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liberstina [14]

4 years ago

7

Factor

=0" align="absmiddle" class="latex-formula"> given that 3 is a zero.
A) $(x+2)(x-3)^2=0$
B) $(x-2)(x-3)^2=0$
C) $(x-2)(x-3)(x+3)=0$
D) $(x+2)(x-\sqrt{3})(x+\sqrt{3})=0$

1 answer:

Nata [24]4 years ago

3 0

Answer:

A) (x +2)(x -3)² = 0

Step-by-step explanation:

Synthetic division by (x-3) gives the quadratic x² -x -6, which has factors (x -3) and (x +2). This is confirmed by another round of synthetic division.

The resulting factorization is ...

... (x +2)(x -3)² = 0

_____

A graph confirms a double zero at x=3 and one at x=-2.

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SCORPION-xisa [38]

Answer:

$x =6$

$y=6\cdot \sqrt{2}$

Step-by-step explanation:

<u>Trigonometric Ratios </u>

The ratios of the sides of a right triangle are called trigonometric ratios. There are six trigonometric ratios: sine, cosine, tangent, cosecant, secant, and cotangent.

The longest side of the right triangle is called the hypotenuse and the other two sides are the legs.

Choosing any of the acute angles as a reference, it has an adjacent side and an opposite side. The trigonometric ratios are defined upon those sides.

The image shows a right triangle where the angle of 45° has x as the opposite leg, 6 as the adjacent leg, and y as the hypotenuse. The trigonometric ratio that applies here is the cosine ratio, defined as:

$\displaystyle \cos\theta=\frac{\text{adjacent leg}}{\text{hypotenuse}}$

$\displaystyle \cos 45^\circ=\frac{6}{y}$

Solving for y:

$\displaystyle y=\frac{6}{\cos 45^\circ}$

$\cos 45^\circ=\frac{\sqrt{2}}{2}=\frac{1}{\sqrt{2}}$

Substituting:

$\displaystyle y=\frac{6}{\frac{1}{\sqrt{2}}}$

$y=6\cdot \sqrt{2}$

Now use the tangent ratio:

$\displaystyle \tan\theta=\frac{\text{opposite leg}}{\text{adjacent leg}}$

$\displaystyle \tan 45^\circ=\frac{x}{6}$

Solving for x:

$x=6\cdot\tan 45^\circ$

$\tan 45^\circ=1$

Substituting:

$x=6\cdot 1$

$x =6$

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10a + -4a + -4a

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Insturctions to care for a plant say to water it with 3/4 cup of water every day.The plant has been getting 25% too much water.

disa [49]

For this case we propose a rule of three. We assume that the excess water received was in one day. So:

$\frac {3} {4}$ cup ------> 100%

x ----------------------------> 25%

Where "x" represents excess water.

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Thus, the excess water is $\frac {3} {16}$ cups.

Then, in one day you receive in total:

$\frac {3} {4} + \frac {3} {16} = \frac {16 * 3 + 4 * 3} {16 * 4} = \frac {48 + 12} {64} = \frac {60 } {64} = \frac {15} {4}$ cups of water.

Answer:

$\frac {15} {4}$ cups of water

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