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Marrrta [24]
4 years ago
15

8+35/w=15a.8b.6c.10d.5​

Mathematics
2 answers:
Nimfa-mama [501]4 years ago
7 0

Answer:

w = 5

Step-by-step explanation:

8 + \frac{35}{w} = 15

subtract 8 from both sides

8 +  \frac{35}{w} - 8 = 15 - 8

simplify

\frac{35}{w} = 7

35/7 = w

w = 5

RoseWind [281]4 years ago
3 0

Answer:

\boxed{ \bold{ \huge{ \bold{w = 5}}}}

Option D is the correct option

Step-by-step explanation:

\sf{8 +  \frac{35}{w}  = 15}

⇒\sf{ \frac{8 \times w + 35}{w}  = 15} { w ÷ 1 = w , So, I wrote 8 × w }

⇒\sf{ \frac{8w + 35}{w}  = 15}

Apply cross product property

⇒\sf{8w + 35 = 15w}

Move 15w to left hand side and change it's sign

Similarly, Move 35 to right hand side and change it's sign

⇒\sf{8w - 15w  =  - 35}

Collect like terms

⇒\sf{ - 7 w =  - 35 }

Divide both sides of the equation by -7

⇒\sf{ \frac{ - 7w}{ - 7}  =  \frac{ - 35}{ - 7} }

Calculate

⇒\sf{w = 5}

Hope I helped!

Best regards!!

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What is the sum of the given polynomials in standard form? (x2 – 3x) (–2x2 5x – 3).
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Sum of standard form polynomials means adding the terms of same variable. The sum of the given polynomial in standard form is,

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<h3>What is polynomial equation?</h3>

A polynomial equation is the equation in which the unknown variable is one and the highest power of the unknown variable is n.

Here, <em>n</em> is any real number.

In the polynomial equation the the terms are added or subtracted from each other only when the power of the variable is same.

Given information-

The first polynomial equations given in the problem is,

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The second polynomial equations given in the problem is,

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To find the sum of these two polynomials, add the two terms together as,

(x^2 - 3x)+(-2x^2+ 5x - 3)

Let the result of the sum is f(x). Thus,

f(x)=(x^2 - 3x)+(-2x^2+ 5x - 3)\\f(x)=x^2-3x-2x^2+5x-3

Separate the similar power variable as,

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juin [17]

5 1/2 - 3/8

Make 5 1/2 into a improper fraction

5 1/2=11/2. Mutiply the whole number with the denominator. 5*2= 10. Add 10 with the numerator. 10+1= 11

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Answer: 41/8 or 5 1/8

5 0
3 years ago
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