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zmey [24]
3 years ago
5

Martin is two years older than Reese, and the same age as Lee. If Lee is 12, how old is Reese?

Mathematics
2 answers:
VLD [36.1K]3 years ago
5 0

Answer: Reese is 10 years old.

Step-by-step explanation:

Let be "m" the age of Martin and "r" tthe age of Reese and "l" the age of Lee.

Based on the information provided in the exercise, you can set up the following equations:

m=r+2     [Equation 1]

m=l         [Equation 2]

l=12        [Equation 3]

Therefore, since:

m=l

You know that:

m=12

Now you must substitute the value of "m" into [Equation 1]:

12=r+2

Finally, you must solve for "r" in order to find its value. This is:

12-2=r\\\\r=10

joja [24]3 years ago
4 0

Answer:

Reese has 10 years old.

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Step-by-step explanation:

it goes straight up

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3 years ago
a circular frame that is 3-inches wide surrounds the mirror what I the combined area in square inches of the circular mirror and
Tcecarenko [31]
Assuming the frame is a perfect circle
and 3 inches wide indicates the diameter


area of a circle=pi times radius^2

diameter=2 times radius
radius=1/2 diameter
diameter=3
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radius=3/2=1.5

area=pi times 1.5^2=2.25
pi is aprox 3.14 so
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Sebuah Prisma Segitiga memiliki tiga sisi masing
Veseljchak [2.6K]

Menjawab:

15cm

Penjelasan langkah demi langkah:

Keliling segitiga adalah jumlah semua sisi segitiga

P = s1 + s2 + s3

Diberikan sisi jika segitiga;

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5 0
3 years ago
D/d{cosec^-1(1+x²/2x)} is equal to​
SIZIF [17.4K]

Step-by-step explanation:

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

\rm :\longmapsto\:\dfrac{d}{dx} {cosec}^{ - 1} \bigg( \dfrac{1 +  {x}^{2} }{2x} \bigg)

Let assume that

\rm :\longmapsto\:y =  {cosec}^{ - 1} \bigg( \dfrac{1 +  {x}^{2} }{2x} \bigg)

We know,

\boxed{\tt{  {cosec}^{ - 1}x =  {sin}^{ - 1}\bigg( \dfrac{1}{x} \bigg)}}

So, using this, we get

\rm :\longmapsto\:y = sin^{ - 1} \bigg( \dfrac{2x}{1 +  {x}^{2} } \bigg)

Now, we use Method of Substitution, So we substitute

\red{\rm :\longmapsto\:x = tanz \: \rm\implies \:z =  {tan}^{ - 1}x}

So, above expression can be rewritten as

\rm :\longmapsto\:y = sin^{ - 1} \bigg( \dfrac{2tanz}{1 +  {tan}^{2} z} \bigg)

\rm :\longmapsto\:y = sin^{ - 1} \bigg( sin2z \bigg)

\rm\implies \:y = 2z

\bf\implies \:y = 2 {tan}^{ - 1}x

So,

\bf\implies \: {cosec}^{ - 1}\bigg( \dfrac{1 +  {x}^{2} }{2x} \bigg) = 2 {tan}^{ - 1}x

Thus,

\rm :\longmapsto\:\dfrac{d}{dx} {cosec}^{ - 1} \bigg( \dfrac{1 +  {x}^{2} }{2x} \bigg)

\rm \:  =  \: \dfrac{d}{dx}(2 {tan}^{ - 1}x)

\rm \:  =  \: 2 \: \dfrac{d}{dx}( {tan}^{ - 1}x)

\rm \:  =  \: 2 \times \dfrac{1}{1 +  {x}^{2} }

\rm \:  =  \: \dfrac{2}{1 +  {x}^{2} }

<u>Hence, </u>

\purple{\rm :\longmapsto\:\boxed{\tt{ \dfrac{d}{dx} {cosec}^{ - 1} \bigg( \dfrac{1 +  {x}^{2} }{2x} \bigg) =  \frac{2}{1 +  {x}^{2} }}}}

<u>Hence, Option (d) is </u><u>correct.</u>

6 0
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