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3 years ago

10

Just question nine …

2 answers:

Semenov [28]3 years ago

7 0

Answer:

A translation always preserves orientation

Step-by-step explanation:

Translations simply shift the shape, it does not affect size, rotation, or any orientations.

Helga [31]3 years ago

3 0

The original place.

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Let P(x) = 3x³-ax+b. If P(1) = P(-1) and P(2)=16, find the values of a and b

kozerog [31]

P(1) = P(-1)

P(1) = 3 - a + b
P(-1) = -3 + a + b

-> 3 - a + b = -3 + a + b
-> 3 - a + b + 3 - a - b = 0
-> 6 - 2a = 0
-> a = 3.

P(2) = 24 - 2a + b -> 24 - (2a - b) = 16 -> 2a - b = 8
-> 6 - b = 8
b = -2.

So, a = 3 and b = -2

Recheck : P(1) = 3 - 3 + (-2) = -2
P(-1) = -3 + 3 + (-2) = -2 => P(1) = P(-1) (true)

P(2) = 24 - 6 + (-2) = 16.

6 0

2 years ago

Help me guys <br> thx so much

navik [9.2K]

Answer:

Option D, $10x^4\sqrt{6} +x^3\sqrt{30x} -10x^4\sqrt{3} -x^3\sqrt{15x}$

Step-by-step explanation:

<u>Step 1: Multiply</u>

<u /> $(\sqrt{10x^4} -x\sqrt{5x^2} )*(2\sqrt{15x^4} + \sqrt{3x^3})\\ (\sqrt{10 * x^2 * x^2} -x\sqrt{5 * x^2} ) * (2\sqrt{15 * x^2 * x^2} +\sqrt{3 * x^2 * x})\\(x^2\sqrt{10} -x^2\sqrt{5} )*(2x^2\sqrt{15} +x\sqrt{3x}) \\\\$

$(x^2\sqrt{10}*2x^2\sqrt{15} )+(x^2\sqrt{10}*x\sqrt{3x} ) + (-x^2\sqrt{5} *2x^2\sqrt{15}) + (-x^2\sqrt{5} *x\sqrt{3x}$

$(2x^4\sqrt{150} ) + (x^3\sqrt{30x}) + (-2x^4\sqrt{75}) + (-x^3\sqrt{15x} )$

$2x^4\sqrt{5^2*6} + x^3\sqrt{30x} -2x^4\sqrt{5^2*3} -x^3\sqrt{15x}$

$10x^4\sqrt{6} +x^3\sqrt{30x} -10x^4\sqrt{3} -x^3\sqrt{15x}$

Answer: Option D, $10x^4\sqrt{6} +x^3\sqrt{30x} -10x^4\sqrt{3} -x^3\sqrt{15x}$

6 0

3 years ago

Read 2 more answers

YOO PLZ HELP ASAP!! MARKING BRAINIEST!!!

Alexeev081 [22]

Answer:

attached the drawing

Step-by-step explanation:

hope it works

5 0

3 years ago

Solving Quadratic Equations using the Square Root Property

Afina-wow [57]

Answer:

$x = -2$ or $x = -5$

Step-by-step explanation:

You need to complete the square before you can take the square root of both sides.

$x^2 + 7x + 10 = 0$

Subtract 10 from both sides.

$x^2 + 7x = -10$

To complete the square, you need to add the square of half of the x-term coefficient to both sides.

The x-term coefficient is 7. Half of that is 7/2. Square it to get 49/4. Now we add 49/4 to both sides of the equation.

$x^2 + 7x + \dfrac{49}{4} = -10 + \dfrac{49}{4}$

$(x + \dfrac{7}{2})^2 = -\dfrac{40}{4} + \dfrac{49}{4}$

$(x + \dfrac{7}{2})^2 = \dfrac{9}{4}$

Now we use the square root property, if

$x^2 = k$ , then

$x = \pm \sqrt{k}$

$x + \dfrac{7}{2} = \pm \sqrt{\dfrac{9}{4}}$

$x + \dfrac{7}{2} = \pm \dfrac{3}{2}$

$x + \dfrac{7}{2} = \dfrac{3}{2}$ or $x + \dfrac{7}{2} = -\dfrac{3}{2}$

$x = -\dfrac{4}{2}$ or $x = -\dfrac{10}{2}$

$x = -2$ or $x = -5$

5 0

3 years ago

Sharon lives in Virginia and expects the taxes on her bond income to be 5.75%

Firlakuza [10]

Well good for Sharon :D

4 0

4 years ago