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4 years ago

Please help math 25 points if you guess your answer will be deleted

Mathematics

2 answers:

Ne4ueva [31]4 years ago

6 0

Because the inequality sign is just greater than ( no equal to), the correct answer would be A) a shaded region above a dashed boundary line.

Anna007 [38]4 years ago

6 0

A. - it would be dashed because its not equal to, and it would be above because its saying y is greater than x + 4

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What is point slope form?

sashaice [31]

Answer:

Point slope is the general form y-y₁=m(x-x₁) for linear equations.

Step-by-step explanation:

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3 years ago

Simplify (3x2 - 5x + 3) + (-3x2 + 5x - 3).

zepelin [54]

Answer:

(9 -5x) + (-9 +5x)

Step-by-step explanation:

-Three times two is six plus three is nine

-Negative three times two is negative six minus three which is negative nine

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3 years ago

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Gerry and his family have moved into a new home with a great backyard. In the yard, they plan to have a garden, a lawn, and a pa

Alina [70]

3 centimetres = 1 yard

Answer 1:

30 cm

Answer 2:

45 cm

Answer 3:

24 cm

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3 years ago

Solve for x if log 9 base x + log 3 base x^2 = 2.5

Y_Kistochka [10]

Not sure if the equation is

$\log_9x+\log_3(x^2)=\dfrac52$

$\log_x9+\log_{x^2}3=\dfrac52$

If it's the first one:

$9^{\log_9x+\log_3(x^2)}=9^{\log_9x}\cdot9^{\log_3(x^2)}$

$9^{\log_9x+\log_3(x^2)}=9^{\log_9x}\cdot(3^2)^{\log_3(x^2)}$

$9^{\log_9x+\log_3(x^2)}=9^{\log_9x}\cdot3^{2\log_3(x^2)}$

$9^{\log_9x+\log_3(x^2)}=9^{\log_9x}\cdot3^{\log_3(x^2)^2}$

$9^{\log_9x+\log_3(x^2)}=9^{\log_9x}\cdot3^{\log_3(x^4)}$

$9^{\log_9x+\log_3(x^2)}=x\cdot x^4$

$9^{\log_9x+\log_3(x^2)}=x^5$

On the other side of the equation, we'd get

$9^{5/2}=(3^2)^{5/2}=3^{2\cdot(5/2)}=3^5$

Then

$x^5=3^5\implies\boxed{x=3}$

If it's the second one instead, you can use the same strategy as above:

$x^{\log_x9+\log_{x^2}3}=x^{\log_x9}\cdot x^{\log_{x^2}3}$

$x^{\log_x9+\log_{x^2}3}=x^{\log_x9}\cdot\left((x^2)^{1/2}\right)^{\log_{x^2}3}$

(Note that this step assume $x>0$ )

$x^{\log_x9+\log_{x^2}3}=x^{\log_x9}\cdot(x^2)^{(1/2)\log_{x^2}3}$

$x^{\log_x9+\log_{x^2}3}=x^{\log_x9}\cdot(x^2)^{\log_{x^2}\sqrt3}$

$x^{\log_x9+\log_{x^2}3}=9\sqrt3$

Then we get

$9\sqrt3=x^{5/2}\implies x=(9\sqrt3)^{2/5}\implies\boxed{x=3}$

6 0

3 years ago

If the solution of an equation is x=-2 what could the original equation be?

Katyanochek1 [597]

There are an infinite number of possibilities.
Here's one:

6x + 107 = -77x - 59

5 0

3 years ago

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