Which is a pair of equal expressions

Martin burns 360 calories running for 30 minutes. Paul burns 165 calories for every 15 minute he runs. If Kara runs for 20 minut

Yanka [14]

Answer: She would have to burn between 220 and 240 calories per minute to have a calorie burn rate between those of Martin and Paul.

Step-by-step explanation:

Martin:

360 calories / 30 min = 12 calories/min

Paul:

165 calories / 15 min = 11 calories/min

Kara:

Needs to burn between 11 and 12 calories/min, so we'll calculate the range:

11 calories/min x 20 min = 220 calories (minimum # calories she can burn)

12 calories/min x 20 min = 240 calories (maximum # calories she can burn)

7 0

2 years ago

Plzz help Solve for x x ÷3 3/10 =2 2/5

Svet_ta [14]

Answer:

$\huge\boxed{x=7\dfrac{23}{25}}$

Step-by-step explanation:

$x\div3\dfrac{3}{10}=2\dfrac{2}{5}\\\\\text{convert the mixed number to the impropper fraction}\\\\3\dfrac{3}{10}=\dfrac{3\cdot10+3}{10}=\dfrac{33}{10}\\\\2\dfrac{2}{5}=\dfrac{2\cdot5+2}{5}=\dfrac{12}{5}\\\\x\div\dfrac{33}{10}=\dfrac{12}{5}\\\\x\times\dfrac{10}{33}=\dfrac{12}{5}\qquad\text{multiply both sides by}\ \dfrac{33}{10}\\\\x\times\dfrac{10\!\!\!\!\!\diagup}{33\!\!\!\!\!\diagup}\times\dfrac{33\!\!\!\!\!\diagup}{10\!\!\!\!\!\diagup}=\dfrac{12}{5}\times\dfrac{33}{10}\\\\x=\dfrac{396}{50}$

$x=\dfrac{198}{25}\\\\x=\dfrac{175+23}{25}\\\\x=\dfrac{175}{25}+\dfrac{23}{25}\\\\x=7\dfrac{23}{25}$

3 0

2 years ago

In a pack of 52 cards, 4 of the cards are kings. the first card is picked, the results are recorded and the card is returned to

Ulleksa [173]

Total cards = 52

Total kings = 4

P(2 kings) = (4/52)(4/52) = 1/169

Answer : 1/169

8 0

2 years ago

If f(x)=2x+sinx and the function g is the inverse of f then g'(2)=

Alexxx [7]

$\bf f(x)=y=2x+sin(x)
\\\\\\
inverse\implies x=2y+sin(y)\leftarrow f^{-1}(x)\leftarrow g(x)
\\\\\\
\textit{now, the "y" in the inverse, is really just g(x)}
\\\\\\
\textit{so, we can write it as }x=2g(x)+sin[g(x)]\\\\
-----------------------------\\\\$

$\bf \textit{let's use implicit differentiation}\\\\
1=2\cfrac{dg(x)}{dx}+cos[g(x)]\cdot \cfrac{dg(x)}{dx}\impliedby \textit{common factor}
\\\\\\
1=\cfrac{dg(x)}{dx}[2+cos[g(x)]]\implies \cfrac{1}{[2+cos[g(x)]]}=\cfrac{dg(x)}{dx}=g'(x)\\\\
-----------------------------\\\\
g'(2)=\cfrac{1}{2+cos[g(2)]}$

now, if we just knew what g(2) is, we'd be golden, however, we dunno

BUT, recall, g(x) is the inverse of f(x), meaning, all domain for f(x) is really the range of g(x) and, the range for f(x), is the domain for g(x)

for inverse expressions, the domain and range is the same as the original, just switched over

so, g(2) = some range value
that means if we use that value in f(x), f( some range value) = 2

so... in short, instead of getting the range from g(2), let's get the domain of f(x) IF the range is 2

thus 2 = 2x+sin(x)

$\bf 2=2x+sin(x)\implies 0=2x+sin(x)-2
\\\\\\
-----------------------------\\\\
g'(2)=\cfrac{1}{2+cos[g(2)]}\implies g'(2)=\cfrac{1}{2+cos[2x+sin(x)-2]}$

hmmm I was looking for some constant value... but hmm, not sure there is one, so I think that'd be it

5 0

2 years ago

What percent of 99.2 is 91

olya-2409 [2.1K]

To find percent, you take 91 and divide it by 99.2 then multiply by 100.

(91/99.2) × 100 = 91.92%

7 0

2 years ago