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2 years ago

True or False? An equiangular triangle can never be scalene. A. True B. False

Mathematics

2 answers:

slavikrds [6]2 years ago

8 0

Its true because scalene doesn't have equal sides

Lubov Fominskaja [6]2 years ago

8 0

Answer:

this is true :)))))))))))))))

Step-by-step explanation:

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Quaaaaastion halp me

Yuki888 [10]

Answer:

the answer is red

Step-by-step explanation:

since the ques has asked to move 9 units to the the right , if you do so , the blue triangle wil be near the red one

7 0

2 years ago

It takes Bert 30 minutes longer to mow a rectangular lawn that measures 30 feet by 25 feet than it takes him to mow a rectangula

nevsk [136]

Answer:

C: 70 Mins

Step-by-step explanation:

1, 20ft*15ft=300ft^2

2, 30ft*25ft=750ft^2

3, 750ft-350ft=450ft^2

4, 450 ft^2 = 30 mins

5, 350ft=750ft=1050ft^2

6, 1050/450=2.3333

7, 30*2.3333=70

8, 70 mins

4 0

2 years ago

It costs Mrs. Smith $4.00 to make a pie. She sells them for $12 each. What is the Markup rate (percent)?

nignag [31]

70% I think beacuse she is making more profit about 70% then she bought it for

4 0

2 years ago

Read 2 more answers

Does this look good??????

VLD [36.1K]

Answer:

It should be RGB that's a L

Step-by-step explanation:

8 0

2 years 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

2 years ago