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3 years ago

14

Explain how to solve this equation. -4n - Sn + 10 = 34

1 answer:

kicyunya [14]3 years ago

8 0

There go the answer

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No explanation required, just a quick answer please!

hram777 [196]

The answer to the question

4 0

3 years ago

How do i cross multiply rational equations? EX: 6/x = 4/3

aliya0001 [1]

You would multiply 6 by 3 and x by 4. You would end up getting 4x = 18 and then you simplify that to get 9/2 or 4.5.

3 0

3 years ago

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Devon made a box with length x+1, width x+3, and height x-3. What is the volume of Devon's box as a function of x?

erica [24]

Answer:

(a)

$V=x^3+x^2-9x-9$

(b)

$x=10$

(c)

$x=\frac{7}{2}$

Step-by-step explanation:

we are given

length is x+1

$L=x+1$

width is x+3

$W=x+3$

height is x-3

$H=x-3$

(a)

we can use volume formula

$V=L\times W\times H$

now, we can plug it

$V=(x+1)\times (x+3)\times (x-3)$

now, we can simplify it

$V=x^3+x^2-9x-9$

(b)

we are given volume =1001

so, we can set V=1001

and then we can solve for x

$V=x^3+x^2-9x-9=1001$

now, we can factor it

$\left(x-10\right)\left(x^2+11x+101\right)=0$

$x-10=0$

$x=10$

(b)

we are given volume

so, we can set

$V=14\frac{5}{8}=\frac{14\times 8+5}{8}$

$V=\frac{117}{8}$

and then we can solve for x

$V=x^3+x^2-9x-9=\frac{117}{8}$

now, we can factor it

$8x^3+8x^2-72x-189=0$

$\left(2x-7\right)\left(4x^2+18x+27\right)=0$

$2x-7=0$

$x=\frac{7}{2}$

8 0

3 years ago

Students in Mr. Clark's class are reading a book that has 180 pages.

Aleonysh [2.5K]

Answer:

Step-by-step explanation:

12 pages I think

4 0

3 years ago

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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