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MrMuchimi
3 years ago
15

Jackson's kitten weighed 2 pounds 3 ounces. A month later the kitten weighed 56 ounces. How much did the kitten gain in the mont

h?
Mathematics
1 answer:
polet [3.4K]3 years ago
6 0
The kitten gained 21 ounces. 2 pounds is equal to 32 ounces, 32 + 3 = 35 and 56 - 35 = 21 ounces
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A model of a volcano has a height of 12 in. , and a diameter of 12 in. What is the approximate volume of the model? Use 3. 14 to
jenyasd209 [6]

The approximate volume of the model will be

v=452.33 inch^{2}

<h3 /><h3>What is the approximate volume of the model? </h3>

It is given that

Height of the volcano =12 inch

Diameter of the volcano= 12inch=r=6inch

So the volume of the model will be =

=\dfrac{1}{3\\} \pi \times r^{2} \times h

= \dfrac{1}{3} \pi \times (6)^{2} \times 12

v=452.33inch^{2}

Thus the approximate volume of the model will be  v=452.33 inch^{2}

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https://brainly.in/question/18480521

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2 years ago
What is the solution to the following system of equations? x+y=5
marta [7]

Answer:

(3,2)

Step-by-step explanation:

Let's solve this via elimination method:

\begin{cases} x+y=5\\x-y=1\end{cases}

By subtracting equation 2 from equation one we obtain:

\begin{cases} x+y=5\\-(x-y=1)\end{cases}\\\\0x+2y=4\\\\y=2

Next we can use any equation either 1 or 2 to determine what x is, I'll use equation 1.  Let y=2 and so:

x+y=5\\\\x+2=5\\x=5-2\\\\x=3

Therefore the solution for the system of equations is:

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7 0
3 years ago
Write a description of how a diagram can help you solve 2×40
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Draw 40 boxes, and in a separate section draw another 40 boxes.
(If you have the time :) )
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3 years ago
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Let f(x) = log(x). Find values of a such that f(kaa) = kf(a).
meriva

Answer:

a = k^{\frac{1}{k-2}}

Step-by-step explanation:

Given:

f(x) = log(x)

and,

f(kaa) = kf(a)

now applying the given function, we get

⇒ log(kaa) = k × log(a)

or

⇒ log(ka²) = k × log(a)

Now, we know the property of the log function that

log(AB) = log(A) + log(B)

and,

log(Aᵇ) = b × log(A)

Thus,

⇒ log(k) + log(a²) = k × log(a)         (using log(AB) = log(A) + log(B) )

or

⇒ log(k) + 2log(a) = k × log(a)            (using log(Aᵇ) = b × log(A) )

or

⇒ k × log(a) - 2log(a) = log(k)

or

⇒ log(a) × (k - 2) = log(k)

or

⇒ log(a) = (k - 2)⁻¹ × log(k)

or

⇒ log(a) = \log(k^{\frac{1}{k-2}})          (using log(Aᵇ) = b × log(A) )

taking anti-log both sides

⇒ a = k^{\frac{1}{k-2}}

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3 years ago
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