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3 years ago

Foina wrote an linear equation y=2/5-5

1 answer:

8 0

Fiona wrote the linear equation y=2/5x-5. When Henry wrote his equation, they discovered that his equation had all the same solutions as Fiona's. Which equation could be Henry's?
A.) x-5/4y=25/4
B.) x-5/2y=25/4
C.) x-5/4y=25/2
D.) x-5/2y=25/2
y=2/5x-5=>5/2y=x-25/2 (times 5/2) => x-/2y=25/2
Answer: x-5/2y=25/2 or D.
I hope this helps!! :)

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(( NEED HELP ))

sergiy2304 [10]

Answer:

C) x/2 +2y = 4

Step-by-step explanation:

Of the equations offered, only equation C has the given point as a possible solution.

___

Since the graph is not given, we don't know if the given point is on the graphed equation.

3 0

3 years ago

What is 26% as a fraction in simplest form?

Eddi Din [679]

26% is equal to

26/100

divide the numerator and denominator by 2

13/50= is your answer!

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4 0

3 years ago

Read 2 more answers

If you calculate sle to be $25,000 and that there will be one occurrence every four years (aro), then what is the ale?

Cloud [144]

If you calculate SLE to be $25,000 and that there will be one occurrence every four years (ARO), then the ALE is $40,000.

<h3>What is Single-loss expectancy (SLE)?</h3>

A expected monetary decline each moment an asset is at risk is referred to as single-loss expectancy (SLE). It is a term that is most frequently used during risk analysis and attempts to assign a monetary value to each individual threat.

Quantitative risk analysis predicts the likelihood of certain risk outcomes as well as their approximate monetary cost using relevant, verifiable data.

IT professionals must consider a wide range of risks, including the following:

Errors caused by humans
Cyber attacks, unauthorised data disclosure, or data misuse are examples of hostile action.
Errors in application
System or network failures
Physical harm caused by fire, natural disasters, or vandalism.

To know more about the Single-loss expectancy (SLE), here

brainly.com/question/14587600

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5 0

1 year ago

Write the equation of a line with an undefined slope and point (1, 7).

Sonja [21]

Answer:

The equation of straight line is x = 1

Step-by-step explanation:

<u>Step </u>:-

The equation of straight line having slope 'm' and passing through the point

$(x_{1} ,y_{1} )$ is $y-y_{1} = m (x-x_{1} )$

Slope of horizontal line(that is x-axis) is m = 0

slope of vertical line (that is y-axis) is m is not defined

The equation of straight line having slope is not defined and passing through the point (1,7).

$y-7 = \frac{1}{0} (x-1)$

cross multiplication we will get equation

$(y-7)(0)=1(x-1)$

x-1 = 0

The equation of straight line is x = 1

4 0

3 years ago

Help. <br>Please its urgent show workings.<br>

laiz [17]

Answer:

see explanation

Step-by-step explanation:

There are 2 possible approaches to differentiating these.

Expand the factors and differentiate term by term, or

Use the product rule for differentiation.

I feel they are looking for use of product rule.

Given

y = f(x). g(x) , then

$\frac{dy}{dx}$ = f(x).g'(x) + g(x).f'(x) ← product rule

(a)

y = (2x - 1)(x + 4)²

f(x) = 2x - 1 ⇒ f'(x) = 2

g(x) = (x + 4)²

g'(x) = 2(x + 4) × $\frac{d}{dx}$ (x + 4) ← chain rule

= 2(x + 4) × 1

= 2(x + 4)

Then

$\frac{dy}{dx}$ = (2x - 1). 2(x + 4) + (x + 4)². 2

= 2(2x - 1)(x + 4) + 2(x + 4)² ← factor out 2(x + 4) from each term

= 2(x + 4) (2x - 1 + x + 4)

= 2(x + 4)(3x + 3) ← factor out 3

= 6(x + 4)(x + 1)

--------------------------------------------------------------------------

(b)

y = x(x² - 1)³

f(x) = x ⇒ f'(x) = 1

g(x) = (x² - 1)³

g'(x) = 3(x² - 1)² × $\frac{d}{dx}$ (x² - 1) ← chain rule

= 3(x² - 1)² × 2x

= 6x(x² - 1)²

Then

$\frac{dy}{dx}$ = x. 6x(x² - 1)² + (x² - 1)³. 1

= 6x²(x² - 1)² + (x² - 1)³ ← factor out (x² - 1)²

= (x² - 1)² (6x² + x² - 1)

= (x² - 1)²(7x² - 1)

----------------------------------------------------------------------

(c)

y = (x² - 1)(x³ + 1)

f(x) = x² - 1 ⇒ f'(x) = 2x

g(x) = (x³ + 1) ⇒ g'(x) = 3x²

Then

$\frac{dy}{dx}$ = (x² - 1). 3x² + (x³ + 1), 2x

= 3x²(x² - 1) + 2x(x³ + 1) ← factor out x

= x[3x(x² - 1) + 2(x³ + 1) ]

= x(3x³ - 3x + 2x³ + 2)

= x(5x³ - 3x + 2) ← distribute

= 5 $x^{4}$ - 3x² + 2x

--------------------------------------------------------------------

(d)

y = 3x³(x² + 4)²

f(x) = 3x³ ⇒ f'(x) = 9x²

g(x) = (x² + 4)²

g'(x) = 2(x² + 4) × $\frac{d}{dx}$ (x² + 4) ← chain rule

= 2(x² + 4) × 2x

= 4x(x² + 4)

Then

$\frac{dy}{dx}$ = 3x³. 4x(x² + 4) + (x² + 4)². 9x²

= 12 $x^{4}$ (x² + 4) + 9x²(x² + 4)² ← factor out 3x²(x² + 4)

= 3x²(x² + 4) [ 4x² + 3(x² + 4) ]

= 3x²(x² + 4)(4x² + 3x² + 12)

= 3x²(x² + 4)(7x² + 12)

5 0

3 years ago