# Module 4 lesson 10

## Module 4 B: Lessons 10-14

### Eureka Math Grade 4 Module 4 Lesson 10 Problem Set Answer Key

Write an equation, and solve for the measure of ∠x. Verify the measurement using a protractor.

Question 1.
∠CBA is a right angle.

45° + ________ = 90°
x° = __________
The value of x is 45°.

Explanation:
Given that <CBA is a right angle and angle B is 45°, so the angle CBA= 90° which is 45°+x= 90°, so the value of x is
x°= 90° – 45°
= 45°
So the value of x° is 45°.

Question 2.
∠GFE is a right angle.

_______ + ________ = _________
x° = __________
The value of x is 70°.

Explanation:
Given that <GFE is a right angle which is 90° and another angle is 20°, so the angle GFE= 90° which is 20°+x= 90°, so the value of x is x°= 90° – 20°
= 70°.
So the value of x° is 70°.

Question 3.
∠IJK is a straight angle

___________ + 70° = 180°
x° = ____________
The value of x is 110°

Explanation:
Given that <IJK is a straight angle which is 180° and another angle is 70°, so the angle IJK= 180° which is 70°+x= 180°, so the value of x is
x°= 180° – 70°
= 110°.
So the value of x° is 110°.

Question 4.
∠MNO is a straight angle

_________ + _________ = __________
x° = ___________
The value of x is 97°

Explanation:
Given that <MNO is a straight angle which is 180° and another angle is 83°, so the angle MNO= 180° which is 83°+x= 180°, so the value of x is
x°= 180° – 83°
= 97°.
So the value of x° is 97°.

Solve for the unknown angle measurements. Write an equation to solve.

Question 5.
Solve for the measurement of ∠TRU. ∠QRS is a straight angle.

The value of <TRU is 144°.

Explanation:
Given that <QRS is a straight angle which is 180° and another angle is 36°, so the angle MNO= 180° which is 36°+x= 180°, so the value of x is
x°= 180° – 36°
= 144°.
So the value of <TRU is 144°.

Question 6.
Solve for the measurement of ∠ZYV. ∠XYZ is a straight angle.

The value of <ZYV is 12°.

Explanation:
Given that <XYZ is a straight angle which is 180° and another two angle is 108°+60° which is 168°, so the angle XYZ= 180° which is 168°+x= 180°, so the value of x is
x°= 180° – 168°
= 12°.
So the angle ZYV is 12°.

Question 7.
In the following figure, ACDE ¡s a rectangle. Without using a protractor, determine the measurement of ∠DEB. Write an equation that could be used to solve the problem.

The value of x° is 63°.

Explanation:
Here, in the above image we can see that a rectangle where every angle is a right angle, so let’s take one angle as 90° and then we should find out the other angle and the other angle be x. So the equation will be 90°= 27°+x° and the value of x is
x= 90° – 27°
= 63°.
So the value of x° is 63°.

Question 8.
Complete the following directions in the space to the right.
a. Draw 2 points: M and N. Using a straightedge, draw .
b. Plot a point O somewhere between points M and N.
c. Plot a point P, which is not on .
d. Draw $$\overline{O P}$$.
e. Find the measure of ∠MOP and ∠NOP.
f. Write an equation to show that the angles add to the measure of a straight angle.
The equation of the angles add to the measure of a straight angle is <MOP+<PON= 180°.

Explanation:
Here, we have plotted 2 points which are M and N using a straightedge, and have constructed .
Then we plotted a point O somewhere between points M and N.
Then we need to plot a point P, which is not on .
And now we will construct $$\overline{O P}$$.
So, now we need to find the measure of ∠MOP and ∠NOP.
As we have plotted a straight line which is <MON is 180°, which is <MOP+<PON= 180°.
So the equation of the angles add to the measure of a straight angle is <MOP+<PON= 180°.

### Eureka Math Grade 4 Module 4 Lesson 10 Exit Ticket Answer Key

Write an equation, and solve for x. ∠TUV is a straight angle.

Equation = ______________
x° = ______________
The value of x is 60°.

Explanation:
Given that <TUV is a straight angle which is 180° and another two angle is 53°+67° which is 120°, as the angle TUV= 180° which is 168°+x= 180°, so the value of x is
x°= 180° – 120°
= 60°.
So the angle x is 60°.

Write an equation, and solve for the measurement of ∠x. Verify the measurement using a protractor
Question 1.
∠DCB is a right angle

___________ + 35° = 90°
x° = ___________
The value of x° is 55°.

Explanation:
Given that <DCB is a right angle and other angle is 35°, so the angle DCB is 90° which is 35°+x= 90°, so the value of x is
x°= 90° – 35°
= 55°
So the value of x° is 55°.

