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2 years ago

Find the value of 24/25 ÷ 4/5. Explain your reasoning

Mathematics

2 answers:

Komok [63]2 years ago

8 0

Answer:

0.048

Step-by-step explanation:

Divided both numbers

Svetach [21]2 years ago

6 0

Answer:

6/5

Step-by-step explanation:

Reduce the expression, if possible, by cancelling the common factors.

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What is this ? 8(m+3)-20+m

KonstantinChe [14]

8m+24-20+m= 9m+4. "9m+4" is the answer

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3 years ago

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A figure has a vertex at (5,2). If the figure has line symmetry about the x axis, what are the coordinates of another vertex of

Fiesta28 [93]

X axis is the ones that goes left and right

(x,y)

symetric about the x axis means that the vertical distance from the x axis is the same for 2 points that are symetric about the x axis

y is height

(x,y)
(5,2)
2 is height
distance from x axis
therefor the other point is -2
(5,-2)

the other vertex is at (5,-2)

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3 years ago

Please help i need to write out my work to so please tell me how to show it

Anarel [89]

For the first one write 9 divided by 2 gives your your radius an it’s 3.5. the formula is 2 x 3.14 x radius and that gives you 21.98.

the 2nd one is 2 x 3.14 x 9 = 56.52

the 3rd is 2 x 3.14 x 5= 31.4

formula is always 2 x 3.14 (or pie) x radius

to get your radius from your diameter you do your diameter divided by 2

3 0

1 year ago

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Derivative of tan(2x+3) using first principle

kodGreya [7K]

$f(x)=\tan(2x+3)$

The derivative is given by the limit

$f'(x)=\displaystyle\lim_{h\to0}\frac{f(x+h)-f(x)}h$

You have

$\displaystyle\lim_{h\to0}\frac{\tan(2(x+h)+3)-\tan(2x+3)}h$
$\displaystyle\lim_{h\to0}\frac{\tan((2x+3)+2h)-\tan(2x+3)}h$

Use the angle sum identity for tangent. I don't remember it off the top of my head, but I do remember the ones for (co)sine.

$\tan(a+b)=\dfrac{\sin(a+b)}{\cos(a+b)}=\dfrac{\sin a\cos b+\cos a\sin b}{\cos a\cos b-\sin a\sin b}=\dfrac{\tan a+\tan b}{1-\tan a\tan b}$

By this identity, you have

$\tan((2x+3)+2h)=\dfrac{\tan(2x+3)+\tan2h}{1-\tan(2x+3)\tan2h}$

So in the limit you get

$\displaystyle\lim_{h\to0}\frac{\dfrac{\tan(2x+3)+\tan2h}{1-\tan(2x+3)\tan2h}-\tan(2x+3)}h$
$\displaystyle\lim_{h\to0}\frac{\tan(2x+3)+\tan2h-\tan(2x+3)(1-\tan(2x+3)\tan2h)}{h(1-\tan(2x+3)\tan2h)}$
$\displaystyle\lim_{h\to0}\frac{\tan2h+\tan^2(2x+3)\tan2h}{h(1-\tan(2x+3)\tan2h)}$
$\displaystyle\lim_{h\to0}\frac{\tan2h}h\times\lim_{h\to0}\frac{1+\tan^2(2x+3)}{1-\tan(2x+3)\tan2h}$
$\displaystyle\frac12\lim_{h\to0}\frac1{\cos2h}\times\lim_{h\to0}\frac{\sin2h}{2h}\times\lim_{h\to0}\frac{\sec^2(2x+3)}{1-\tan(2x+3)\tan2h}$

The first two limits are both 1, and the single term in the last limit approaches 0 as $h\to0$ , so you're left with

$f'(x)=\dfrac12\sec^2(2x+3)$

which agrees with the result you get from applying the chain rule.

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2 years ago

14) When f(x) = x + 3 is transformed to g(x) = f(x) -1, what will be the new

iren [92.7K]

Step-by-step explanation:

14) g(x)= x + 3 - 1

= x + 2

15) g(x)= x - 2 + 5

= x + 3

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2 years ago

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