Enter your responses in the blank.

How many triangles can be formed with segments measuring 4. 7 m, 1. 6 m, and 2. 9 m? none one more than one.

Svet_ta [14]

The sum of the length of the two sides of the triangle must be greater than the length of the third side. None of the triangles is formed. Then option A is correct.

<h3>What is the triangle?</h3>

Triangle is a polygon that has three sides and three angles. The sum of the angle of the triangle is 180 degrees.

The sides of the triangle are 4.7 m, 1.6 m, and 2.9 m.

The condition will be

$1.6 + 2.9 \ngtr 4.7$

We know that the sum of the length of two sides of the triangle must be greater than the length of the third side.

Hence, none of the triangles is formed.

Thus, option A is correct.

More about the triangle link is given below.

brainly.com/question/25813512

4 0

2 years ago

WORTH 30 POINTS! Replace ... with <, ≤, =, >, or ≥ so that the inequality will be true for any value of x.

Len [333]

Hope this helps a bit

8 0

2 years ago

Read 2 more answers

Which point on the number line below best represent 30

san4es73 [151]

Where’s the number line?

5 0

3 years ago

Read 2 more answers

Find e^cos(2+3i) as a complex number expressed in Cartesian form.

ozzi

Answer:

The complex number $e^{\cos(2+31)} = \exp(\cos(2+3i))$ has Cartesian form

$\exp\left(\cosh 3\cos 2\right)\cos(\sinh 3\sin 2)-i\exp\left(\cosh 3\cos 2\right)\sin(\sinh 3\sin 2)$ .

Step-by-step explanation:

First, we need to recall the definition of $\cos z$ when $z$ is a complex number:

$\cos z = \cos(x+iy) = \frac{e^{iz}+e^{-iz}}{2}$ .

Then,

$\cos(2+3i) = \frac{e^{i(2+31)} + e^{-i(2+31)}}{2} = \frac{e^{2i-3}+e^{-2i+3}}{2}$ . (I)

Now, recall the definition of the complex exponential:

$e^{z}=e^{x+iy} = e^x(\cos y +i\sin y)$ .

So,

$e^{2i-3} = e^{-3}(\cos 2+i\sin 2)$

$e^{-2i+3} = e^{3}(\cos 2-i\sin 2)$ (we use that $\sin(-y)=-\sin y)$ .

Thus,

$e^{2i-3}+e^{-2i+3} = e^{-3}\cos 2+ie^{-3}\sin 2 + e^{3}\cos 2-ie^{3}\sin 2)$

Now we group conveniently in the above expression:

$e^{2i-3}+e^{-2i+3} = (e^{-3}+e^{3})\cos 2 + i(e^{-3}-e^{3})\sin 2$ .

Now, substituting this equality in (I) we get

$\cos(2+3i) = \frac{e^{-3}+e^{3}}{2}\cos 2 -i\frac{e^{3}-e^{-3}}{2}\sin 2 = \cosh 3\cos 2-i\sinh 3\sin 2$ .

Thus,

$\exp\left(\cos(2+3i)\right) = \exp\left(\cosh 3\cos 2-i\sinh 3\sin 2\right)$

$\exp\left(\cos(2+3i)\right) = \exp\left(\cosh 3\cos 2\right)\left[ \cos(\sinh 3\sin 2)-i\sin(\sinh 3\sin 2)\right]$ .

5 0

3 years ago

Denise has $270 in her savings adcount. She wants to save at least $800.

Nadusha1986 [10]

Answer:

c) d is greater than or equal to 530

Step-by-step explanation:

Hope this helps!

6 0

3 years ago