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Lady_Fox [76]
1 year ago
7

Andrew has 8 cassettes. Mary has x cassettes and Jim has twice as many as Andrew. Together they have four times as Mary has. For

m an equation and find how many cassettes Mary has.
Mathematics
1 answer:
kupik [55]1 year ago
4 0

Answer:

mary=x

andrew=8

jim=2*8=16

together=4*mary

total=8+16=24

24/4=x

x=6

Step-by-step explanation:

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Jayson takes a total of 5 hours of swimming lessons per week.if each lesson is 5/6 of an hour how many lessons does he take per
IRINA_888 [86]
5/1 / 5/6 = 5/1 x 6/5 = 6
3 0
4 years ago
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The average temperatures for two of the world's hottest places are given. Determine which location has the warmer average temper
zubka84 [21]

Answer: City A, 0.28^{\circ}C

Step-by-step explanation:

Given

The temperature of city A is 86.7\ ^{\circ}F

The temperature of city B is 30.1^{\circ}C

We know, 32^{\circ}F=0^{\circ}C

So, 86.7^{\circ}F=30.38^{\circ}C

30.38>30.1

Clearly, the average temperature of city A is more than city B

The difference in temperature is 30.38-30.1=0.28^{\circ}C

5 0
3 years ago
What is the unknown value? that makes the statement below true? ______ is 40 percent of 280
kolezko [41]
The answer is 112. 4/10 of 280. 280 divided by 10 equals 28. 28 x 4 = 112
4 0
2 years ago
Find x on this special right triangle
tigry1 [53]

Answer:

The answer is 7

Step-by-step explanation:

We know than Tan45=1

Tanx=opp/adj

tan45=y/(7\sqrt{2})/2

Thus y=7\sqrt{2}/2

We know by pythagoras theorem that hyp^2=side^2+side^2

x^{2}=((7\sqrt{2} /2)^{2}+(7\sqrt{2}/2)^{2}

x^{2}=(49(2)/4)+(49(2)/4)

x^{2}=(49/2)+(49/2)

x^{2}=(98/2)

x^{2}=49

x=7

6 0
3 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
3 years ago
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