All categories

3 years ago

2a + 4c = 38.00 solve for c

Mathematics

2 answers:

Dominik [7]3 years ago

5 0

Step-by-step explanation:

2 (a+2c)=38

a+2c=19

c=(19-a)/2

irakobra [83]3 years ago

4 0

Answer:

c = 19/2 - 1/2a

Step-by-step explanation:

Step 1: Write equation

2a + 4c = 38.00

Step 2: Solve for c

Subtract 2a on both sides: 4c = 38 - 2a

Divide both sides by 4: c = 19/2 - 1/2a

You might be interested in

13. Here is one way Maleka proved that 2(5 + 2) + 2s is equivalent to

Kaylis [27]

Malaria proved that the equation you need to add the parenthesis first

3 0

3 years ago

Solve 11 2/5 = q - 4 2/7 + 2 1/7

RideAnS [48]

I turn the mixed numbers into decimals before doing anything else.
11 2/5----> 57/5----> 11.4
4 2/7----> 30/7----> 4.29
2 1/7----> 15/7----> 2.14

11.4=q-4.29+2.14
11.4=q-2.15
13.55= q

13.55 converted to a fraction is 13 11⁄20
q=13 11⁄20

6 0

4 years ago

Uh. I don't understand this at all

kvv77 [185]

Answer: BCD

Step-by-step explanation:

A is incorrect since there is no angle Y to exist

Hope this helps

3 0

3 years ago

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

How much do u need to subtract from 41/6 to make 6

dem82 [27]

41/6-5/6 = 36/6
36/6= 6

So 5/6 would need to be subtracted to develop a fraction that divides to 6

4 0

3 years ago