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3 years ago

Solve the equation given in Exercise 15 with and without using the LCD of the fractions. Are your answers the same?

1 answer:

Colt1911 [192]3 years ago

3 0

Without LCD :
1/2(4x + 6) = 1/3(9x - 24)
2x + 3 = 3x - 8
2x - 3x = -8 - 3
-x = - 11
x = 11

with LCD :
1/2(4x + 6) = 1/3(9x - 24) ...multiply by 6
3(4x + 6) = 2(9x - 24)
12x + 18 = 18x - 48
12x - 18x = -48 - 18
-6x = - 66
x = -66/-6
x = 11

Yes, the answers are the same :)

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A 4-column table with 3 rows. The first column has no label with entries before 10 p m, after 10 p m, total. The second column i

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Answer:

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Step-by-step explanation:

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What is 3×10^-2 in standerd notation??

Gemiola [76]

Answer: 0.03

Step-by-step explanation:

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3 years ago

How do u round 33 and 89 to estimate their sum

Zolol [24]

33 would round down. (30)

This is because the second 3 is smaller than 5 and when your rounding, if number your rounding is smaller then 5 you will round down

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3 years ago

The maximum height of a vehicle that can safely pass under a bridge is 12 feet 5 inches. A truck measures 162 inches in height.

Bumek [7]

<h2>Answer:</h2>

<em><u>The truck cannot pass safely under the bridge. The truck is 13 inches taller than the maximum height.</u></em>

<h2>Step-by-step explanation:</h2>

In the question,

The maximum height of the vehicle which is capable of passing under the bridge is 12 feet and 5 inches.

So,

Now we know that,

1 feet = 12 inches

So,

12 feet = 12 x 12 = 144 inches

So,

Total height of the vehicle which is permissible to pass under the bridge is,

12 feet 5 inches = 144 + 5 = 149 inches

Also,

Height of the truck = 162 inches

Therefore, we can see that the permissible height is smaller than the height of the vehicle.

Height of vehicle which is more than permissible height is by,

162 - 149 = 13 inches

<em><u>Therefore, the truck cannot pass safely under the bridge. The truck is 13 inches taller than the maximum height.</u></em>

7 0

4 years ago

The sum of first three terms of a finite geometric series is -7/10 and their product is -1/125. [Hint: Use a/r, a, and ar to rep

Alchen [17]

Ooh, fun

geometric sequences can be represented as
$a_n=a(r)^{n-1}$
so the first 3 terms are
$a_1=a$
$a_2=a(r)$
$a_2=a(r)^2$

the sum is -7/10
$\frac{-7}{10}=a+ar+ar^2$
and their product is -1/125
$\frac{-1}{125}=(a)(ar)(ar^2)=a^3r^3=(ar)^3$

from the 2nd equation we can take the cube root of both sides to get
$\frac{-1}{5}=ar$
note that a=ar/r and ar²=(ar)r
so now rewrite 1st equation as
$\frac{-7}{10}=\frac{ar}{r}+ar+(ar)r$
subsituting -1/5 for ar
$\frac{-7}{10}=\frac{\frac{-1}{5}}{r}+\frac{-1}{5}+(\frac{-1}{5})r$
which simplifies to
$\frac{-7}{10}=\frac{-1}{5r}+\frac{-1}{5}+\frac{-r}{5}$
multiply both sides by 10r
-7r=-2-2r-2r²
add (2r²+2r+2) to both sides
2r²-5r+2=0
solve using quadratic formula
for $ax^2+bx+c=0$
$x=\frac{-b \pm \sqrt{b^2-4ac}}{2a}$
so
for 2r²-5r+2=0
a=2
b=-5
c=2

$r=\frac{-(-5) \pm \sqrt{(-5)^2-4(2)(2)}}{2(2)}$
$r=\frac{5 \pm \sqrt{25-16}}{4}$
$r=\frac{5 \pm \sqrt{9}}{4}$
$r=\frac{5 \pm 3}{4}$
so
$r=\frac{5+3}{4}=\frac{8}{4}=2$ or $r=\frac{5-3}{4}=\frac{2}{4}=\frac{1}{2}$

use them to solve for the value of a
$\frac{-1}{5}=ar$
$\frac{-1}{5r}=a$
try for r=2 and 1/2
$a=\frac{-1}{10}$ or $a=\frac{-2}{5}$

test each
for a=-1/10 and r=2
a+ar+ar²= $\frac{-1}{10}+\frac{-2}{10}+\frac{-4}{10}=\frac{-7}{10}$
it works

for a=-2/5 and r=1/2
a+ar+ar²= $\frac{-2}{5}+\frac{-1}{5}+\frac{-1}{10}=\frac{-7}{10}$
it works

both have the same terms but one is simplified

the 3 numbers are $\frac{-2}{5}$ , $\frac{-1}{5}$ , and $\frac{-1}{10}$

6 0

3 years ago