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3 years ago

I’m having a hard time help me

Mathematics

2 answers:

gtnhenbr [62]3 years ago

6 0

Look at the attached picture ⤴

I have shown the steps there....

So,Hope it will help u ....:)

✴Have a great day✴

miss Akunina [59]3 years ago

5 0

Answer:

The answer is 2881

Step-by-step explanation:

You might be interested in

A solution of the equation x^2/4=9 is

abruzzese [7]

Answer:

x = ±6

Step-by-step explanation:

x^2/4=9

Multiply by 4 on each side

x^2/4 *4=9*4

x^2 = 36

Take the square root of each side

sqrt(x^2) = ±sqrt(36)

x = ±6

5 0

3 years ago

What's the answer for 24

scZoUnD [109]

X= -2 or X= 3 either one works.

3 0

3 years ago

I need help with these 2 questions.

Usimov [2.4K]

The first one is 32m^15n^10
The second one is:
(h^12/h^4)^5
(h^6/h)^5
h^30/h^5
h^25/h

Hope this helps :)

3 0

3 years ago

PLZ HELP ME ☻ <img src="https://tex.z-dn.net/?f=%5C%5B%5Cfrac%7Bxy%7D%7Bx%20%2B%20y%7D%20%3D%201%2C%20%5Cquad%20%5Cfrac%7Bxz%7D%

Yanka [14]

Answer:

$x=\frac{12}{7} \\y=\frac{12}{5} \\z=-12$

Step-by-step explanation:

Let's re-write the equations in order to get the variables as separated in independent terms as possible \:

First equation:

$\frac{xy}{x+y} =1\\xy=x+y\\1=\frac{x+y}{xy} \\1=\frac{1}{y} +\frac{1}{x}$

Second equation:

$\frac{xz}{x+z} =2\\xz=2\,(x+z)\\\frac{1}{2} =\frac{x+z}{xz} \\\frac{1}{2} =\frac{1}{z} +\frac{1}{x}$

Third equation:

$\frac{yz}{y+z} =3\\yz=3\,(y+z)\\\frac{1}{3} =\frac{y+z}{yz} \\\frac{1}{3}=\frac{1}{z} +\frac{1}{y}$

Now let's subtract term by term the reduced equation 3 from the reduced equation 1 in order to eliminate the term that contains "y":

$1=\frac{1}{y} +\frac{1}{x} \\-\\\frac{1}{3} =\frac{1}{z} +\frac{1}{y}\\\frac{2}{3} =\frac{1}{x} -\frac{1}{z}$

Combine this last expression term by term with the reduced equation 2, and solve for "x" :

$\frac{2}{3} =\frac{1}{x} -\frac{1}{z} \\+\\\frac{1}{2} =\frac{1}{z} +\frac{1}{x} \\ \\\frac{7}{6} =\frac{2}{x}\\ \\x=\frac{12}{7}$

Now we use this value for "x" back in equation 1 to solve for "y":

$1=\frac{1}{y} +\frac{1}{x} \\1=\frac{1}{y} +\frac{7}{12}\\1-\frac{7}{12}=\frac{1}{y} \\ \\\frac{1}{y} =\frac{5}{12} \\y=\frac{12}{5}$

And finally we solve for the third unknown "z":

$\frac{1}{2} =\frac{1}{z} +\frac{1}{x} \\\\\frac{1}{2} =\frac{1}{z} +\frac{7}{12} \\\\\frac{1}{z} =\frac{1}{2}-\frac{7}{12} \\\\\frac{1}{z} =-\frac{1}{12}\\z=-12$

8 0

3 years ago

2. G.GPE.7 Consider ∆ABC below. What is the perimeter of ∆ABC? * A. 16.9 B. 12.9 C. 18.6 D. 17.4

dsp73

Answer:B

Step-by-step explanation:

3 0

2 years ago