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4 years ago

Estimate the sum to the nearest hundreds 166+427

Mathematics

2 answers:

quester [9]4 years ago

5 0

The answer to your question is 600

mr_godi [17]4 years ago

4 0

166 estimated to the nearest hundreds would be 200 (6 is greater than 5, and so you round up. & since you are rounding to the nearest hundred, you have to look at the digit in the tens place)

427 estimated to the nearest hundreds would be 400

400 + 200 = 600

your estimated sum is 600

hope this helps

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A tank contains 300 liters of fluid in which 40 grams of salt is dissolved. Brine containing 1 gram of salt per liter is then pu

bonufazy [111]

Answer:

A(t) = 300 -260e^(-t/50)

Step-by-step explanation:

The rate of change of A(t) is ...

A'(t) = 6 -6/300·A(t)

Rewriting, we have ...

A'(t) +(1/50)A(t) = 6

This has solution ...

A(t) = p + qe^-(t/50)

We need to find the values of p and q. Using the differential equation, we ahve ...

A'(t) = -q/50e^-(t/50) = 6 - (p +qe^-(t/50))/50

0 = 6 -p/50

p = 300

From the initial condition, ...

A(0) = 300 +q = 40

q = -260

So, the complete solution is ...

A(t) = 300 -260e^(-t/50)

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6 0

3 years ago

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How fast is somebody running if they run 26.2 miles in 2 hours

lianna [129]

They are running 13.1 miles per hour since half of 26.2 is 13.1

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3 years ago

If you place a 17-foot ladder against the top of a 8-foot building, how many feet will the bottom of the ladder be from the bott

Goshia [24]

Answer:

15 feet

Step-by-step explanation:

This problem involves using the Pythagorean theorem, since the figure made with the ladder, building, and ground would make a right triangle. You are given the values 17ft and 8ft, which is enough to plug into the Pythagorean theorem.

The ladder, 17ft, would be the longest side (hypotenuse). The 8ft building would be one of the legs of the right triangle.

1. Plug your given values correctly into the Pythagorean Theorem.

$a^{2} + b^{2} = c^{2}$

$8^{2} + b^{2} = 17^{2}$

2. Now solve for b, which is your unknown distance (the distance the bottom of the ladder is from the bottom of the building).

$8^{2} + b^{2} = 17^{2}$ --> Square 8 and 17

$64 + b^{2} = 289$ --> Subtract 64 from both sides

$b^{2} = 225$ --> Square root both sides to get b by itself

b = 15

3. The distance is 15 feet

*Note: to make solving this problem easier, try drawing out the given situation, namely the building and the ladder

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3 years ago

The perimeter of a 5 sided figure is 45.56 meters. Two of the sides have the same length. The sum of the other three side length

klio [65]

Let

x---------> the length of the sides that have the same length in meters

we know that

the perimeter of the figure is equal to

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<u>solve for x</u>

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3 0

4 years ago

<img src="https://tex.z-dn.net/?f=%5Cfrac%7Ba%5E%7B2%7D-1%7D%7B2-5a%7D%20times%20%5Cfrac%7B15a-6%7D%7Ba%5E%7B2%7D%2B5a-6%7D" id=

stepan [7]

$\bf \cfrac{a^2-1}{2-5a}\times \cfrac{15-6}{a^2+5a-6}\\\\
-----------------------------\\\\
recall\quad \textit{difference of squares}
\\ \quad \\
(a-b)(a+b) = a^2-b^2\qquad \qquad 
a^2-b^2 = (a-b)(a+b)\\\\
thus\quad a^2-1\iff a^2-1^2\implies (a-1)(a+1)
\\\\\\
now\quad a^2+5a-6\implies (a+6)(a-1)\\\\
-----------------------------\\\\
thus
\\\\\\
\cfrac{a^2-1}{2-5a}\times \cfrac{15-6}{a^2+5a-6}\implies \cfrac{(a-1)(a+1)}{2-5a}\times \cfrac{3(5a-2)}{(a+6)(a-1)}\\\\
-----------------------------\\\\
$

$\bf now\quad 3(5a-2) \iff -3(2-5a)\\\\
-----------------------------\\\\
thus
\\\\\\
\cfrac{\underline{(a-1)}(a+1)}{\underline{2-5a}}\times \cfrac{-3\underline{(2-5a)}}{(a+6)\underline{(a-1)}}\implies \cfrac{-3(a+1)}{a+6}$

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3 years ago