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jeka57 [31]
3 years ago
5

Help!!!!!!!!!!!!!!!!!!!!

Mathematics
1 answer:
lapo4ka [179]3 years ago
7 0
Well...   the segment FG = 4x + 3, the segment GH = 7x - 12, now, the
 
segment FH = FG + GH, those two fellows added up, and we know that FG and GH are both equal halves because G is the midpoint.

thus, FG = GH
or

\bf 4x+3=7x-12\implies 3+12=7x-4x\implies 15=3x
\\\\\\
\cfrac{15}{3}=x\implies \boxed{5=x}\\\\
-------------------------------\\\\
thus\qquad 
\begin{cases}
\overline{FG}=4x+3\\
4(5)+3\\
23\\
------\\
\overline{GH}=7x-12\\
7(5)-12\\
23
\end{cases}\impliedby equal\ halves
\\\\\\
\overline{FH}=23+23\implies \overline{FH}=46\implies 6y-2=46
\\\\\\
6y=46+2\implies y=\cfrac{48}{6}

and fairly sure, you know how much that is.
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a machine can fill 650 bottles in 25 minutes. What is the unit rate at which the machine fills the bottles?
4vir4ik [10]
unit\ rate=\frac{amount\ of\ bottle}{time}\\\\
unit\ rate=\frac{650}{25}=26\frac{bottle}{min}\\\\Unit\ rate\ is\ 26\ bottle\ per\ minute.
8 0
3 years ago
13 = -7 + 9x - 5x<br> X =
brilliants [131]

Answer:

x = 5

Step-by-step explanation:

Given

13 = - 7 + 9x - 5x , that is

13 = - 7 + 4x ( add 7 to both sides )

20 = 4x ( divide both sides by 4 )

5 = x

3 0
3 years ago
Really need an answer to this please help.
Alex

Answer:

a

Step-by-step explanation:

\sqrt[4]{144a^{12}b^{3}} = \sqrt[4]{4^{2}*3^{2}a^{12}b^{3}}=\\= \sqrt[4]{2^{4}*3^{2}a^{12}b^{3}}=2a^{3}\sqrt[4]{3^{2}b^{3}} =\\}=2a^{3}\sqrt[4]{9b^{3}}

7 0
3 years ago
ANSWER ASAP MUST BE CORRECT
Galina-37 [17]
Solve the second equation for y so that you can substitute it in for y in the first equation..

y = 5x - 4

Substitute:

3x + (5x - 4) = 1


3 0
3 years ago
Read 2 more answers
Suppose a tub has the shape of an elliptical paraboloid given by z = ax2+by2 (where a, b are some positive constants). If a marb
Umnica [9.8K]

Answer:

It would roll in this direction.

\nu  =  (-a/\sqrt{a^2+b^2},-b/\sqrt{a^2+b^2})

Step-by-step explanation:

It would roll to the direction of maximum decrease, which is the -1 times the direction of maximum increase, which is given by the gradient of the function.  

Since  

z =  ax^2  + by^2

For this case, the gradient of your function would be

\nabla z  = (2ax , 2by)

And  -1  times the gradient of your function would be

-\nabla z  = (-2ax , -2by)\\

Then, at

 (1,1,a+b),\\x = 1 \\y = 1

So it would go towards

v = (-2a,-2b)

The magnitud of that vector is

|v| =  2\sqrt{a^2+b^2}

and to conclude it would roll in this direction.

\nu  =  (-a/\sqrt{a^2+b^2},-b/\sqrt{a^2+b^2})

6 0
3 years ago
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