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2 years ago

PLEASE HELP!!!!how to solve 9x^3-16x^2=0 by factoring WILL GIVE BRAINLIEST IF CORRECT!!!!

Mathematics

1 answer:

andreev551 [17]2 years ago

7 0

Answer:

x=5.3

Step-by-step explanation:

Graph each side of the equation. The solution is the x-value of the point of intersection.

≈

5.3

Hope this helps, can you now give me brainliest??

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Solve for X, will give you brainliest and a like

grigory [225]

Answer

Step-by-step explanation:

One solution :

x = -1/3 = -0.333

Rearrange:

Rearrange the equation by subtracting what is to the right of the equal sign from both sides of the equation :

-5*(3*x+1)+4*x-(10*x+2)=0

Step by step solution :

Step 1 :

Equation at the end of step 1 :

((0 - 5 • (3x + 1)) + 4x) - (10x + 2) = 0

Step 2 :

Step 3 :

Pulling out like terms :

3.1 Pull out like factors :

-21x - 7 = -7 • (3x + 1)

Equation at the end of step 3 :

-7 • (3x + 1) = 0

Step 4 :

Equations which are never true :

4.1 Solve : -7 = 0

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2 years ago

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dsp73

Answer: I belive the answer is C

Step-by-step explanation:

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2 years ago

I dont know what a complementary angle is, can someone help me with this quistion please?

Allisa [31]

Answer:

A, complementary angles sum up to 90°

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3 years ago

Use the slope formula to determine the slope of the line.

lukranit [14]

Answer:

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3 years ago

Prove that the segments joining the midpoint of consecutive sides of an isosceles trapezoid form a rhombus.

sergiy2304 [10]

Answer:

See explanation

Step-by-step explanation:

a) To prove that DEFG is a rhombus, it is sufficient to prove that:

All the sides of the rhombus are congruent: $|DG|\cong |GF| \cong |EF| \cong |DE|$
The diagonals are perpendicular

Using the distance formula; $d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

$|DG|=\sqrt{(0-(-a-b))^2+(0-c)^2}$

$\implies |DG|=\sqrt{a^2+b^2+c^2+2ab}$

$|GF|=\sqrt{((a+b)-0)^2+(c-0)^2}$

$\implies |GF|=\sqrt{a^2+b^2+c^2+2ab}$

$|EF|=\sqrt{((a+b)-0)^2+(c-2c)^2}$

$\implies |EF|=\sqrt{a^2+b^2+c^2+2ab}$

$|DE|=\sqrt{(0-(-a-b))^2+(2c-c)^2}$

$\implies |DE|=\sqrt{a^2+b^2+c^2+2ab}$

Using the slope formula; $m=\frac{y_2-y_1}{x_2-x_1}$

The slope of EG is $m_{EG}=\frac{2c-0}{0-0}$

$\implies m_{EG}=\frac{2c}{0}$

The slope of EG is undefined hence it is a vertical line.

The slope of DF is $m_{DF}=\frac{c-c}{a+b-(-a-b)}$

$\implies m_{DF}=\frac{0}{2a+2b)}=0$

The slope of DF is zero, hence it is a horizontal line.

A horizontal line meets a vertical line at 90 degrees.

Conclusion:

Since $|DG|\cong |GF| \cong |EF| \cong |DE|$ and $DF \perp FG$ , DEFG is a rhombus

b) Using the slope formula:

The slope of DE is $m_{DE}=\frac{2c-c}{0-(-a-b)}$

$m_{DE}=\frac{c}{a+b)}$

The slope of FG is $m_{FG}=\frac{c-0}{a+b-0}$

$\implies m_{FG}=\frac{c}{a+b}$

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3 years ago