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4 years ago

Simplify the following expression:

1 answer:

5 0

Distributive property
a(b+c+d)=ab+ac+ad

-7(3e-2f+4)=-7(3e)-7(-2f)-7(4)=-21e+14f-28

now we have

-21e+14f-24+6e-2
group like terms
-21e+6e+14f-24-2
-15e+14f-26

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Write a real word problem that can be represented by the equation a - 6 = 15

chubhunter [2.5K]

Answer: your friend is trying to find out how many canys are in a box(a) and he gives you 6 and has 15 left

Step-by-step explanation:

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3 years ago

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three bags of sweets weigh 27/4 kg. two of them have the same weight and the third bag is heavier than each of the bags of equal

Nana76 [90]

I'm here buddy,

so, let's take the value of the two bags with equal weight as x.

= x + x + (x + $\frac{6}{5}$ ) = $\frac{27}{4}$

= 3x + $\frac{6}{5}$ = $\frac{27}{4}$

= 3x = $\frac{27}{4}$ - $\frac{6}{5}$

( let's take the LCM of 4 and 5 = 20

= 3x = $\frac{135}{20}$ - $\frac{24}{20}$

= 3x = $\frac{111}{20}$

= x = $\frac{111}{20}$ ÷ $\frac{3}{1}$ = $\frac{111}{20}$ × $\frac{1}{3}$ = $\frac{37}{20}$

So, the weight of the equal bags are $\frac{37}{20}$ and the weight of the third bag ( heavy one ) is $\frac{37}{20}$ + $\frac{6}{5}$ = $\frac{37}{20}$ + $\frac{24}{20}$ = $\frac{61}{20}$

1st bag = $\frac{37}{20}$ kg

2nd bag = $\frac{37}{20}$ kg

3rd bag = $\frac{61}{20}$ kg

Hope it helps...

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3 years ago

What is the length of the missing side of the triangle? The figure is not drawn to scale.

NARA [144]

Answer:

Step-by-step explanation:

add all the numbers together to get your answer and divide it by two.

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3 years ago

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2 years ago

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Solve the equation. StartFraction dy Over dx EndFraction equals5 x Superscript 4 Baseline (1 plus y squared )Superscript three h

dangina [55]

Answer:

Step-by-step explanation:

To solve the differential equation

dy/dx = 5x^4(1 + y²)^(3/2)

First, separate the variables

dy/(1 + y²)^(3/2) = 5x^4 dx

Now, integrate both sides

To integrate dy/(1 + y²)^(3/2), use the substitution y = tan(u)

dy = (1/cos²u)du

So,

dy/(1 + y²)^(3/2) = [(1/cos²u)/(1 + tan²u)^(3/2)]du

= (1/cos²u)/(1 + (sin²u/cos²u))^(3/2)

Because cos²u + sin²u = 1 (Trigonometric identity),

The equation becomes

[1/(1/cos²u)^(3/2) × 1/cos²u] du

= cos³u/cos²u

= cosu

Integral of cosu = sinu

But y = tanu

Therefore u = arctany

We then have

cos(arctany) = y/√(1 + y²)

Now, the integral of the equation

dy/(1 + y²)^(3/2) = 5x^4 dx

y/√(1 + y²) = x^5 + C

y - (x^5 + C)√(1 + y²) = 0

is the required implicit solution

5 0

3 years ago