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Marta_Voda [28]

3 years ago

13

Linear equation 2x + y = 3?

1 answer:

mote1985 [20]3 years ago

6 0

Answer:

Yes,Linear

Step-by-step explanation:

Find the degree of the equation to determine if linear.

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Scott wants to tip his taxi driver 20%. if his commute costs $15 what is the total cost

Vlad [161]

The total cost is $18

8 0

4 years ago

What is 3.41 ((with 41 repeating at the end)) as a simplified fraction?

Andre45 [30]

<u>3 41/99. </u>

3.4141414...

100x= 341.4141414...

341.414141414...

- 3.41414141414..= 338

99x= 338/99 or 3 44/99

5 0

3 years ago

Using long division to find the quotient, what changes do you first

lora16 [44]

Answer:

$\frac{125x^3 - 8}{5x - 2} = 25x^2 + 10x +4$

Step-by-step explanation:

Given

$Dividend = 125x^3 - 8$

$Divisor = 5x - 2$

Required

Determine the quotient

See attachment for complete process.

First, divide 125x^3 by 5x

$\frac{125x^3}{5x} =25x^2$

Write $25x^2$ at the top

Multiply $5x - 2$ by $25x^2$

$= 125x^3 - 50x^2$

Subtract from 125x^3 - 8

i.e.

$125x^3 - 8 - (125x^3 - 50x^2) = 50x^2 - 8$

Step 2:

Divide 50x^2 by 5x

$\frac{50x^2}{5x} = 10x$

Write $10x$ at the top

Multiply $5x - 2$ by $10x$

$= 50x^2 - 20x$

Subtract from 50x^2 - 8

i.e.

$50x^2 - 8 - (50x^2 - 20x) = 20x - 8$

Step 3:

Divide 20x by 5x

$\frac{20x}{5x} = 4$

Write $4$ at the top

Multiply $5x - 2$ by $4$

$= 20x - 8$

Subtract from 20x - 8

i.e.

$20x - 8 - (20x - 8) = 0$

Hence:

$\frac{125x^3 - 8}{5x - 2} = 25x^2 + 10x +4$

8 0

3 years ago

Which of the following represents the zeros of f(x) = 2x3 − 5x2 − 28x + 15?

likoan [24]

$\mathrm{Use\:the\:rational\:root\:theorem}$

$a_0=15,\:\quad a_n=2$

$\mathrm{The\:dividers\:of\:}a_0:\quad 1,\:3,\:5,\:15,\:\quad \mathrm{The\:dividers\:of\:}a_n:\quad 1,\:2$

$\mathrm{Therefore,\:check\:the\:following\:rational\:numbers:\quad }\pm \frac{1,\:3,\:5,\:15}{1,\:2}$

$-\frac{3}{1}\mathrm{\:is\:a\:root\:of\:the\:expression,\:so\:factor\:out\:}x+3$

$-\frac{3}{1}\mathrm{\:is\:a\:root\:of\:the\:expression,\:so\:factor\:out\:}x+3$

$\mathrm{Compute\:}\frac{2x^3-5x^2-28x+15}{x+3}\mathrm{\:to\:get\:the\:rest\:of\:the\:eqution:\quad }2x^2-11x+5$

$=\left(x+3\right)\left(2x^2-11x+5\right)$

$Factor: 2x^2-11x+5$

$2x^2-11x+5=\left(2x^2-x\right)+\left(-10x+5\right)$

$=x\left(2x-1\right)-5\left(2x-1\right)$

$2x^3-5x^2-28x+15=\left(x+3\right)\left(x-5\right)\left(2x-1\right)$

$\left(x+3\right)\left(x-5\right)\left(2x-1\right)=0$

$\mathrm{Using\:the\:Zero\:Factor\:Principle:}$

thus zeros of f(x) is

$x=-3,\:x=5,\:x=\frac{1}{2}$

5 0

3 years ago

Read 2 more answers

Choose the equation that represents the line that passes through the point (2, 6) and has a slope of −5.

Luden [163]

Answer:

y=-5x+16

Step-by-step explanation:

Plug in the slope and point coordinates into point-slope form (attached in image)

(y-6)=-5(x-2)

1) Distribute -5 to x and -2:

y-6=-5x+10

2) Add 6 to both sides:

y=-5x+16

4 0

3 years ago

Read 2 more answers