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3 years ago

Help fast please

Mathematics

2 answers:

Novosadov [1.4K]3 years ago

6 0

Answer:

D) Acute and Obtuse angles

Step-by-step explanation:

The pink quadrilateral has obtuse angles and I believe it has 2 acute angles.

The orange one has 2 acute and 2 obtuse angles

And the green one has both acute and obtuse angles.

Lady_Fox [76]3 years ago

3 0

Answer:

Acute and Obtuse..I cant see any right

Step-by-step explanation:

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sweet-ann [11.9K]

Answer:

90°

Step-by-step explanation:

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8 0

3 years ago

Please solve this WILL GIVE BRAINLIEST AND 50+ points

nata0808 [166]

105.9965:700 is what I worked out. Hope it's right.

7 0

3 years ago

Read 2 more answers

Please help asap

Shkiper50 [21]

I am not to sure but I think that it is 111.804 inches^3 I hope I am right if not I am really sorry.

6 0

3 years ago

Read 2 more answers

Sketch the graph by hand using asymptotes and intercepts, but not derivatives. Then use your sketch as a guide to producing grap

Arisa [49]

The local minima of $f\left(x\right)=\ \frac{\left(2x+3\right)^2\left(x\ -2\right)^5}{x^3\left(x-5\right)^2}$ are (x, f(x)) = (-1.5, 0) and (7.980, 609.174)

<h3>How to determine the local minima?</h3>

The function is given as:

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

See attachment for the graph of the function f(x)

From the attached graph, we have the following minima:

Minimum = (-1.5, 0)

Minimum = (7.980, 609.174)

The above means that, the local minima are

(x, f(x)) = (-1.5, 0) and (7.980, 609.174)

Read more about graphs at:

brainly.com/question/20394217

#SPJ1

7 0

1 year ago

Y''+y'+y=0, y(0)=1, y'(0)=0

mars1129 [50]

Answer:

$y=e^{\frac{-t}{2}}\left ( \cos\left ( \frac{\sqrt{3}t}{2} \right )+\frac{1}{\sqrt{3}}\sin \left ( \frac{\sqrt{3}t}{2} \right ) \right )$

Step-by-step explanation:

A second order linear , homogeneous ordinary differential equation has form $ay''+by'+cy=0$ .

Given: $y''+y'+y=0$

Let $y=e^{rt}$ be it's solution.

We get,

$\left ( r^2+r+1 \right )e^{rt}=0$

Since $e^{rt}\neq 0$ , $r^2+r+1=0$

{ we know that for equation $ax^2+bx+c=0$ , roots are of form $x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$ }

We get,

$y=\frac{-1\pm \sqrt{1^2-4}}{2}=\frac{-1\pm \sqrt{3}i}{2}$

For two complex roots $r_1=\alpha +i\beta \,,\,r_2=\alpha -i\beta$ , the general solution is of form $y=e^{\alpha t}\left ( c_1\cos \beta t+c_2\sin \beta t \right )$

i.e $y=e^{\frac{-t}{2}}\left ( c_1\cos\left ( \frac{\sqrt{3}t}{2} \right )+c_2\sin \left ( \frac{\sqrt{3}t}{2} \right ) \right )$

Applying conditions y(0)=1 on $e^{\frac{-t}{2}}\left ( c_1\cos\left ( \frac{\sqrt{3}t}{2} \right )+c_2\sin \left ( \frac{\sqrt{3}t}{2} \right ) \right )$ , $c_1=1$

So, equation becomes $y=e^{\frac{-t}{2}}\left ( \cos\left ( \frac{\sqrt{3}t}{2} \right )+c_2\sin \left ( \frac{\sqrt{3}t}{2} \right ) \right )$

On differentiating with respect to t, we get

$y'=\frac{-1}{2}e^{\frac{-t}{2}}\left ( \cos\left ( \frac{\sqrt{3}t}{2} \right )+c_2\sin \left ( \frac{\sqrt{3}t}{2} \right ) \right )+e^{\frac{-t}{2}}\left ( \frac{-\sqrt{3}}{2} \sin \left ( \frac{\sqrt{3}t}{2} \right )+c_2\frac{\sqrt{3}}{2}\cos\left ( \frac{\sqrt{3}t}{2} \right )\right )$

Applying condition: y'(0)=0, we get $0=\frac{-1}{2}+\frac{\sqrt{3}}{2}c_2\Rightarrow c_2=\frac{1}{\sqrt{3}}$

Therefore,

$y=e^{\frac{-t}{2}}\left ( \cos\left ( \frac{\sqrt{3}t}{2} \right )+\frac{1}{\sqrt{3}}\sin \left ( \frac{\sqrt{3}t}{2} \right ) \right )$

3 0

3 years ago