Multiply. write the product in lowest terms. (2/3) (- 5/7)

If you subtract 12<br> from my number and<br> multiply the difference by -3,<br> the result is -54.

worty [1.4K]

Answer:

the answer is 30

Step-by-step explanation:

Let the number be x

- 3 (x - 12) = - 54

-3x + 36 = -54

-3x = -54 - 36

-3x = -90

x = -90/-3

x = 30

4 0

3 years ago

Find the slope and y intercept <br> (Please show work)

zepelin [54]

Answer:

slope = - 6, y- intercept = 5

Step-by-step explanation:

The equation of a line in slope- intercept form is

y = mx + c ( m is the slope and c the y- intercept )

Given

0 = 2x - $\frac{5}{3}$ + $\frac{1}{3}$ y ( multiply through by 3 to clear the fractions )

0 = 6x - 5 + y ( subtract 6x - 5 from both sides )

- 6x + 5 = y , that is

y = - 6x + 5 ← in slope- intercept form

with slope m = - 6 and y- intercept c = 5

4 0

3 years ago

What is the slope of a line that is perpendicular to the line whose equation is 3x+2y=6?

LekaFEV [45]

first, we can find the slope from the equation that is given buy solving the equation for y

3x+2y = 6

2y = 6-3x

y = 3-3/2x

y = -3/2x+3

now that the equation is in slope-intercept form, we can easily see that the slope of the given line is -3/2

perpendicular lines have slopes that are negative reciprocals, so we can just take the negative reciprocal of the slope we have

-3/2 → 3/2 → 2/3

the slope of the perpendicular line is 2/3

hope this helped

7 0

3 years ago

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A flagpole at a right angle to the horizontal is located on a slope that makes an angle of 15° with the horizontal. The flagpole

Veseljchak [2.6K]

Answer is 5 its because ummmm uhhhhhhh hahaha i really dont know how to explain it but the answer is 5

3 0

2 years ago

What is the solution for the equation StartFraction 5 Over 3 b cubed minus 2 b squared minus 5 EndFraction = StartFraction 2 Ove

wolverine [178]

Answer:

The solutions are:

$b=0,\:b=4$

Step-by-step explanation:

Considering the expression

$\frac{5}{3b^3-2b^2-5}=\frac{2}{b^3-2}$

Solving the expression

$\frac{5}{3b^3-2b^2-5}=\frac{2}{b^3-2}$

$\mathrm{Apply\:fraction\:cross\:multiply:\:if\:}\frac{a}{b}=\frac{c}{d}\mathrm{\:then\:}a\cdot \:d=b\cdot \:c$

$5\left(b^3-2\right)=\left(3b^3-2b^2-5\right)\cdot \:2$

$5b^3-10=6b^3-4b^2-10$

$\mathrm{Switch\:sides}$

$6b^3-4b^2-10=5b^3-10$

$6b^3-4b^2-10+10=5b^3-10+10$

$6b^3-4b^2=5b^3$

$\mathrm{Subtract\:}5b^3\mathrm{\:from\:both\:sides}$

$6b^3-4b^2-5b^3=5b^3-5b^3$

$b^3-4b^2=0$

$Using\:the\:Zero\:Factor\:Principle:$ $if\:\mathrm ab=0\:\mathrm{then}\:a=0\:\mathrm{or}\:b=0\:\left(\mathrm{or\:both}\:a=0\:\mathrm{and}\:b=0\right)$

So,

$b=0,b-4=0$

$b=0,b=4$

Therefore, the solutions are:

$b=0,\:b=4$

4 0

3 years ago

Read 2 more answers