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3 years ago

13

Find the constant of proportionality and write an equation for the relationship.

1 answer:

Ad libitum [116K]3 years ago

5 0

Answer:

I have no clue but i have an examm sorry

Step-by-step explanation:

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1/5 (4-3x) = 1/7 (3x - 4) pls solve

natulia [17]

$\huge \boxed{\mathfrak{Question} \downarrow}$

$\large\sf\frac{1}{5} (4 - 3x) = \frac{1}{7} (3x - 4) \\$

$\large \boxed{\mathfrak{Answer \: with \: Explanation} \downarrow}$

$\frac{1}{5} (4 - 3x) = \frac{1}{7} (3x - 4) \\$

Use the distributive property to multiply 1/5 by 4-3x.

$\frac{1}{5}\times 4+\frac{1}{5}\left(-3\right)x=\frac{1}{7}\left(3x-4\right) \\$

Multiply 1/5 and 4 to get 4/5.

$\frac{4}{5}+\frac{1}{5}\left(-3\right)x=\frac{1}{7}\left(3x-4\right) \\$

Multiply 1/5 and -3 to get -3/5.

$\frac{4}{5}+\frac{-3}{5}x=\frac{1}{7}\left(3x-4\right) \\$

Use the distributive property to multiply 1/7 by 3x-4.

$\frac{4}{5}-\frac{3}{5}x=\frac{1}{7}\times 3x+\frac{1}{7}\left(-4\right) \\$

Multiply 1/7 and 3 to get 3/7 & 1/7 × -4 to get -4/7.

$\frac{4}{5}-\frac{3}{5}x=\frac{3}{7}x-\frac{4}{7} \\$

Subtract $\frac{3}{7}x\\$ from both sides.

$\frac{4}{5}-\frac{3}{5}x-\frac{3}{7}x=-\frac{4}{7} \\$

Combine $-\frac{3}{5}x\\$ and $-\frac{3}{7}x\\$ to get $-\frac{36}{35}x\\$ .

$\frac{4}{5}-\frac{36}{35}x=-\frac{4}{7} \\$

Subtract 4/5 from both sides.

$-\frac{36}{35}x=-\frac{4}{7}-\frac{4}{5} \\$

The least common multiple of 7 and 5 is 35. Convert -4/7 and 4/5 to fractions with denominator 35.

$-\frac{36}{35}x=-\frac{20}{35}-\frac{28}{35} \\$

Because $-\frac{20}{35} \\$ and $\frac{28}{35}\\$ have the same denominator, subtract them by subtracting their numerators.

$-\frac{36}{35}x=\frac{-20-28}{35} \\$

Subtract 28 from -20 to get -48.

$-\frac{36}{35}x=-\frac{48}{35} \\$

Multiply both sides by $-\frac{35}{36}\\$ , the reciprocal of $-\frac{36}{35}\\$ .

$x=-\frac{48}{35}\left(-\frac{35}{36}\right) \\$

Multiply $-\frac{48}{35}\\$ by $-\frac{35}{36}\\$ by multiplying the numerator by the numerator and the denominator by the denominator.

$x=\frac{-48\left(-35\right)}{35\times 36} \\$

Carry out the multiplications in the fraction $\frac{-48\left(-35\right)}{35\times 36}\\$ .

$x=\frac{1680}{1260} \\$

Reduce the fraction 1680/1260 to its lowest terms by extracting and cancelling out 420.

$\huge \boxed{ \boxed{ \bf \: x=\frac{4}{3} \approx \: 1.33}}$

3 0

3 years ago

Read 2 more answers

How to put this in an algebraic expression

likoan [24]

8%w-16%z
I think that this is the answer.

3 0

4 years ago

Hey i need help with number 8 and 9 thanks!

snow_lady [41]

Answer:

8 9.8 - 6.4n

9 15 - j/2

Step-by-step explanation:

8

2.8 - 4.4n - 2n +7

= 2.8+7 -4.4n -2n

= 9.8 - 6.4n

9

11 + (-3) - 1/8 j - 3/8 j +7

= 11 - 3 - 1/8 j - 3/8 j + 7

= 11 - 3 + 7 - 1/8 j - 3/8 j

= 15 - 4/8 j

= 15 - j/2

5 0

3 years ago

Find the volume V of the solid obtained by rotating the region bounded by the given curves about the specified line. x

miskamm [114]

Answer:

$\frac{784}{15} \pi$

Step-by-step explanation:

According to the given situation, the calculation of volume of the solid is shown below:-

Here we will consider the curves that is

$x = 7y^2, x = 7$

Now, rotating the line for the line x which is equals to 7

$7y^2 = 7\\\\y^2 = 1\\\\ y = \pm1$

So, the inner radio is

7 - 7 = 0

and the outer radius is

$7y^2 - 7\\\\ = 7(y^2 - 1)$

Now, the area of cross section is

$A(y) = \pi(7(y^2 - 1))^2\\\\ = 49\pi(y^4 - 2y^2 + 1)$

The volume is

$V = \int\limits^1_{-1} A(y)dy$

now we will put the values into the above formula

$= \int\limits^1_{-1} 49\pi(y^4 - 2y^2 + 1)dy\\\\ = 49\pi(\frac{y^5}{5} - \frac{2y^3}{3} + y)^{-1}\\\\ = 49\pi(\frac{1}{5} - \frac{2}{3} + 1 + \frac{1}{5} - \frac{2}{3} + 1)\\\\ = 49\pi(2 + \frac{2}{5} - \frac{4}{3} )\\\\ = 49\pi(\frac{30+6-20}{15} )\\\\ = \frac{49\pi}{15} (16)$

After solving the above equation we will get

$= \frac{784}{15} \pi$

6 0

4 years ago

Write 5.42 as a mixed number in simplest form.<br> 5.42 =

dmitriy555 [2]

Answer: 5 21/50

Step-by-step explanation: 5.42 = 271/50 = 521/50 as a fraction

8 0

4 years ago