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2 years ago

NEED HELP FOR A TEST, PLEASE :) (30 POINTS)

Mathematics

1 answer:

Nadya [2.5K]2 years ago

6 0

Answer:

Hey!

Your Answer Is,

No Triangles: A, C

One Triangle, F, D

Many Triangles: B,E

I hope this helps you!

Step-by-step explanation:

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-5(x+7)<10. What does x equal?

gulaghasi [49]

Answer:

X>-9

Step-by-step explanation:

Distribute

-5x-35<10

add 35 on both sides

-5x<45

Divide -5 on both sides and flip the sign since your dividing by a NEGATIVE number

x>-9

4 0

4 years ago

Read 2 more answers

Gabby is buying flowers for a friend. She can choose daisies, carnations, or sunflowers. She can put the flowers in a clear, red

In-s [12.5K]

Answer:

Daisies: Clear vase

Carnations: Red vase

Sunflowers: Blue vase

Step-by-step explanation:

Your using all the objects and is is in all the combinations.

8 0

3 years ago

Read 2 more answers

PLEASE HELP ASAP!! WORDED QUADRATIC QUESTIONS!!!

Roman55 [17]

Answer:

x² - 5x = 14

x = -2 or x = 7

Step-by-step explanation:

x² - 5x = 14

x² - 5x - 14 = 0

(x+2)(x-7) = 0

x = -2 or x = 7

4 0

3 years ago

I NEED HELP ASAP!! Subtract. -30-27= ? -5-(-4)= ?

pashok25 [27]

-30-27= -57
When both numbers are negative, they are “added” in a sense, but remain negative.

-5-(-4)= -1
You have to multiply the (-4) by the “invisible” -1 on the outside before you add/subtract.

5 0

4 years ago

1) Use power series to find the series solution to the differential equation y'+2y = 0 PLEASE SHOW ALL YOUR WORK, OR RISK LOSING

iogann1982 [59]

$y=\displaystyle\sum_{n=0}^\infty a_nx^n$

then

$y'=\displaystyle\sum_{n=1}^\infty na_nx^{n-1}=\sum_{n=0}^\infty(n+1)a_{n+1}x^n$

The ODE in terms of these series is

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

$\displaystyle\sum_{n=0}^\infty\bigg(a_{n+1}+2a_n\bigg)x^n=0$

$\implies\begin{cases}a_0=y(0)\\(n+1)a_{n+1}=-2a_n&\text{for }n\ge0\end{cases}$

We can solve the recurrence exactly by substitution:

$a_{n+1}=-\dfrac2{n+1}a_n=\dfrac{2^2}{(n+1)n}a_{n-1}=-\dfrac{2^3}{(n+1)n(n-1)}a_{n-2}=\cdots=\dfrac{(-2)^{n+1}}{(n+1)!}a_0$

$\implies a_n=\dfrac{(-2)^n}{n!}a_0$

So the ODE has solution

$y(x)=\displaystyle a_0\sum_{n=0}^\infty\frac{(-2x)^n}{n!}$

which you may recognize as the power series of the exponential function. Then

$\boxed{y(x)=a_0e^{-2x}}$

7 0

3 years ago