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Digiron [165]
3 years ago
10

The segments shown below could form a triangle 9,4,15 true or false

Mathematics
1 answer:
yawa3891 [41]3 years ago
3 0
No - there is a rule that state that the sum of  two sides of a triangle must be greater than the 3rd side.
While 9+15 >4
and 4+15 >9
9+4 is not greater than 15 - so this triangle cannot be made.
You might be interested in
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
How do I find the values of these
skad [1K]
Since the triangles are similar, then 32 = x+7 those two angles are too
solve for "x"

now using proportions   \bf \cfrac{\textit{small triangle}}{\textit{large triangle}}\qquad \cfrac{JK}{18}=\cfrac{12}{12+10}

solve for JK

\bf \cfrac{\textit{small triangle}}{\textit{large triangle}}\qquad \cfrac{12}{12+10}=\cfrac{9}{9+KG}\implies 12(9+KG)=(22)9
\\\\\\
108+9KG=198

solve for KG

the scale factor of the perimeter of both triangles is just the same as the ratio of the sides, or \bf \cfrac{12}{12+10}

3 0
3 years ago
Enter an equation in point-slope form for the line.
MissTica

Answer:

y+1=4(x+2)

Step-by-step explanation:

Hi there!

Point-slope form: y-y_1=m(x-x_1) where m is the slope and (x₁,y₁) is the given point

y-y_1=m(x-x_1)

Plug in the slope and the point

y-(-1)=4(x-(-2))\\y+1=4(x+2)

I hope this helps!

6 0
3 years ago
Given mZ1= x° and m Z2 = (x +50)".
Tju [1.3M]

Given:

m∠1= x° and m∠2 = (x +50)°.

To find:

The value of x, m∠1 and m∠2.

Solution:

From the figure it is clear that, transversal side intersect the opposite sides which are parallel and form ∠1 and ∠2.

Sum of same sided interior angles is 180 degrees. So,

m\angle 1+m\angle 2=180^\circ

x^\circ+(x+50)^\circ=180^\circ

2x^\circ+50^\circ=180^\circ

2x^\circ=180^\circ-50^\circ

2x^\circ=130^\circ

Divide both sides by 2.

x^\circ=65^\circ

x=65

Now,

m\angle 1=65^\circ

m\angle 2=(65+50)^\circ=115^\circ

Therefore, the correct option is C.

5 0
3 years ago
What is 3.4 × 10^-6 in standered notation
Schach [20]
0.0000034 hope this helps
3 0
3 years ago
Read 2 more answers
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