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3 years ago

Are all right triangles similar explain your answer

2 answers:

8 0

The triangles are similar because they are both right triangles. The triangles are not similar because they are not the same size. Explanation: ... Given that the angle between the two legs is a right angle in each triangle, these angles are congruent.

kolbaska11 [484]3 years ago

3 0

I think the answer is yes

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If it is about 41 miles around how many square miles does it cover

vodka [1.7K]

Answer:

133,7697

Step-by-step explanation:

$C=2*r*\pi 41 = 2*r*\pi r= 6,5253\\S=r^2*\pi = 133,7697$

8 0

3 years ago

Question is in the picture. Please help

irakobra [83]

BD = AC Given. BD cuts AC in half. That's what bisectors do.

BD = BD Reflexive property of a line (it is always equal to itself).

<ADB = <ADC = 90o

DB is perpendicular to AC Given.

<ADB + <ADC = 180o The two are supplementary.

Each angle = 90o Perpendicular means that one line meets another at 90o

2x = 180

x = 90o

Since each of the enclosed angles = 90, two have two triangles congruent by SAS <<<< Answer

4 0

3 years ago

Ratio of two complementary angles is 3:7. What's the measures of both angles

Advocard [28]

The total 'parts' of the ratio is 10 (3+7).
Complementary angles add to 90 degrees.
Divide 90 by 10 and you get 9.
Each 'part' is 9.
3:7 * 9 = 27:63
One angle is 27 degrees and the other is 63 degrees.

3 0

3 years ago

The solution to 4p+2<2(p+5) is

gizmo_the_mogwai [7]

Answer:

p<4

Step-by-step explanation:

4p+2< 2(p+5)

Distribute: 4p+2<2p+10

Subtract 2 from both sides: 4p<2p+8

Subtract 2p from both sides: 2p<8

Divide and we get: p<4

7 0

3 years ago

Find the work done by the force field F(x, y) = xi + (y + 5)j in moving an object along an arch of the cycloid r(t) = (t − sin(t

Nimfa-mama [501]

Integrate the force field along the given path (call it <em>C</em>):

$W=\displaystyle\int_C\mathbf F(x,y)\cdot\mathrm d\mathbf r=\int_0^{2\pi}\mathbf F(x(t),y(t))\cdot\frac{\mathrm d\mathbf r}{\mathrm dt}\,\mathrm dt$

$=\displaystyle\int_0^{2\pi}\bigg((t-\sin t)\,\mathbf i+(6-\cos t)\,\mathbf j\bigg)\cdot\bigg((1-\cos t)\,\mathbf i+\sin t\,\mathbf j\bigg)\,\mathrm dt$

$=\displaystyle\int_0^{2\pi}(t-t\cos t+5\sin t)\,\mathrm dt=\boxed{2\pi^2}$

5 0

3 years ago