthesis, strt superscript, 2, end superscript, divided by, 300, end fraction, plus, 20", I hope your model equation is this :

$T = \frac{(d-15)^{2}}{300} + 20$

Hope this is your question, if not I think you will, still be able to find an answer of your question based on this solution

As we have to find lowest average temperature, So for minimum of a function its derivative is equal to 0 there.

So lets find derivative of T function first

So first expand $(d-15)^{2}$ as (d-15)(d-15)

we will use FOIL to multiply these

so $(d-15)(d-15) = d^{2} -15d -15d +225$

$= d^{2} -30d +225$

so we have $T = \frac{d^{2} -30d +225}{300} +20$

Now we will derivate each term here,300 in denominator is constant so that will come as it in in denominator.

To derivate terms in $dx^{2} -30d +225$ we will use power rule formula:

$(x^{n} )'= nx^{n-1}$

so derivative of $(d^{2} )'= 2d^{2-1} = 2d^{1} = 2d$

Then derivative of d will be 1

so that of -30d will be -30

then derivate of constant -225 will be 0

so we will have derivative as $\frac{2d-30}{300}$ for the fraction part and then derivative of +20 is again 0 as its constant term

$T' = \frac{2d-30}{300}$

For minimum we will put this derivative =0

$0 = \frac{2d-30}{300}$

Now solve for d

times both sides by 300

$0 \times 300 = \frac{2d-30}{300} \times 300$

$0 = 2d-30$

$0 +30 = 2d -30 +30$

$30 = 2d$

$\frac{30}{2} = \frac{2d}{2}$

$15 = d$

So now we have to find value of lowest temperature.

For that simply plug 15 in d place in original T function equation

$T = \frac{(d-15)^{2}}{300} + 20$

$T = \frac{(15-15)^{2}}{300} + 20$

$T = 20$

So T = 20 °C is the lowest average temperature and the answer.

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Margo deposited $100 into a savings account earning 4.5% simple annual interest. At the end of each year, she adds $100 to her a

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Answer:

principal interest total money in account

at the end of year

Year one $100 $4.5 $104.5

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Step-by-step explanation:

Simple interest for any principal is given by

I = p* r* t/100

I = interest rate accrued on principle amount

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r is the rate of interest

t is the time period of saving

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For year one

p = $100

r = 4.5%

t=1

I = 100*4.5*1/100 = 4.5

_______________________________________________

For year two $100 more is added to already existing $100 in account.

p = 100 +100 = $200

r = 4.5%

t=1

I = 200*4.5*1/100 = 9

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For year two $100 more is added to already existing $200 in account after two years.

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I = 300*4.5*1/100 = 13.5

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There fore total money in Margo account is

$300 saving deposited by her

$4.5 + $9 + $13.5 = $27 (interest accrued in three time)

Formulating the results in tabular form

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