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3 years ago

Write a description of this inequality: -5a+3>1

Mathematics

1 answer:

iren [92.7K]3 years ago

8 0

The sum of -5a and 3 is greater than 1.

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Someone please help

OLEGan [10]

Answer:

e= -2

Step-by-step explanation:

1) distribute so you get 6+0.75e=2-1.25e

2) move the variable to one side so you get 6+2e=2

3) subtract the 6: 2e=-4

4) divide by 2: e= -2

4 0

3 years ago

The sum of 4 consecutive intergers is 50, what is the first interger in this relationship

Vesnalui [34]

Answer:

x = 11

Step-by-step explanation:

Consecutive means in a row

Let x be the first integer

x+1 is the second

x+2 is the third

x+3 is the 4th

The sum is 50

x+ x+1 + x+2 + x+3 = 50

Combine like terms

4x+6 = 50

Subtract 6 from each side

4x+6-6 = 50-6

4x = 44

Divide each side by 4

4x/4 = 44/4

x = 11

7 0

3 years ago

Round the number 2,787 to the nearest hundred

miskamm [114]

Answer:

2,800

Step-by-step explanation:

5 0

3 years ago

Hey, I need help with questions 1 and 2, if anyone can help, please? thanks!

Eva8 [605]

The correct works are:

$Blue(s + h) = 2s^2 + 4sh + 2h^2 + 3$ .
$\frac{Blue(s + h) - Blue(s)}{h} = 4s + 2h$

<h3>Function Notation</h3>

The function is given as:

$Blue(s) = 2s^2 + 3$

The interpretation when Steven is asked to calculate Blue(s + h) is that:

Steven is asked to find the output of the function Blue, when the input is s + h

So, we have:

$Blue(s + h) = 2(s + h)^2 + 3$

Evaluate the exponent

$Blue(s + h) = 2(s^2 + 2sh + h^2) + 3$

Expand the bracket

$Blue(s + h) = 2s^2 + 4sh + 2h^2 + 3$

So, the correct work is:

$Blue(s + h) = 2s^2 + 4sh + 2h^2 + 3$

<h3>Simplifying Difference Quotient</h3>

In (a), we have:

$Blue(s + h) = 2s^2 + 4sh + 2h^2 + 3$

$Blue(s) = 2s^2 + 3$

The difference quotient is represented as:

$\frac{f(x + h) - f(x)}{h}$

So, we have:

$\frac{Blue(s + h) - Blue(s)}{h} = \frac{2s^2 + 4sh + 2h^2 + 3 - 2s^2 - 3}{h}$

Evaluate the like terms

$\frac{Blue(s + h) - Blue(s)}{h} = \frac{4sh + 2h^2}{h}$

Evaluate the quotient

$\frac{Blue(s + h) - Blue(s)}{h} = 4s + 2h$

Hence, the correct work is:

$\frac{Blue(s + h) - Blue(s)}{h} = 4s + 2h$

Read more about function notations at:

brainly.com/question/13136492

4 0

2 years ago

A ladder 25 feet long is leaning against the wall of a house. The base of the ladder is pulled away from the wall at a rate of 2

Alika [10]

As the ladder is pulled away from the wall, the area and the height with the

wall are decreasing while the angle formed with the wall increases.

The correct response are;

(a) The velocity of the top of the ladder = <u>1.5 m/s downwards</u>

<u />

(b) The rate the area formed by the ladder is changing is approximately <u>-75.29 ft.²/sec</u>

<u />

Reasons:

The given parameter are;

Length of the ladder, <em>l</em> = 25 feet

Rate at which the base of the ladder is pulled, $\displaystyle \frac{dx}{dt}$ = 2 feet per second

(a) Let <em>y</em> represent the height of the ladder on the wall, by chain rule of differentiation, we have;

$\displaystyle \frac{dy}{dt} = \mathbf{\frac{dy}{dx} \times \frac{dx}{dt}}$

25² = x² + y²

y = √(25² - x²)

$\displaystyle \frac{dy}{dx} = \frac{d}{dx} \sqrt{25^2 - x^2} = \frac{x \cdot \sqrt{625-x^2} }{x^2- 625}$

