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3 years ago

What is 27/36 as a mixed number

Mathematics

2 answers:

Ivan3 years ago

7 0

This fraction cannot be turned into a mixed number but it can be reduced to $\frac{3}{4}$ or 0.75
hope this helps

Lana71 [14]3 years ago

3 0

Hey there :)

This fraction $( \frac{27}{36} )$ cannot be written as a mixed number the numerator is not greater than the denominator

However, it can be further simplified

27 and 36 had a common multiple: 9
$\frac{21/9}{36/9} = \frac{3}{4}$

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What is the sum? (g2 – 4g4 + 5g + 9) + (–3g3 + 3g2 – 6)

levacccp [35]

B. –4g4 – 3g3 + 4g2 + 5g + 3

Hope it helped!

4 0

4 years ago

Read 2 more answers

If a is real and the product of (a+6i)(5-3i) is a real number, the find a.

irina [24]

Answer:

5a−3ai+30i+18

Step-by-step explanation:

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2 years ago

DochEvi [55]

Answer:

12pi cm

Step-by-step explanation:

The Perimeter of the full shape is the sum of the lengths of the edges of the parts. For convenience in referencing them, we'll call the large curve " $curve_{big}$ " and the three smaller curves " $curve_1$ " " $curve_2$ " " $curve_3$ " in order from left to right.

Thus, the Perimeter of the full shape can be written as an equation:

$P_{overall} = Length(curve_{big})+Length(curve_1)+Length(curve_2)+Length(curve_3)$ Since all of those edge lengths are curves, and the question states that all of the curves are made from parts of circles, then we need to know how to find the length of the edge of a circle.

Parts of a circle

Since values in the diagram are diameters, use the formula for the Perimeter of a circle $P=\pi d$ (where d is the diameter).

Let's call the diameters of each of our curves " $d_{big}$ " " $d_1$ " " $d_2$ " " $d_3$ ", with the subscripts denoting which curve we're referring to.

Note that for each curve, the curve only represents half of a circle. So, to find the length of each curve, we'll need half of the full perimeter of each circle.

So for instance: $Length(curve_{big})=\frac{1}{2} \pi d_{big}$

Substituting back into the main equation above:

$P_{overall} = Length(curve_{big})+Length(curve_1)+Length(curve_2)+Length(curve_3)$ $P_{overall}=\frac{1}{2} \pi d_{big} + \frac{1}{2} \pi d_{1} + \frac{1}{2} \pi d_{2} + \frac{1}{2} \pi d_{3}$

Note that all terms have common factors of "one-half" and "pi" in them. These can be factored out:

$P_{overall}=\frac{1}{2} \pi (d_{big} + d_{1} + d_{2} +d_{3})$

The diameter for the large Curve, is the sum of the three small diameters, so $d_{big}=12cm$ , and $d_{1}=d_{2}=d_{3}=4cm$

Substituting and simplifying (in terms of pi):

$P_{overall}=\frac{1}{2} \pi ( (12cm) + (4cm) + (4cm) + (4cm) )\\P_{overall}=\frac{1}{2} \pi ( 24cm)\\P_{overall}=12 \pi cm$

Additional Understanding

Interesting for this problem, since the diameters of the 3 small curves formed the diameter of the large curve $d_{1} + d_{2} + d_{3} =d_{big}$ , one could make a different substitution into one of our formulas above:

$P_{overall}=\frac{1}{2} \pi (d_{big} + d_{1} + d_{2} +d_{3})$

$P_{overall}=\frac{1}{2} \pi (d_{big} + (d_{big}))$

$P_{overall}=\frac{1}{2} \pi (2d_{big})$

$P_{overall}=\pi d_{big}$

Notice that $\pi d_{big}$ is just the full perimeter of a circle with the big diameter.

So, if one imagined starting with a full circle with the big diameter, even though the bottom half of the circle was turned into a bunch of smaller half circles, since they were in a line along the diameter of the large circle, the full perimeter of the new shape didn't change.

The number of smaller circles doesn't need to be 3 either... as long as it goes the full distance across, right along the diameter.

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2 years ago

Solve this two column proof! (25 pts!)

ahrayia [7]

Answer:

1.) Given

2.)Given

3.)Angle bisector

4.)Perp. Segments

5.)substitution

6.)Reflexive

7.)ASA postulate

8.)CPCTC

9.)Def of isosceles triangle

Step-by-step explanation:

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3 years ago

The graph of y= (x + 2)(x-2)(x + 1) is shown.

In-s [12.5K]

Answer:

Point C

Step-by-step explanation:

At point C, the graph is changing from decreasing to increasing.

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2 years ago