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2 years ago

11

Select the values that make the inequality h < 1 true.

1 answer:

kari74 [83]2 years ago

4 0

H = 0 and all negative numbers

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(7z^3+42z^2-15z+1)- (36z^2+12z-21)

My name is Ann [436]

Answer:

7z^3+6z^2-27z+22

Step-by-step explanation:

The given question is of subtraction:

So,

(7z^3+42z^2-15z+1)- (36z^2+12z-21)

=7z^3+42z^2-15z+1-36z^2-12z+21

Combining alike terms to simplify

=7z^3+42z^2-36z^2-15z-12z+1+21

=7z^3+6z^2-27z+22

So the answer in simplified form is:

7z^3+6z^2-27z+22 ..

6 0

2 years ago

Read 2 more answers

PLEASE HELP Meeeeeeeeeeeeeeeeee

klemol [59]

The answer is A.Use a compound interest calculator. Multiple sites to use. For me, I had to use my head for a while

3 0

3 years ago

X^2-21.75x=-15.75 what is one solution

Jlenok [28]

Well, there are two solutions.

Add both sides by 15.75 then from here, you can use quadratic formula.

So your answer should be x = -2.44949 or 2.44949.

5 0

3 years ago

Solve y ' ' + 4 y = 0 , y ( 0 ) = 2 , y ' ( 0 ) = 2 The resulting oscillation will have Amplitude: Period: If your solution is A

Vlad [161]

Answer:

$y(x)=sin(2x)+2cos(2x)$

Step-by-step explanation:

$y''+4y=0$

This is a homogeneous linear equation. So, assume a solution will be proportional to:

$e^{\lambda x} \\\\for\hspace{3}some\hspace{3}constant\hspace{3}\lambda$

Now, substitute $y(x)=e^{\lambda x}$ into the differential equation:

$\frac{d^2}{dx^2} (e^{\lambda x} ) +4e^{\lambda x} =0$

Using the characteristic equation:

$\lambda ^2 e^{\lambda x} + 4e^{\lambda x} =0$

Factor out $e^{\lambda x}$

$e^{\lambda x}(\lambda ^2 +4) =0$

Where:

$e^{\lambda x} \neq 0\\\\for\hspace{3}any\hspace{3}\lambda$

Therefore the zeros must come from the polynomial:

$\lambda^2+4 =0$

Solving for $\lambda$ :

$\lambda =\pm2i$

These roots give the next solutions:

$y_1(x)=c_1 e^{2ix} \\\\and\\\\y_2(x)=c_2 e^{-2ix}$

Where $c_1$ and $c_2$ are arbitrary constants. Now, the general solution is the sum of the previous solutions:

$y(x)=c_1 e^{2ix} +c_2 e^{-2ix}$

Using Euler's identity:

$e^{\alpha +i\beta} =e^{\alpha} cos(\beta)+ie^{\alpha} sin(\beta)$

$y(x)=c_1 (cos(2x)+isin(2x))+c_2(cos(2x)-isin(2x))\\\\Regroup\\\\y(x)=(c_1+c_2)cos(2x) +i(c_1-c_2)sin(2x)\\$

Redefine:

$i(c_1-c_2)=c_1\\\\c_1+c_2=c_2$

Since these are arbitrary constants

$y(x)=c_1sin(2x)+c_2cos(2x)$

Now, let's find its derivative in order to find $c_1$ and $c_2$

$y'(x)=2c_1 cos(2x)-2c_2sin(2x)$

Evaluating $y(0)=2$ :

$y(0)=2=c_1sin(0)+c_2cos(0)\\\\2=c_2$

Evaluating $y'(0)=2$ :

$y'(0)=2=2c_1cos(0)-2c_2sin(0)\\\\2=2c_1\\\\c_1=1$

Finally, the solution is given by:

$y(x)=sin(2x)+2cos(2x)$

5 0

3 years ago

I need help pleaseeeeee

-BARSIC- [3]

Answer:

30

Step-by-step explanation:

total area of triangle is 180

so if 97+53= 150 180-150= 30

hope this helped

plz mark brainlest

i wanna rank up!!!

6 0

3 years ago