Which gives;

$\displaystyle \frac{dy}{dt} = \frac{x \cdot \sqrt{625-x^2} }{x^2- 625}\times \frac{dx}{dt} = \frac{x \cdot \sqrt{625-x^2} }{x^2- 625}\times2$

$\displaystyle \frac{dy}{dt} = \mathbf{ \frac{x \cdot \sqrt{625-x^2} }{x^2- 625}\times2}$

When x = 15, we get;

$\displaystyle \frac{dy}{dt} = \frac{15 \times \sqrt{625-15^2} }{15^2- 625}\times2 = \mathbf{-1.5}$

The velocity of the top of the ladder = <u>1.5 m/s downwards</u>

When x = 20, we get;

$\displaystyle \frac{dy}{dt} = \frac{20 \times \sqrt{625-20^2} }{20^2- 625}\times2 = -\frac{8}{3} = -2.\overline 6$

The velocity of the top of the ladder = $\underline{-2.\overline{6} \ m/s \ downwards}$

When x = 24, we get;

$\displaystyle \frac{dy}{dt} = \frac{24 \times \sqrt{625-24^2} }{24^2- 625}\times2 = \mathbf{-\frac{48}{7}} \approx -6.86$

The velocity of the top of the ladder ≈ <u>-6.86 m/s downwards</u>

(b) $\displaystyle The \ area\ of \ the \ triangle, \ A =\mathbf{\frac{1}{2} \cdot x \cdot y}$

Therefore;

$\displaystyle The \ area\ A =\frac{1}{2} \cdot x \cdot \sqrt{25^2 - x^2}$

$\displaystyle \frac{dA}{dx} = \frac{d}{dx} \left (\frac{1}{2} \cdot x \cdot \sqrt{25^2 - x^2}\right) = \mathbf{\frac{(2 \cdot x^2- 625)\cdot \sqrt{625-x^2} }{2\cdot x^2 - 1250}}$

$\displaystyle \frac{dA}{dt} = \mathbf{ \frac{dA}{dx} \times \frac{dx}{dt}}$

Therefore;

$\displaystyle \frac{dA}{dt} = \frac{(2 \cdot x^2- 625)\cdot \sqrt{625-x^2} }{2\cdot x^2 - 1250} \times 2$

When the ladder is 24 feet from the wall, we have;

x = 24

$\displaystyle \frac{dA}{dt} = \frac{(2 \times 24^2- 625)\cdot \sqrt{625-24^2} }{2\times 24^2 - 1250} \times 2 \approx \mathbf{ -75.29}$

The rate the area formed by the ladder is changing, $\displaystyle \frac{dA}{dt}$ ≈ <u>-75.29 ft.²/sec</u>

$\displaystyle sin(\theta) = \frac{x}{25}$

$\displaystyle \theta = \mathbf{arcsin \left(\frac{x}{25} \right)}$

$\displaystyle \frac{d \theta}{dt} = \frac{d \theta}{dx} \times \frac{dx}{dt}$

$\displaystyle\frac{d \theta}{dx} = \frac{d}{dx} \left(arcsin \left(\frac{x}{25} \right) \right) = \mathbf{ -\frac{\sqrt{625-x^2} }{x^2 - 625}}$

Which gives;

$\displaystyle \frac{d \theta}{dt} = -\frac{\sqrt{625-x^2} }{x^2 - 625}\times \frac{dx}{dt}= \mathbf{ -\frac{\sqrt{625-x^2} }{x^2 - 625} \times 2}$

When x = 24 feet, we have;

$\displaystyle \frac{d \theta}{dt} = -\frac{\sqrt{625-24^2} }{24^2 - 625} \times 2 \approx \mathbf{ 0.286}$

Rate at which the angle between the ladder and the wall of the house is changing when the base of the ladder is 24 feet from the wall is $\displaystyle \frac{d \theta}{dt}$ ≈ <u>0.286 rad/sec</u>

Learn more about the chain rule of differentiation here:

brainly.com/question/20433457

3 0

3 years ago