Find the value of xx, yy, and zz in the rhombus below.

Question

88°
(2x+10)°
(-3y-2)°
(-6z+2)°
Answer
Attempt 1 out of 2

x, equalsx=
y, equalsy=
z, equalsz=

Answer 1

In a rhombus, opposite angles are equal, and the sum of all angles in any quadrilateral is ${360}^{\circ }$.

Given the angles:

1. ${88}^{\circ }$
2. $(2x+10{)}^{\circ }$
3. $(-3y-2{)}^{\circ }$
4. $(-6z+2{)}^{\circ }$

Step 1: Set Up Equation for Angles

Since opposite angles are equal, we know:

• $(2x+10{)}^{\circ }={88}^{\circ }$
• $(-3y-2{)}^{\circ }={88}^{\circ }$

Step 2: Solve for $x$

From the first equation: $$2x+10=88$$ Subtract 10 from both sides: $$2x=78$$ Now divide by 2: $$x=39$$

Step 3: Solve for $y$

From the second equation: $$-3y-2=88$$ Add 2 to both sides: $$-3y=90$$ Divide by -3: $$y=-30$$

Step 4: Find the Fourth Angle

Now, let's find the fourth angle using the sum of the angles in a quadrilateral: $$88+(2x+10)+(-3y-2)+(-6z+2)=360$$ Substituting the solved values:

• $x=39$ so $2x+10=2(39)+10=88+10=98$
• $y=-30$ so $-3y-2=-3(-30)-2=90-2=88$

Now, plug these in: $$88+98+88+(-6z+2)=360$$ Combine the known angles: $$274+(-6z+2)=360$$ Now simplify: $$-6z+2=360-274$$ $$-6z+2=86$$ Subtract 2 from both sides: $$-6z=84$$ Divide by -6: $$z=-14$$

Final Answers

Thus, the values are: $$x=39,y=-30,z=-14$$

Answer 2

it say your answer is wrong

Answer 3

I apologize for any confusion with my previous solution. Let's re-evaluate the problem.

Given the angles in a rhombus:

1. ${88}^{\circ }$
2. $(2x+10{)}^{\circ }$
3. $(-3y-2{)}^{\circ }$
4. $(-6z+2{)}^{\circ }$

Properties of a Rhombus

• Opposite angles are equal.
• The sum of all angles in the rhombus is ${360}^{\circ }$.

From this, we have:

1. One angle is ${88}^{\circ }$.
2. Therefore, the opposite angle is also ${88}^{\circ }$.

Let’s analyze the other angles:

• The angle $(2x+10{)}^{\circ }$ must also equal ${88}^{\circ }$.
• The angle $(-3y-2{)}^{\circ }$ will be the same as the other complementary angle, which can be computed as follows: ${360}^{\circ }-{88}^{\circ }-{88}^{\circ }={184}^{\circ }$.

This leads to:

• Thus, we have: $$2x+10=88$$ $$-3y-2=184$$ $$-6z+2=184$$

Step 1: Solve for $x$

Starting with $2x+10=88$: $$2x=88-10$$ $$2x=78$$ $$x=\frac{78}{2}=39$$

Step 2: Solve for $y$

Next, solve for $y$: $$-3y-2=184$$ Add 2 to both sides: $$-3y=184+2=186$$ Divide by -3: $$y=\frac{-186}{3}=-62$$

Step 3: Solve for $z$

Now solve for $z$: $$-6z+2=184$$ Subtract 2 from both sides: $$-6z=184-2=182$$ Divide by -6: $$z=\frac{-182}{6}=-30.33\text{(approximately)}$$

Final Answers

The values of $x$, $y$, and $z$ are: $$x=39,y=-62,z\approx -30.33.$$

Please verify these calculations and check any angles or information that may need adjustment. If there's still an error in the problem setup, kindly let me know how the angles are structured or given.