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2 years ago

12

What is another way to say 14 over 16 in fractions ( preferably in lowest number example 4 over 8 equals 1 over 2 both are halve

s)

1 answer:

Degger [83]2 years ago

4 0

Answer:

7/8

Step-by-step explanation:

The greatest common factor in each is two so you divide both the numerator and the denominator by two to simplify the fraction

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Mike makes a commission of 10% on each TV set sold at store. Each TV costs $120. How much

GrogVix [38]

Answer:

$300

Step-by-step explanation:

10% = .1 and each TV costs 120 so .1*120=12 then you take 12*25 since 12 is the money he makes for seeling the tv multiplied by the total TV's sold and you get 300 so Mike makes $300

5 0

3 years ago

Can anyone help with the last problem about The Difference Quotient of a Function

Marysya12 [62]

Answer:

tbh i really dont know

Step-by-step explanation:

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2 years ago

What is the area of a right triangle with a base of 25 centimeters and a height of 13 centimeters?

sdas [7]

Answer:

<h3>162.5 cm²</h3><h3>Step-by-step explanation:</h3>

Area of a triangle = (BH)/2

Base= 25

Height=13

(25×13)/2= 162.5 cm²

6 0

3 years ago

I need help ASAP!!!!

Ad libitum [116K]

Step-by-step explanation:

Kindly find 4 attached files

The proportionality is determined by the lines that only passes through the origin

And all the lines drawn are passing through the origin

So by either one of the 2 ways

By using rule of direct variation y=kx that has no y intercept

Or by finding the equation of straight line and find the slope with neglecting the y intercept since it's 0

5 0

2 years ago

Suppose a, b denotes of the quadratic polynomial x² + 20x - 2022 & c, d are roots of x² - 20x + 2022 then the value of ac(a

Alja [10]

<h3><u>Correct Question :- </u></h3>

$\sf\:a,b \: are \: the \: roots \: of \: {x}^{2} + 20x - 2020 = 0 \: and \: \\ \sf \: c,d \: are \: the \: roots \: of \: {x}^{2} - 20x + 2020 = 0 \: then \:$

$\sf \: ac(a - c) + ad(a - d) + bc(b - c) + bd(b - d) =$

(a) 0

(b) 8000

(c) 8080

(d) 16000

$\large\underline{\sf{Solution-}}$

Given that

$\red{\rm :\longmapsto\:a,b \: are \: the \: roots \: of \: {x}^{2} + 20x - 2020 = 0}$

We know

$\boxed{\red{\sf Product\ of\ the\ zeroes=\frac{Constant}{coefficient\ of\ x^{2}}}}$

$\rm \implies\:ab = \dfrac{ - 2020}{1} = - 2020$

And

$\boxed{\red{\sf Sum\ of\ the\ zeroes=\frac{-coefficient\ of\ x}{coefficient\ of\ x^{2}}}}$

$\rm \implies\:a + b = - \dfrac{20}{1} = - 20$

Also, given that

$\red{\rm :\longmapsto\:c,d \: are \: the \: roots \: of \: {x}^{2} - 20x + 2020 = 0}$

$\rm \implies\:c + d = - \dfrac{( - 20)}{1} = 20$

and

$\rm \implies\:cd = \dfrac{2020}{1} = 2020$

Now, Consider

$\sf \: ac(a - c) + ad(a - d) + bc(b - c) + bd(b - d)$

$\sf \: = {ca}^{2} - {ac}^{2} + {da}^{2} - {ad}^{2} + {cb}^{2} - {bc}^{2} + {db}^{2} - {bd}^{2}$

$\sf \: = {a}^{2}(c + d) + {b}^{2}(c + d) - {c}^{2}(a + b) - {d}^{2}(a + b)$

$\sf \: = (c + d)( {a}^{2} + {b}^{2}) - (a + b)( {c}^{2} + {d}^{2})$

$\sf \: = 20( {a}^{2} + {b}^{2}) + 20( {c}^{2} + {d}^{2})$

$\sf \: = 20\bigg[ {a}^{2} + {b}^{2} + {c}^{2} + {d}^{2}\bigg]$

We know,

$\boxed{\tt{ { \alpha }^{2} + { \beta }^{2} = {( \alpha + \beta) }^{2} - 2 \alpha \beta \: }}$

So, using this, we get

$\sf \: = 20\bigg[ {(a + b)}^{2} - 2ab + {(c + d)}^{2} - 2cd\bigg]$

$\sf \: = 20\bigg[ {( - 20)}^{2} + 2(2020) + {(20)}^{2} - 2(2020)\bigg]$

$\sf \: = 20\bigg[ 400 + 400\bigg]$

$\sf \: = 20\bigg[ 800\bigg]$

$\sf \: = 16000$

Hence,

$\boxed{\tt{ \sf \: ac(a - c) + ad(a - d) + bc(b - c) + bd(b - d) = 16000}}$

<em>So, option (d) is correct.</em>

4 0

2 years ago