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3 years ago

Can someone help with this, or just give a example of one???

Mathematics

1 answer:

Gala2k [10]3 years ago

4 0

Really simple, just do what they tell you to

example 10
1.divide y by 7
so 7/y
2. then add x

7/y+x

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I answered it but I just need someone to answer just to make sure mine are right

Artemon [7]

<h2>The missing measures of the angles are:</h2>

$m\angle 1= 60^{\circ}\\\\m\angle 2= 39^{\circ}\\\\m\angle 3= 21^{\circ}\\\\m\angle 4 = 39^{\circ}\\\\m\angle 5 = 21^{\circ}$

<u>Given</u>:

$m\angle Z = 138^{\circ}$

<em><u>Note the following:</u></em>

An equilateral triangle has all its three angles equal to each other. Each angle = $60^{\circ}$ .

This implies that, since $\triangle WXY$ is equilateral, therefore, $m\angle Y = m\angle XWY = m\angle XYW = 60^{\circ}$

Base angles of an isosceles triangle are congruent to each other.

This implies that, since $\triangle WZY$ is isosceles, therefore, $m\angle 3 = \angle 5$

Applying the above stated, let's find the measure of each angle:

Find $m\angle 1$

$m\angle 1 = 60^{\circ}$ (an angle in an equilateral triangle equals 60 degrees)

Find $m\angle 2$

$m\angle 2 = 60 - m \angle 3$

$m\angle 2 = 60 -\frac{1}{2}(180 - 138)$ $(Note: \frac{1}{2}(180 - 138) = 1 $ base $ angle $ of $ \triangle WZY)$

$m\angle 2 = 60 -21\\\\m\angle 2 = 39^{\circ}$

Find $m\angle 3$ and $m\angle 5$ (base angles of isosceles triangle WZY)

$m\angle 3 = \frac{1}{2}(180 - 138) (1 $ base $ angle $ of $ \triangle WZY)$

$m\angle 3 = \frac{1}{2}(42) \\\\m\angle3 = 21^{\circ}$

$m\angle3 = m\angle 5$ (base angles of isosceles triangle are congruent)

Therefore,

$m\angle5 = 21^{\circ}$

Find $m\angle 4$

$m\angle 4 = 60 - m\angle 5$

Substitute

$m\angle 4 = 60 - 21\\\\m\angle 4 = 39^{\circ}$

The missing measures of the angles are:

$m\angle 1= 60^{\circ}\\\\ m\angle 2= 39^{\circ}\\\\m\angle 3= 21^{\circ}\\\\m\angle 4 = 39^{\circ}\\\\m\angle 5 = 21^{\circ}$

Learn more here:

brainly.com/question/2944195

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2 years ago

Find the derivative of the function y = sin(tan 4x)

Virty [35]

$\bf y=sin[tan(4x)]\implies \cfrac{dy}{dx}=\stackrel{chain~rule}{cos[tan(4x)]\cdot sec^2(4x)\cdot 4}
\\\\\\
\cfrac{dy}{dx}=4cos[tan(4x)]sec^2(4x)$

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andreev551 [17]

Answer:

AAS

Step-by-step explanation:

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A line of symmetry basically is like a reflection of a shape, so your answer is D.

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Two intersecting segments form the diagonals of a quadrilateral. What additional information do you need to

Aleksandr-060686 [28]

Answer:

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