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Triss [41]
3 years ago
14

Rewrite the equation below so that it does not have fractions.

Mathematics
1 answer:
Eddi Din [679]3 years ago
8 0
Since the 3 and 7 have 21 as a common multiple you can divide 21 by the denominator of each term and then multiply it by the numerator which get you while numbers
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I am very confused please solve and explain.
e-lub [12.9K]

Answer:

c = \frac{b^2-R^2}{4a}

Step-by-step explanation:

Given

R= \sqrt{b^2-4ac}

Clear the radical by squaring both sides

R² = b² - 4ac ( subtract b² from both sides )

R² - b² = - 4ac ( multiply all terms by - 1 )

b² - R² = 4ac ( divide both sides by 4a )

\frac{b^2-R^2}{4a} = c

6 0
3 years ago
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WILL GIVE BRAINLIEST AND 20 POINTS<br><br> Please solve this page
guajiro [1.7K]
Search it up, you’ll get some answers. it maybe work
5 0
3 years ago
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The total resistance in a circuit with two parallel resistors is 2 ohms and R1 is 6 ohms. Using the equation for R2, in terms of
ivanzaharov [21]
<h2>Answer:</h2>

R_2  is 3 ohms

<h2>Step-by-step explanation:</h2>

In a circuit containing two resistors R_1 and R_2 connected together in parallel, the total resistance R_T is given by;

\frac{1}{R_T} = \frac{1}{R_1}  + \frac{1}{R_2}              ---------(i)

<em>Make </em>R_2<em> subject of the formula;</em>

=> \frac{1}{R_2} = \frac{1}{R_T}  - \frac{1}{R_1}  

=> \frac{1}{R_2} = \frac{R_1 - R_T}{R_TR_1}

=>   {R_2} = \frac{R_TR_1}{R_1 - R_T}       ---------------(ii)

From the question,

R_1 = 6Ω

R_T = 2Ω

Substitute these values into equation (ii) as follows;

=> {R_2} = \frac{2*6}{6 - 2}

{R_2} = \frac{12}{4}

R_2 = 3Ω

Therefore, the value of R_2 = 3 ohms or R_2 = 3Ω

3 0
3 years ago
Somebody please help with this problem
Mrrafil [7]

Step-by-step explanation:

We have been given that AE=BE and \angle1\cong \angle2.

We can see that angle CEA is vertical angle of angle DEB, therefore, m\angle CEA=m\angle DEB as vertical angles are congruent.

We can see in triangles CEA and DEF that two angles and included sides are congruent.

\angle 1\cong \angle 2

AE=BE

\angle CEA\cong\angle DEB or \angle 3\cong \angle 4  

Therefore, \Delta CEA\cong \Delta DEB by ASA postulate.

Since corresponding parts of congruent triangles are congruent, therefore CE must be congruent to DE.

5 0
3 years ago
Give to examples of units that can be used on a graph
lbvjy [14]

<u>Metric:                                                           </u>

Distance: Centimeters (cm), Meters (m), Kilometers (km)

Weight: Milligrams (mg), Grams (g), Kilograms (kg)

Volume: Milliliters (mL), Liters (L)

<u>Customary:                                                   </u>

Distance: Inches (in), Feet (ft), Yards (yd), Miles (mi)

Weight: Ounces (oz), Pounds (lbs), Tons (T)

Volume: Cups (c), Pints (pt), Quarts (qt), Gallons (gal)

3 0
3 years ago
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