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2 years ago

SORRY BUT I HAVE ANOTHER PROBLEM!!!!! WILL GIVE BRAINLIEST & 50 POINTS

Mathematics

2 answers:

11111nata11111 [884]2 years ago

8 0

Just factor out

$\\ \rm\hookrightarrow \dfrac{x^2-6x+9}{6x-18}$

$\\ \rm\hookrightarrow \dfrac{x^2-3x-3x+9}{6(x-3)}$

$\\ \rm\hookrightarrow \dfrac{x(x-3)-3(x-3)}{6(x-3)}$

$\\ \rm\hookrightarrow \dfrac{(x-3)(x-3)}{6(x-3)}$

$\\ \rm\hookrightarrow \dfrac{x-3}{6}$

liberstina [14]2 years ago

7 0

Answer:

The correct option is G.Hope this help!!

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The variable z is inversely proportional to x. When x is 6, z has the value 0.5. What is the value of z when x= 14?

Lilit [14]

Z is inversely proportional to X is the same as
z is directly proportional to 1/X
which can be written like this:-
y=k/X
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0.5=k/6
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2 years ago

Defining congruent triangles

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Answer:

E -->R

D-->K

O-->O

Step-by-step explanation:

Assuming you are asking for congruent parts, correspondence is the part which corresponds and is also equal.

Here Angle E corresponds and is congruent to Angle R in these congruent triangles.

Angle D is congruent and corresponds to Angle K for these congruent triangles.

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3 years ago

As a first step in solving the system shown, Yumiko multiplies both sides of the equation 2x – 3y = 12 by 6. By what factor shou

GenaCL600 [577]

Answer:

Yumiko should multiply the other equation by 3.

If she adds the two equations she would be left with the variable 'x'.

Step-by-step explanation:

Given the two equations are as follows:

$2x - 3y = 12 \hspace{5mm} \hdots (1)$

$5x + 6y = 18 \hspace{5mm} \hdots (2)$

It is given that she multiplies the first equation by 6. Therefore, (1) becomes

$12x - 18y = 72 \hspace{15mm} \hdots (a)$

Now, note that the sign of the variable 'y' is negative. So, if we make the co-effecient of 'y' equal in both the cases, add them it would result in the elimination of the variable 'y'.

The co-effecient of y in Equation (2) is 6. To make it 18 like it is in Equation (1), we multiply throughout by 3.

Therefore, Equation (2) becomes:

$15x + 18y = 54 \hspace{5mm} \hdots (b)$