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3 years ago

What is the answer for 3\5=a+5\25

Mathematics

2 answers:

Zina [86]3 years ago

7 0

Answer: $a=\frac{2}{5}$

Step-by-step explanation:

$\frac{3}{5} =a+\frac{5}{25} \\\\\frac{3}{5} =a+\frac{1}{5} \\\\\frac{3}{5} - \frac{1}{5}=a\\\\a=\frac{2}{5}$

Hope this helps, now you know the answer and how to do it. HAVE A BLESSED AND WONDERFUL DAY! As well as a great Valentines Day! :-)

- Cutiepatutie ☺❀❤

IceJOKER [234]3 years ago

4 0

Answer:

a = 2/5 = 0.400

Step-by-step explanation:

Step 1 :

Simplify —

Equation at the end of step 1 :

3 1

— - (a + —) = 0

5 5

Step 2 :

Rewriting the whole as an Equivalent Fraction :

2.1 Adding a fraction to a whole

Rewrite the whole as a fraction using 5 as the denominator :

a a • 5

a = — = —————

1 5

Equivalent fraction : The fraction thus generated looks different but has the same value as the whole

Common denominator : The equivalent fraction and the other fraction involved in the calculation share the same denominator

Adding fractions that have a common denominator :

2.2 Adding up the two equivalent fractions

Add the two equivalent fractions which now have a common denominator

Combine the numerators together, put the sum or difference over the common denominator then reduce to lowest terms if possible:

a • 5 + 1 5a + 1

————————— = ——————

5 5

Equation at the end of step 2 :

3 (5a + 1)

— - ———————— = 0

5 5

Step 3 :

Simplify —

Equation at the end of step 3 :

3 (5a + 1)

— - ———————— = 0

5 5

Step 4 :

Adding fractions which have a common denominator :

4.1 Adding fractions which have a common denominator

Combine the numerators together, put the sum or difference over the common denominator then reduce to lowest terms if possible:

3 - ((5a+1)) 2 - 5a

———————————— = ——————

5 5

Equation at the end of step 4 :

2 - 5a

—————— = 0

Step 5 :

When a fraction equals zero :

5.1 When a fraction equals zero ...

Where a fraction equals zero, its numerator, the part which is above the fraction line, must equal zero.

Now,to get rid of the denominator, Tiger multiplys both sides of the equation by the denominator.

Here's how:

2-5a

———— • 5 = 0 • 5

Now, on the left hand side, the 5 cancels out the denominator, while, on the right hand side, zero times anything is still zero.

The equation now takes the shape :

2-5a = 0

Solving a Single Variable Equation :

5.2 Solve : -5a+2 = 0

Subtract 2 from both sides of the equation :

-5a = -2

Multiply both sides of the equation by (-1) : 5a = 2

Divide both sides of the equation by 5:

a = 2/5 = 0.400

One solution was found :

a = 2/5 = 0.400

Processing ends successfully

plz mark me as brailiest :)

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mariarad [96]

Answer:

a. $\frac{35}{51}$

b. $\frac{51}{100}$

c. $\frac{1}{5}$

Step-by-step explanation:

Suppose cities represented by C', suburbs represented by S and rural represented by R,

Let x be the total number of bonds issued throughout the US,

According to the question,

n(A) = 70% of x = 0.7x,

n(B) = 10% of x = 0.1x,

n(C) = 20% of x = 0.2x,

n(A∩C') = 50% of n(A) = 0.5 × 0.7x = 0.35x,

n(A∩S) = 20% of n(A) = 0.2 × 0.7x = 0.14x,

n(A∩R) = 30% of n(A) = 0.3 × 0.7x = 0.21x,

n(B∩C') = 40% of n(B) = 0.4 × 0.1x = 0.04x,

n(B∩S) = 30% of n(B) = 0.3 × 0.1x = 0.03x,

n(B∩R) = 30% of n(B) = 0.3 × 0.1x = 0.03x,

n(C∩C') = 60% of n(C) = 0.6 × 0.2x = 0.12x,

n(C∩S) = 15% of n(C) = 0.15 × 0.2x = 0.03x,

n(C∩R) = 25% of n(C) = 0.25 × 0.2x = 0.05x,

n(C') = n(A∩C') + n(B∩C') + n(C∩C') = 0.35x + 0.04x + 0.12x = 0.51x

n(S) = n(A∩S) + n(B∩S) + n(C∩S) = 0.14x + 0.03x + 0.03x = 0.20x

a. The probability that it will receive an A rating, if a new municipal bond is to be issued by a city,

$P(\frac{A}{C'})=\frac{P(A\cap C')}{P(C')}=\frac{0.35x/x}{0.51x/x}=\frac{0.35}{0.51}=\frac{35}{51}$

b. The proportion of municipal bonds are issued by cities = $\frac{n(C')}{x}$

$=\frac{0.51x}{x}$

$=\frac{51}{100}$

c. The proportion of municipal bonds are issued by suburbs = $\frac{n(S)}{x}$

$=\frac{0.20x}{x}$

$=\frac{20}{100}$

$=\frac{1}{5}$

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Ilia_Sergeevich [38]

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Rate of change also represents slope of a linear function.

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Plugging values of m and b in the above slope-intercept form of a linear equation.We get,

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We get,

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