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3 years ago

What id the height of the triangle?

Mathematics

2 answers:

Triss [41]3 years ago

5 0

Answer:

6.5 cm

Step-by-step explanation:

Serhud [2]3 years ago

4 0

The answer to this question is 6.5

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NEED IT NOW

DaniilM [7]

Answer:

Mia is correct.

Step-by-step explanation:

You can see this if you write 5/11 in a calculator, you get 0.454545454545 infinitely. In other cases, you would write it like Malik said if the answer were to be 0.455555555, but it isn't.

6 0

3 years ago

Read 2 more answers

HELP ME PLEASE PLEASE IM BEGGING

suter [353]

Answer:

FALSE, (2, 9) is not a solution to the set of inequalities given.

Step-by-step explanation:

Simply replace x by 2 and y by 9 in the inequalities and see if the inequality is true or not:

irst inequality:

$y\geq 4x\\9\geq 4\,(2)\\9\geq 8$

so thi inequality is verified as true since 9 is larger or equal than 8

Now the second inequality:

$y$

This is FALSE since 9 is larger than 4 (not smaller)

Therefore the answer to the question is FALSE, (2, 9) is not a solution to the set of inequalities given.

6 0

3 years ago

Solutions for 2x^2+18x=20, worth a TON, please only 100% answers ! Thank u !

victus00 [196]

Answer:

The solution of the given quadratic equation is 1 and (-10).

Step-by-step explanation:

$2x^2+18x=20$

$x^2+9x=\frac{20}{2}$

$x^2+9x-10=0$

$ax^2+bx+c=0$

Solving equation by quadratic formula:

Here , a = 1,b = 9, c = (-10)

$D=b^2-4ac=(9)^2-4(1)(-10)=121$

$x=\frac{-b+{\sqrt{D}}}{2a}=\frac{(-9)+\sqrt{121}}{2(1)}=1$

$x=\frac{-b-{\sqrt{D}}}{2a}=\frac{(-9)-\sqrt{121}}{2(1)}=(-10)$

The solution of the given quadratic equation is 1 and (-10).

5 0

3 years ago

564000 in scientific notation

Mkey [24]

Answer:

5.64 × 10^5

Step-by-step explanation:

8 0

3 years ago

Find two power series solutions of the given differential equation about the ordinary point x = 0. compare the series solutions

monitta

I don't know what method is referred to in "section 4.3", but I'll suppose it's reduction of order and use that to find the exact solution. Take $z=y'$ , so that $z'=y''$ and we're left with the ODE linear in $z$ :

$y''-y'=0\implies z'-z=0\implies z=C_1e^x\implies y=C_1e^x+C_2$

Now suppose $y$ has a power series expansion

$y=\displaystyle\sum_{n\ge0}a_nx^n$
$\implies y'=\displaystyle\sum_{n\ge1}na_nx^{n-1}$
$\implies y''=\displaystyle\sum_{n\ge2}n(n-1)a_nx^{n-2}$

Then the ODE can be written as

$\displaystyle\sum_{n\ge2}n(n-1)a_nx^{n-2}-\sum_{n\ge1}na_nx^{n-1}=0$

$\displaystyle\sum_{n\ge2}n(n-1)a_nx^{n-2}-\sum_{n\ge2}(n-1)a_{n-1}x^{n-2}=0$

$\displaystyle\sum_{n\ge2}\bigg[n(n-1)a_n-(n-1)a_{n-1}\bigg]x^{n-2}=0$

All the coefficients of the series vanish, and setting $x=0$ in the power series forms for $y$ and $y'$ tell us that $y(0)=a_0$ and $y'(0)=a_1$ , so we get the recurrence

$\begin{cases}a_0=a_0\\\\a_1=a_1\\\\a_n=\dfrac{a_{n-1}}n&\text{for }n\ge2\end{cases}$

We can solve explicitly for $a_n$ quite easily:

$a_n=\dfrac{a_{n-1}}n\implies a_{n-1}=\dfrac{a_{n-2}}{n-1}\implies a_n=\dfrac{a_{n-2}}{n(n-1)}$

and so on. Continuing in this way we end up with

$a_n=\dfrac{a_1}{n!}$

so that the solution to the ODE is

$y(x)=\displaystyle\sum_{n\ge0}\dfrac{a_1}{n!}x^n=a_1+a_1x+\dfrac{a_1}2x^2+\cdots=a_1e^x$

We also require the solution to satisfy $y(0)=a_0$ , which we can do easily by adding and subtracting a constant as needed:

$y(x)=a_0-a_1+a_1+\displaystyle\sum_{n\ge1}\dfrac{a_1}{n!}x^n=\underbrace{a_0-a_1}_{C_2}+\underbrace{a_1}_{C_1}\displaystyle\sum_{n\ge0}\frac{x^n}{n!}$

4 0

3 years ago

What id the height of the triangle?​

What id the height of the triangle?