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romanna [79]
2 years ago
3

Sherry had 36 thousands-blocks in her group, and then added 54 more thousands-blocks. If she traded them in for ten thousands-bl

ocks, how many would she receive?
A.9000
B.900
C.90
D.9 ​
Mathematics
1 answer:
masya89 [10]2 years ago
7 0

Answer:

A.9000

Step-by-step explanation:

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<span>-Both box plots show the same interquartile range.
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</span><span>-The smallest class size was 24 students.
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Five years ago an alligator was to 2 1/6 feet long today the alligator three times longer how long is the alligator now
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I hope this helps you

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2 years ago
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A rock is thrown in the air from the edge of a seaside cliff. Its height in feet is represented by f(x) = –16(x^2 – 8x – 9), whe
algol13
Since x is the number of seconds it took the rock to hit the water, x is the value that we are looking for. When the rock hits the water, it's height is 0, therefore, f(x) = 0. So, we can set the expression on the right equal to 0, and solve for 0. Let's do just that:
0 = -16(x^2-8x-9)

We can divide by -16, nothing will happen because 0 divided by anything is 0:
0 = x^2-8x-9

To factor any trinomial in standard form (given below), we must find two numbers that add to "b" and multiply to "c":
0 = ax^2 + bx + c

So, for this specific question, let's find two numbers that add to -8 and multiply to 9. These are clearly 9 and -1. We can write it in this way now:
0 = (x-9)(x+1)

Both answers would be valid, but there can't be negative time, so your answer is "It takes 9 seconds for the rock to reach the water."
4 0
2 years ago
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Which expression is equivalent to StartFraction RootIndex 7 StartRoot x squared EndRoot Over RootIndex 5 StartRoot y cubed EndRo
Lera25 [3.4K]

Answer:

Option A.

Step-by-step explanation:

The given expression is

\dfrac{\sqrt[7]{x^2}}{\sqrt[5]{y^3}}

where, y\neq 0.

We need to find the expression which is equivalent to the given expression.

The given expression can be rewritten as

\dfrac{(x^2)^{\frac{1}{7}}}{(y^3)^{\frac{1}{5}}}      [\because \sqrt[n]{x}=x^{\frac{1}{n}}]

\dfrac{x^{\frac{2}{7}}}{y^{\frac{3}{5}}}      [\because (a^m)^n=a^{mn}]

x^{\frac{2}{7}}y^{-\frac{3}{5}}      [\because a^{-n}=\dfrac{1}{a^n}]

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2 years ago
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Orin solved an equation and justified his steps as shown in the table.
sergij07 [2.7K]

Answer:

Step-by-step explanation:

Given the equation as

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apply multiplication property of equality where you multiply every term by 5

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3x/3=75/3

x=25

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