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3 years ago

What is 117/320 simplified to a mixed fraction

Mathematics

2 answers:

Alex3 years ago

8 0

Answer:

hope this helps 229 999 0523

117 /320 ≈ 0.366

Step-by-step explanation:

Step 1 of 1: Simplify.

Simplify

117 over 320

117

320

Step 1 of 1: Simplify, sub-step a: Reduce fraction to lowest terms.

Reduce fraction to lowest terms

1 is the greatest common divisor of 117 and 320. The result can't be further reduced.

pshichka [43]3 years ago

3 0

Answer:

0.5087

Step-by-step explanation:

i just know it...

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What are the solutions of the quadratic equation? 4x^2+34x+60=0

insens350 [35]

Answer:

Option 3) -6,-5/2

Step-by-step explanation:

we know that

The formula to solve a quadratic equation of the form

$ax^{2} +bx+c=0$

is equal to

$x=\frac{-b(+/-)\sqrt{b^{2}-4ac}} {2a}$

in this problem we have

$4x^{2} +34x+60=0$

$a=4\\b=34\\c=60$

substitute in the formula

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3 years ago

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A direct relationship requires a change in the same direction of both variables, answer c.

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What can you do when the unit prices for two comparable items are listed using different units of measure?

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When you have two unit prices for two comparable items, you have to convert one of the prices to the other one.

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Find the value of x. When a leg is 6 units and the hypotenuse is 12 units.

DIA [1.3K]

Answer:

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Step-by-step explanation:

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3 years ago

What would be the coordinates of triangle A'B'C' if triangle ABC was dilated by a factor of

notsponge [240]

Answer:

$A(0,0)\rightarrow A'(0,0) \\ B(2,0)\rightarrow B'(\frac{2}{3},0) \\ C(0,2)\rightarrow C'(0,\frac{2}{3})$

Step-by-step explanation:

The question is as following:

The verticies of a triangle on the coordinate plane are

A(0, 0), B(2, 0) and C(0, 2).

What would be the coordinates of triangle A'B'C' if triangle ABC was dilated by a factor of 1/3 ?

=============================================

Given: the vertices of a triangle ABC are A(0, 0), B(2, 0) and C(0, 2).

IF the triangle is dilated by a factor of k about the origin, then

(x,y) → (kx , ky)

that triangle ABC was dilated by a factor of 1/3 to create the triangle A'B'C'.

It is given that triangle ABC was dilated by a factor of 1/3 to create the triangle A'B'C'.

If a figure dilated by a factor of 1/3 about the origin

So, $(x,y)\rightarrow (\frac{1}{3}x,\frac{1}{3}y)$

So, The coordinates of the triangle A'B'C' are:

$A(0,0)\rightarrow A'(0,0) \\ B(2,0)\rightarrow B'(\frac{2}{3},0) \\ C(0,2)\rightarrow C'(0,\frac{2}{3})$

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3 years ago