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ExtremeBDS [4]
3 years ago
14

Calcula el volumen. Largo: 50cm Largo: 35cm Largo: 40cm Ancho: 30cm Ancho: 20cm Ancho: 60cm Alto: 20cm Alto: 50cm Alto: 30cm Vol

umen: ___________ Volumen:______ Volumen:_______
Mathematics
1 answer:
Ronch [10]3 years ago
5 0

Answer:

Volumen A = 30000 cm³

Volumen B = 35000 cm³

Volumen C = 72000 cm³

Step-by-step explanation:

a) Volumen A

Largo = 50 cm, Ancho = 30 cm Alto = 20 cm

Volumen = Largo × Ancho × Alto

= 50 cm × 30 cm × 20 cm

= 30000 cm³

b) Volumen B

Largo = 35 cm, Ancho = 20 cm Alto = 50 cm

Volumen = Largo × Ancho × Alto

= 35 cm × 20 cm × 50 cm

= 35000 cm³

c) Volumen C

Largo = 40 cm, Ancho = 60 cm Alto = 30 cm

Volumen = Largo × Ancho × Alto

= 40 cm × 60 cm × 30 cm

= 72000 cm³

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Please help fast.
podryga [215]

\huge{\underline{\boxed{\tt{Answer:}}}}

Let AB be a chord of the given circle with centre and radius 13 cm.

Then, OA = 13 cm and ab = 10 cm

From O, draw OL⊥ AB

We know that the perpendicular from the centre of a circle to a chord bisects the chord.

∴ AL = ½AB = (½ × 10)cm = 5 cm

From the right △OLA, we have

OA² = OL² + AL²

==> OL² = OA² – AL²

==> [(13)² – (5)²] cm² = 144cm²

==> OL = √144cm = 12 cm

Hence, the distance of the chord from the centre is 12 cm.

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8 0
3 years ago
Which value completes the table?
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Alex bikes 2 3/4 miles. Then, she runs 1 2/4 miles. The distance she bikes is how many times the distance she runs?
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Answer:

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6 0
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Solve for x and y.<br><br> 4x-3y=16<br> -2x+4y=-2<br><br> DUE 2/7/17!
Trava [24]
\left \{ {{4x-3y=16\:(I)} \atop {-2x+4y=-2\:(II)}} \right.

<span>Multiply the 2nd equation by (-4). This will eliminate the x's when you add the two new equations together.
</span>\left \{ {{4x-3y=16\:\:\:\:\:\:\:\:\:\:} \atop {-2x+4y=-2\:*(-4)}} \right.
\left \{ {{\diagup\!\!\!\!4x-3y=16} \atop {-\diagup\!\!\!\!4x+8y=-4}} \right.
--------------------
5y = 12
\boxed{\boxed{y =  \frac{12}{5}}}\end{array}}\qquad\quad\checkmark

Let's replace the value found in the first equation:
4x-3y=16\:(I)
4x-3*( \frac{12}{5}) =16
4x -  \frac{36}{5} = 16
least common multiple (5)
\frac{20x}{\diagup\!\!\!\!5} - \frac{36}{\diagup\!\!\!\!5} = \frac{80}{\diagup\!\!\!\!5}
20x - 36 = 80
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x = \frac{116}{20} \frac{\div4}{\div4} \to\: \boxed{\boxed{x = \frac{29}{5} }}\end{array}}\qquad\quad\checkmark



4 0
3 years ago
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