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4 years ago

Can u please help me with these two questions .

Mathematics

1 answer:

scoundrel [369]4 years ago

3 0

Question 1
2x + 4 + x - 7 = 3(x-7)
Add like terms, x-terms add
3x + 4 - 7 = 3(x-7)
Subtract 7 from 4
3x - 3 = 3(x-7)
Multiply out right side with distributive property, 3 times x, and 3 times -7
3x -3 = 3x - 21
Add 21 on each side to cancel -21 on right
3x + 18 = 3x
This is impossible, 3x + 18 cannot equal 3x, plug value in to check, 3(1) + 18 = 3(1), 21 not equal to 3
Question 1 Answer : No Answer

Question 2
4(m-2) = -3(m-16)
Multiply using distributive property on each side
4m - 8 = -3m + 48
Add 8 on each side to cancel -8 on left
4m = -3m + 56
Add 3m on each side to cancel -3m on right
7m = 56
Divide 7 on each side to make 7m : m
m = 56/7
m = 8
Answer to Question 2 : 8

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Help me with my test

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Answer:

the answer is 0.0183

Step-by-step explanation:

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3 years ago

A large farm has 75 acres of wheat and 62.5 acres of corn. The farm crew can harvest the the wheat from 12 acres and the corn fr

Leviafan [203]

In a Farm, there are:
=> 75 acres of wheat
=> 62.5 acres of corn
In each day, the farm crew can harvest:
=> 12 acres of wheat
=> 10 acres of corn
Find how many days can the farm crew harvest all of the plants,
=> 75 / 12
=> 6.25, thus the farm crew will take 6.25 days to be able to harvest acres of wheat
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4 years ago

Find the partial fraction decomposition of the rational expression with prime quadratic factors in the denominator

SpyIntel [72]

$\dfrac{5x^4-7x^3-12x^2+6x+21}{(x-3)(x^2-2)^2}=\dfrac{a_1}{x-3}+\dfrac{a_2x+a_3}{x^2-2}+\dfrac{a_4x+a_5}{(x^2-2)^2}$
$\implies 5x^4-7x^3-12x^2+6x+21=a_1(x^2-2)^2+(a_2x+a_3)(x-3)(x^2-2)+(a_4x+a_5)(x-3)$

When $x=3$ , you're left with

$147=49a_1\implies a_1=\dfrac{147}{49}=3$

When $x=\sqrt2$ or $x=-\sqrt2$ , you're left with

$\begin{cases}17-8\sqrt2=(\sqrt2a_4+a_5)(\sqrt2-3)&\text{for }x=\sqrt2\\17+8\sqrt2=(-\sqrt2a_4+a_5)(-\sqrt2-3)\end{cases}\implies\begin{cases}-5+\sqrt2=\sqrt2a_4+a_5\\-5-\sqrt2=-\sqrt2a_4+a_5\end{cases}$

Adding the two equations together gives $-10=2a_5$ , or $a_5=-5$ . Subtracting them gives $2\sqrt2=2\sqrt2a_4$ , $a_4=1$ .

Now, you have

$5x^4-7x^3-12x^2+6x+21=3(x^2-2)^2+(a_2x+a_3)(x-3)(x^2-2)+(x-5)(x-3)$
$5x^4-7x^3-12x^2+6x+21=3x^4-11x^2-8x+27+(a_2x+a_3)(x-3)(x^2-2)$
$2x^4-7x^3-x^2+14x-6=(a_2x+a_3)(x-3)(x^2-2)$

By just examining the leading and lagging (first and last) terms that would be obtained by expanding the right side, and matching these with the terms on the left side, you would see that $a_2x^4=2x^4$ and $a_3(-3)(-2)=6a_3=-6$ . These alone tell you that you must have $a_2=2$ and $a_3=-1$ .

So the partial fraction decomposition is

$\dfrac3{x-3}+\dfrac{2x-1}{x^2-2}+\dfrac{x-5}{(x^2-2)^2}$

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3 years ago

I need help on it, I don’t know what I’m supposed to do

bogdanovich [222]

Answer:

The length of river frontage for each lot are 96.55 ft. 98.85 ft, 101.15 ft and 103.45 ft.

Step-by-step explanation:

See the attached diagram.

The river frontage of 400 ft will be divided into 84 : 86 : 88 : 90 for each lot as AP, BQ, CR, DS and ET all are parallel.

Therefore, PQ : QR : RS : ST = 84 : 86 : 88 : 90 = 42 : 43 : 44 : 45

Let, PQ = 42x, QR = 43x, RS = 44x and ST = 45x

So, (42x + 43x + 44x + 45x) = 400

⇒ 175x = 400

⇒ x = 2.2988.

So, PQ = 42x = 96.55 ft.

QR = 43x = 98.85 ft.

RS = 44x = 101.15 ft and

ST = 45x = 103.45 ft

(Answer)

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3 years ago

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Answer:

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3 years ago