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1 year ago

2 feet to 2 yards ratio

Mathematics

1 answer:

vredina [299]1 year ago

8 0

Answer:

$\frac{1}{3}$

Step-by-step explanation:

Three feet make a yard, so 2 yards is 6 feet

$\frac{2}{6}$ If we divide the top (numerator) and the bottom (denominator) by 2 we get

$\frac{2}{6}$ ÷ $\frac{2}{2}$ = $\frac{1}{3}$

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Bruce owns a business that produces widgets. He must bring in more in revenue than he pays out in costs in order to turn a profi

Aliun [14]

I believe the answer is 267

If each is worth $25 and it costs $10 then he is making $15 profit.

4000/15 is 266.6666667

So you would need 267

8 0

3 years ago

Plz plz plz plz plz plzzzzzzz

laiz [17]

Answer:

3 7/24

Step-by-step explanation

but is the last fraction 2 1/8?

4 0

3 years ago

Read 2 more answers

25 POINTS

Semenov [28]

If the slope of AB = CD and BC = AD it's a parallelogram:

Slope of AB = 6+1 / -9+5 = -7/4

CD = -2-5 / 3+1 = -74

These are equal.

BC = 5-6 / -1 +9 = -1/8

AD = -2 +1 / 3+5 = -1/8

These are also equal so it is a parallelogram.

Now to find if the diagonals are perpendicular find the slope of the perpensicular points:

AC = 5 +1 / -1 +5 = 6/4 = 3/2

BD = 6+2 / -9 -3 = 8/-12 = -2/3

Because BD is the reciprocal of AC, this means they are perpendicular.

And because AB is not perpendicular to AD ( AB and AD are not reciprocals) it is a rhombus.

5 0

3 years ago

ASAP - WILL MARK BRIANIST Please explain

marissa [1.9K]

Answer:

D 26 batches

Step-by-step explanation:

for each cup of cranberries, 4 batches of scones can be made

6 1/2 * 4 = 24 + 2 = 26

5 0

3 years ago

Read 2 more answers

Each of these extreme value problems has a solution with both a maximum value and a minimum value. Use Lagrange multipliers to f

tino4ka555 [31]

Answer:

The minimum value of the given function is f(0) = 0

Step-by-step explanation:

Explanation:-

Extreme value :- f(a, b) is said to be an extreme value of given function 'f' , if it is a maximum or minimum value.

i) the necessary and sufficient condition for f(x) to have a maximum or minimum at given point.

ii) find first derivative $f^{l} (x)$ and equating zero

iii) solve and find 'x' values

iv) Find second derivative $f^{ll}(x) >0$ then find the minimum value at x=a

v) Find second derivative $f^{ll}(x)$ then find the maximum value at x=a

Problem:-

Given function is f(x) = log ( x^2 +1)

step1:- find first derivative $f^{l} (x)$ and equating zero

$f^{l}(x) = \frac{1}{x^2+1} \frac{d}{dx}(x^2+1)$

$f^{l}(x) = \frac{1}{x^2+1} (2x)$ ……………(1)

$f^{l}(x) = \frac{1}{x^2+1} (2x)=0$

the point is x=0

step2:-

Again differentiating with respective to 'x', we get

$f^{ll}(x)=\frac{x^2+1(2)-2x(2x)}{(x^2+1)^2}$

on simplification , we get

$f^{ll}(x) = \frac{-2x^2+2}{(x^2+1)^2}$

put x= 0 we get $f^{ll}(0) = \frac{2}{(1)^2}$ > 0

$f^{ll}(x) >0$ then find the minimum value at x=0

Final answer:-

The minimum value of the given function is f(0) = 0

5 0

3 years ago