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3 years ago

Show your work - 1=3r+2r

Mathematics

2 answers:

hram777 [196]3 years ago

6 0

Answer:

r=-1/5

Step-by-step explanation:

erastova [34]3 years ago

4 0

Answer:

-1/5

Step-by-step explanation:

-1=3r+2r

-1=5r

-1/5=5r/5

-1/5=r

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The table represents the function f (x) = 3x-1

Ann [662]

Answer:

a=-16, b=11, c=23

Step-by-step explanation:

The function whose table is drawn is

$f(x) = 3x - 1$

To find the value of a, we substitute x=-5.

$f( - 5) = 3( - 5) - 1 \\ f( - 5) = - 15 - 1 \\ f( - 5) = - 16$

$\therefore \: a = - 16$

To find the value of b, we substitute x=4.

$f(4) =3(4) - 1 \\ f(4) = 12 - 1 \\ f(4) = 11$

To find the value of c, we substitute x=8,

$f(8) = 3(8) - 1 \\ f(8) = 24 - 1 \\ f(8) = 23$

$c = 23$

8 0

3 years ago

PLS HELP ASAP THANKS ILL GIVE BRAINLKEST PLS THANKS

Jlenok [28]

Answer:

its B

Step-by-step explanation:

6 0

2 years ago

Read 2 more answers

How do i do this?!?! its due tomorrow helpppppp

3241004551 [841]

Consider for each point how far it is above (or below) the horizontal line y=2. Then plot its image the same distance below (or above) that line.

6 0

2 years ago

Please please please help and solve this with steps, much help needed thank you :) 20 points for this!

leonid [27]

Answer:

$y=\sqrt{\frac{x^4-6x^2+12x+c_1}{2}},\:y=-\sqrt{\frac{x^4-6x^2+12x+c_1}{2}}$

(Please vote me Brainliest if this helped!)

Step-by-step explanation:

$\frac{dy}{dx}y=x^3+2x-5x+3$

$\mathrm{First\:order\:separable\:Ordinary\:Differential\:Equation}$

$\mathrm{A\:first\:order\:separable\:ODE\:has\:the\:form\:of}\:N\left(y\right)\cdot y'=M\left(x\right)$

1. $\mathrm{Substitute\quad }\frac{dy}{dx}\mathrm{\:with\:}y'\:$

$y'\:y=x^3+2x-5x+3$

2. $\mathrm{Rewrite\:in\:the\:form\:of\:a\:first\:order\:separable\:ODE}$

$yy'\:=x^3-3x+3$

$N\left(y\right)\cdot y'\:=M\left(x\right)$

$N\left(y\right)=y,\:\quad M\left(x\right)=x^3-3x+3$

3. $\mathrm{Solve\:}\:yy'\:=x^3-3x+3:\quad \frac{y^2}{2}=\frac{x^4}{4}-\frac{3x^2}{2}+3x+c_1$

4. $\mathrm{Isolate}\:y:\quad y=\sqrt{\frac{x^4-6x^2+12x+4c_1}{2}},\:y=-\sqrt{\frac{x^4-6x^2+12x+4c_1}{2}}$

5. $\mathrm{Simplify}$

$y=\sqrt{\frac{x^4-6x^2+12x+c_1}{2}},\:y=-\sqrt{\frac{x^4-6x^2+12x+c_1}{2}}$

7 0

3 years ago

Read 2 more answers

The area of a rectangle is 27ft squared, and the length of the rectangle is 3ft less than twice the width. Find the dimensions o

oksano4ka [1.4K]

Answer:

width =4.5ft

length=6ft

Step-by-step explanation:

w = x

L = 2x-3

area = L*w

27 = x(2x-3)

$27=2x^{2} -3x$

$2x^{2} -3x-27=0$

$2x^{2} +6x-9x-27=0$

2x(x+3)-9(x+3)=0

(2x-9)(x+3)=0

2x-9=0

2x=9

x=9/2

x=4.5

L=2x-3=2*4.5-3 =9-3=6

6 0

3 years ago