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3 years ago

What is the root of -9 ? Thank you for help

Mathematics

1 answer:

stiks02 [169]3 years ago

7 0

I don't think this is possible...

But if you mean 9 then the root is 3

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HELP! WILL GIVE BRAINLIEST! DO NOT TAKE ANSWERS OFF ONLINE, THEY ARE WRONG!!!! The following is an incomplete paragraph proving

ohaa [14]

Answer:

The correct option is B.

Step-by-step explanation:

Given information: AB\parallel DCAB∥DC and BC\parallel ADBC∥AD .

Draw a diagonal AC.

In triangle BCA and DAC,

AC\cong ACAC≅AC (Reflexive Property of Equality)

\angle BAC\cong \angle DCA∠BAC≅∠DCA ( Alternate Interior Angles Theorem)

\angle BCA\cong \angle DAC∠BCA≅∠DAC ( Alternate Interior Angles Theorem)

The ASA (Angle-Side-Angle) postulate states that two triangles are congruent if two corresponding angles and the included side of are congruent.

By ASA postulate,

\triangle BCA\cong \triangle DAC△BCA≅△DAC

Therefore option B is correct

4 0

3 years ago

Find , , and if and terminates in quadrant .

ioda

sin2x =12/13

cos2x = 5/13

tan2x = 12/5

STEP - BY - STEP EXPLANATION

What to find?

• sin2x

• cos2x

• tan2x

Given:

tanx = 2/3 = opposite / adjacent

We need to first make a sketch of the given problem.

Let h be the hypotenuse.

We need to find sinx and cos x, but to find sinx and cosx, first determine the value of h.

Using the Pythagoras theorem;

hypotenuse² = opposite² + adjacent²

h² = 2² + 3²

h² = 4 + 9

h² =13

Take the square root of both-side of the equation.

h =√13

This implies that hypotenuse = √13

We can now proceed to find the values of ainx and cosx.

Using the trigonometric ratio;

$\sin x=\frac{opposite}{\text{hypotenuse}}=\frac{2}{\sqrt[]{13}}$ $\cos x=\frac{adjacent}{\text{hypotenuse}}=\frac{3}{\sqrt[]{13}}$

And we know that tanx =2/3

From the trigonometric identity;

sin 2x = 2sinxcosx

Substitute the value of sinx , cosx and then simplify.

$\sin 2x=2(\frac{2}{\sqrt[]{13}})(\frac{3}{\sqrt[]{13}})$ $=\frac{12}{13}$

Hence, sin2x = 12/13

cos2x = cos²x - sin²x

Substitute the value of cosx, sinx and simplify.

$\begin{gathered} \cos 2x=(\frac{3}{\sqrt[]{13}})^2-(\frac{2}{\sqrt[]{13}})^2 \\ \\ =\frac{9}{13}-\frac{4}{13} \\ =\frac{5}{13} \end{gathered}$

Hence, cos2x = 5/13

tan2x = 2tanx / 1- tan²x

$\tan 2x=\frac{2\tan x}{1-\tan ^2x}$ $=\frac{2(\frac{2}{3})}{1-(\frac{2}{3})^2}$ $=\frac{\frac{4}{3}}{1-\frac{4}{9}}$ $=\frac{\frac{4}{3}}{\frac{9-4}{9}}$ $=\frac{\frac{4}{3}}{\frac{5}{9}}$ $=\frac{4}{3}\times\frac{9}{5}=\frac{4}{1}\times\frac{3}{5}=\frac{12}{5}$