All categories

3 years ago

How do I solve this equation: 7x-4=3+8x

Mathematics

2 answers:

marusya05 [52]3 years ago

6 0

Answer:

x=-7

Step-by-step explanation:

Step 1: Simplify both sides of the equation.

7x−4=3+8x

7x+−4=3+8x

7x−4=8x+3

Step 2: Subtract 8x from both sides.

7x−4−8x=8x+3−8x

−x−4=3

Step 3: Add 4 to both sides.

−x−4+4=3+4

−x=7

Step 4: Divide both sides by -1.

−x/−1=7/−1

x=−7

Vlad [161]3 years ago

3 0

The answer to this is no solution

Step by step:
7x - 4 = 3 + 8x
-7x -7x
-4 = 3 + 1x
+4 +4
7 + 1x = no solution

You might be interested in

Long’s Coffee Shop sells a refill mug for $8.95. Each refill costs $1.50. Last month,

dem82 [27]

Answer: 12 refills

Step-by-step explanation:

y = 8.95 + 1.50x

26.95= 8.95 + 1.50x

26.95 - 8.95 = 1.50x

18 = 1.50X

18 / 1.50 = x

X = 12

8 0

3 years ago

F(x) = x^2 What is g(x)?

neonofarm [45]

graph becomes slimmer, so definitely be multipled 1/2

7 0

3 years ago

Help me with this question?

svlad2 [7]

9. Is 37 since you take PR+QR
10. Is 7 since it's PR-QR
11. Is 6 since you take PR-PQ

8 0

3 years ago

The ratio of female birds to male birds in a habitat is 2.4 to 1. If there are 123 female birds in the habitat, and approximatel

dusya [7]

So you will first need to divide 123 by 2.4 to get how many male birds are there. Then you need to multiply that answer by 0.4. That will be your final answer.

8 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