All categories

3 years ago

Please help me solve them

Mathematics

2 answers:

deff fn [24]3 years ago

8 0

Ok so the first thing to do would be get rid of the fraction. To do that multiply the ENTIRE equation but the denominator. Once all the numbers are multiplyed I think it’ll be easier for u to solve them :) or if u wanna solve them quick, download Photomath :)

gavmur [86]3 years ago

6 0

1) x= 22

2) x= -2

3) x= -5

4) x= 7

You might be interested in

Consider the differential equationy′′+3y′−10y=0.(a) Find the general solution to this differential equation.(b) Now solve the in

RSB [31]

Answer:

Y= 2e^(5t)

Step-by-step explanation:

Taking Laplace of the given differential equation:

s^2+3s-10=0

s^2+5s-2s-10=0

s(s+5)-2(s+5) =0

(s-2) (s+5) =0

s=2, s=-5

Hence, the general solution will be:

Y=Ae^(-2t)+ Be^(5t)………………………………(D)

Put t = 0 in equation (D)

Y (0) =A+B

2 =A+B……………………………………… (i)

Now take derivative of (D) with respect to "t", we get:

Y=-2Ae^(-2t)+5Be^(5t) ....................... (E)

Put t = 0 in equation (E) we get:

Y’ (0) = -2A+5B

10 = -2A+5B ……………………………………(ii)

2(i) + (ii) =>

2A+2B=4 .....................(iii)

-2A+5B=10 .................(iv)

Solving (iii) and (iv)

7B=14

B=2

Now put B=2 in (i)

A=2-2

A=0

By putting the values of A and B in equation (D)

Y= 2e^(5t)

6 0

3 years ago

No Spam. I will mark Brainliest, if you get it right. A tank's length is 4 feet longer than its width. The height is 2 feet more

dmitriy555 [2]

Answer: 3

Step-by-step explanation:

Let say x the width of the tank,

x+4 its length,

x+2 its height

$Volume=105=x(x+4)(x+2)\\\\P(x)=x^3+6x^2+8x-105=0\\\\105=3*5*7\\\\P(1)\neq 0\\P(-1)\neq 0\\P(3)=0\\\\\begin{array}{|c|cccc|}&x^3&x^2&x&1\\&1&6&8&-105\\x=3&&3&27&105\\----&--&--&--&--\\&1&9&35&0\\\end{array}\\\\\\P(x)=x^3+6x^2+8x-105=(x-3)(x^2+9x+35)\\And \ x^2+9x+35\ is\ not \ factorizable\ (\Delat=9^2-4*35 < 0)\\$

<u>width=3</u>

length=3+4=7

height=3+2=5

4 0

2 years ago

Ex 3.6<br> 6. find the area enclosed between the curve y= -2x²-5x+3 and the x-axis

Xelga [282]

When y=0,

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

$\\ \\ \therefore \quad x=\frac { 1 }{ 2 } \\ \\ \therefore \quad x=-3$

--------------------

$\int _{ -3 }^{ \frac { 1 }{ 2 } }{ -2{ x }^{ 2 } } -5x+3dx$

$\\ \\ ={ \left[ -\frac { { 2x }^{ 2+1 } }{ 2+1 } -\frac { 5{ x }^{ 1+1 } }{ 1+1 } +3x \right] }_{ -3 }^{ \frac { 1 }{ 2 } }$

$\\ \\ ={ \left[ -\frac { 2{ x }^{ 3 } }{ 3 } -\frac { 5{ x }^{ 2 } }{ 2 } +3x \right] }_{ -3 }^{ \frac { 1 }{ 2 } }$

$\\ \\ \\ =\left\{ -\frac { 2 }{ 3 } { \left( \frac { 1 }{ 2 } \right) }^{ 3 }-\frac { 5 }{ 2 } { \left( \frac { 1 }{ 2 } \right) }^{ 2 }+3\left( \frac { 1 }{ 2 } \right) \right\} -\left\{ -\frac { 2 }{ 3 } { \left( -3 \right) }^{ 3 }-\frac { 5 }{ 2 } { \left( -3 \right) }^{ 2 }+3\left( -3 \right) \right\}$

$\\ \\ \\ =-\frac { 2 }{ 3 } \cdot \frac { 1 }{ 8 } -\frac { 5 }{ 2 } \cdot \frac { 1 }{ 4 } +\frac { 3 }{ 2 } -\left\{ -\frac { 2 }{ 3 } \left( -27 \right) -\frac { 5 }{ 2 } \cdot 9-9 \right\}$

$\\ \\ =-\frac { 2 }{ 24 } -\frac { 5 }{ 8 } +\frac { 3 }{ 2 } -\left\{ \frac { 54 }{ 3 } -\frac { 45 }{ 2 } -9 \right\}$

$\\ \\ =-\frac { 2 }{ 24 } -\frac { 15 }{ 24 } +\frac { 36 }{ 24 } -\frac { 54 }{ 3 } +\frac { 45 }{ 2 } +9$

$\\ \\ =\frac { 19 }{ 24 } -\frac { 54 }{ 3 } +\frac { 45 }{ 2 } +\frac { 18 }{ 2 } \\ \\ =\frac { 19 }{ 24 } -\frac { 54 }{ 3 } +\frac { 63 }{ 2 }$

$\\ \\ =\frac { 343 }{ 24 }$

Answer: 343/24 units squared.

6 0

3 years ago

Read 2 more answers

What x values would be good to use so she does not get fractions for her y values when making the table of ordered pairs.

Ber [7]

Answer:

-3, 0, 3

Step-by-step explanation:

Given, $y = -\frac{4}{3}x + 2$ , the set of x values that will make Kayda not to get y values that are not fraction are -3, 0, 3.