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2 years ago

3z + 6 (m-5) =-48 solve for z

Mathematics

2 answers:

Troyanec [42]2 years ago

4 0

Answer:

$z=2m+6$

Step-by-step explanation:

$3z+6(m-5)=-48\\6(m-5)+48=3z\\6m-30+48=3z\\6m+18=3z\\$

$\frac{6m+18}{3} =\frac{3z}{3}$

$2m+6=z$

deff fn [24]2 years ago

3 0

Answer:

z = -2(m + 3)

Step-by-step explanation:

Given :

3z + 6(m - 5) = -48

Divide throughout by 3 :

1/3 x (3z + 6(m - 5)) = -48 x 1/3
z + 2(m - 5) = -16

Subtract 2(m - 5) on each side :

z + 2(m - 5) - 2(m - 5) = -16 - 2(m - 5)
z = -16 - 2(m - 5)

Simplify the RHS :

z = -16 - 2m + 10
z = -2m - 6
z = -2(m + 3)

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3 years ago

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>

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

Given:

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

Note the following:

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

5 0

3 years ago

If the equation of a circle is (x - 2)2 + (y-6)2 = 4, it passes through point

san4es73 [151]

(x - 2)² + (y - 6)² = 4

You can be certain about one thing just by looking at the equation (2, 6) is the center, so, obviously the circle isnt going through this point

Since the radius is 2 if we don't move from y = 6 we have points in (0, 6) and (4, 6)

So alternative b.

To be more certain just subs the point in the x and y, if its equal, it pass through

(x - 2)² + (y - 6)² = 4

To point (4, 6)

(4 - 2)² + (6 - 6)² = 4

(2²) + 0² = 4

4 = 4

Thats right

4 0

3 years ago