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fenix001 [56]
4 years ago
11

$500, 3.75, 4 months

Mathematics
1 answer:
mario62 [17]4 years ago
6 0

Answer:

theres no question????????????

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If 56% of the 125 fifth-grade students have a skateboard, how many of the students have a skateboard?
olasank [31]

Answer:

70

Step-by-step explanation:

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3 years ago
Solve the Inequality and express each solution in set interval notation.
solniwko [45]
The correct answer is x=5
3 0
3 years ago
The quadrilateral ABCD has area of 58 in2 and diagonal AC = 14.5 in. Find the length of diagonal BD if AC ⊥ BD.
azamat

Answer:

(look in the the Step by step)

Step-by-step explanation:

When the diagonals of a quadrilateral are perpendicular, the area of that quadrilateral is half the product of their lengths.

.. A = (1/2)*d₁*d₂

Substituting the given information, this becomes

.. 58 in² = (1/2)*(14.5 in)*d₂

.. 2*58/14.5 in = d₂ = 8 in

The length of diagonal BD is 8 in.

4 0
3 years ago
I need help plzzzzzzzzzzzzzzzzzzzzz
blagie [28]

Answer:

Step-by-step explanation:

Start box

180 = 89+42+x

180-89-42=x

49 = x

2nd box

180 = 84+58+x

180-84-58=x

38 = x

3rd box

180=74+2x

180-74=2x

106/2 = x

53 = x

4th box

180=102+2x

180-102=2x

78/2 = x

39 = x

4th box

180 = 73+81 +x

180-73-81=x

26 = x

5th box

180=2*54+x

180 - 2*54 = x

72 = x

6th box

180=62-2x

180-62=2x

118=2x

118/2=x

59 = x

7th box

180 = 2*68 + x

180-2*68=x

44 = x

3 0
3 years ago
Which choice is equivalent to the expression below when y2 0?<br> √y^2 + √16y^3 – 4y√y
DanielleElmas [232]

Answer:

Option C.

Step-by-step explanation:

We start with the expression:

\sqrt{y^3}  + \sqrt{16*y^3} - 4*y\sqrt{y}

where y > 0. (this allow us to have y inside a square root, so we don't mess with complex numbers)

We want to find the equivalent expression to this one.

Here, we can do the next two simplifications:

\sqrt{16*y^3} = \sqrt{16} \sqrt{y^3} = 4*\sqrt{y^3}

And:

y*\sqrt{y} = \sqrt{y^2} *\sqrt{y} = \sqrt{y^2*y} = \sqrt{y^3}

If we apply these two to our initial expression, we can rewrite it as:

\sqrt{y^3}  + \sqrt{16*y^3} - 4*y\sqrt{y}

\sqrt{y^3}  + 4*\sqrt{y^3} - 4\sqrt{y^3} = \sqrt{y^3}

Here we can use the second simplification again, to rewrite:

\sqrt{y^3} = y*\sqrt{y}

So, concluding, we have:

\sqrt{y^3}  + \sqrt{16*y^3} - 4*y\sqrt{y} = y*\sqrt{y}

Then the correct option is C.

8 0
3 years ago
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