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3 years ago

Please help me!!!!!

Mathematics

2 answers:

kirill [66]3 years ago

8 0

Answer:

-3 < x < -2

Step-by-step explanation:

We want where the y value is negative

-3 < x < -2 the y values are negative

-1 < x < 0 the y values are positive

0 < x < 1 the y values are positive

a_sh-v [17]3 years ago

3 0

Answer:

Step-by-step explanation:

Since f(x) is graphed, you need to look at the specified intervals and see which of them have the entire portion graphed below the x axis.

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Jeff is shopping for a new T-shirt at a clothing store. The graph shows the total price for the shirts depending on how many T-s

Artyom0805 [142]

Answer:

$ 63

Step-by-step explanation:

All you have to do is see what point of the line falls right above 7, and then see what number corresponds to that point.

Hope it helped,

BioTeacher101

6 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

3x + 2 = -13. Plz help

mote1985 [20]

The answer is -5 hope this helps

3 0

3 years ago

Read 2 more answers

I thought of a number, divided it by 1 1/2 , then added 2 1/2 to the result, then multiplied the sum by 1 2/3 . Then I subtracte

vekshin1

Answer:

Step-by-step explanation:

7 0

4 years ago

Please find the SURFACE AREA of this figure... Write the formula and plug in the values... Write your answer with units and box

ad-work [718]

Answer:

124(cm)^2

Step-by-step explanation:

total surface area of a cuboid= $2(LB+BH+HL)$

length=7cm

Breadth=5cm

Height=4cm

2{(2*7)+(5*4)+(4*7)}

2{14+20+28}

2*62

124(cm)^2

5 0

3 years ago

Please help me!!!!!​

Please help me!!!!!