Question 2.
∠HGF is a right angle

___________ + ___________ = ___________
x° = ___________

The value of x° is 28°.

Explanation:
Given that <HGF is a right angle and other angle is 62°, so the angle HGF is 90° which is 62°+x= 90°, so the value of x is
x°= 90° – 62°
= 28°
So the value of x° is 28°.

Question 3.
∠JKL is a straight angle.

145° + ___________ = 180°
x° = ___________
The value of x is 35°.

Explanation:
Given that <JKL is a straight angle which is 180° and another angle is 145° and the angle JKL is 180° which is 145°+ x= 180°, so the value of x is
x°= 180° – 145°
= 35°.
So the angle x is 35°.

Question 4.
∠PQR is a straight angle.

___________ + ___________ = ___________
x° = ___________
The value of x is 164°.

Explanation:
Given that <PQR is a straight angle which is 180° and another angle is 16° as the angle PQR is 180° which is 16°+x= 180°, so the value of x is
x°= 180° – 16°
= 164°.
So the angle x is 164°.

Write an equation, and solve for the unknown angle measurements.

Question 5.
Solve for the measurement of ∠USW. ∠RST is a straight angle.

The value of x is 75°.

Explanation:
Given that <RST is a straight angle which is 180° and another two angle is 70°+35° which is 105°, as the angle RST= 180° and the equation is 105°+x= 180°, so the value of x is
x°= 180° – 105°
= 75°.
So the angle x is 75°.

Question 6.
Solve for the measurement of ∠OML. ∠LMN is a straight angle.

The value of <OML is 35°.

Explanation:
Given that <LMN is a straight angle which is 180° and another two angle is 72°+73° which is 145°, as the angle LMN= 180° and the equation is 145°+x= 180°, so the value of x is
x°= 180° – 145°
= 35°.
So the angle <OML is 35°.

Question 7.
In the following figure, DEFH is a rectangle. Without using a protractor, determine the measurement of ∠GEF Write an equation that could be used to solve the problem.

The value of x° is 16° and the equation is 90°= 74°+x°.

Explanation:
Here, in the above image we can see that a rectangle where every angle is a right angle, so let’s take one angle as 90° and then we should find out the other angle and the other angle be x. So the equation will be 90°= 74°+x° and the value of x is
x= 90° – 74°
= 16°.
So the value of x° is 16° and the equation is 90°= 74°+x°.

Question 8.
Complete the following directions in the space to the right.
a. Draw 2 points: Q and R. Using a straightedge, draw .
b. Plot a point S somewhere between points Q and R.
c. Plot a point T, which is not on
d. Draw $$\overline{T S}$$.
e. Find the measure of ∠QST and ∠RST.
f. Write an equation to show that the angles add to the measure of a straight angle.
The equation of the angles add to the measure of a straight angle is <QST+<STR= 180°.

Explanation:
Here, we need to draw 2 points which is Q and R using a straightedge and then we need to construct .
Then we need to plot a point S somewhere between points Q and R.
And then we need to plot a point T, which is not on
Then we will construct $$\overline{T S}$$.
Then we need to find the measure of ∠QST and ∠RST.
As we have plotted a straight line which is <QSR is 180°, which is <QST+<STR= 180°.
So the equation of the angles add to the measure of a straight angle is <QST+<STR= 180°.

### Grade 7 Mathematics Module 4, Topic B, Lesson 10

Student Outcomes

• Students solve simple interest problems using the formula I = Prt, where I = interest, P = principal, r = interest rate, and t = time.
• When using the formula I = Prt, students recognize that units for both interest rate and time must be compatible; students convert the units when necessary.

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### Common Core Learning Standards

CCLS State Standard
7.RP.3 Use proportional relationships to solve multistep ratio and percent problems. Examples: simple...
7.EE.3 Solve multi-step real-life and mathematical problems posed with positive and negative rational...
Grade K Module 4 Lesson 10

Then she buried her nose in his shoulder and burst into tears. Quietly and hopelessly, mourning what was not and what will never be. Kirill looked up awkwardly. It was quiet in the Wedding Palace.

## 10 module 4 lesson

Unbuttoned jacket, as I remember now, red, and really stroking my dick. No DUCK !!. I looked closer and saw that he was doing this in women's panties. antennae !!.

EngageNY Grade 5 Module 4 Lesson 10

The chest was very small, it stood out weakly on the thin chest, only large nipples and halos indicated that it was a. Woman's breast. But the ass and legs were just space.

### Now discussing:

Long time. He was intended, rather for flirting, teased with his openness, so when the nipples began to harden rapidly, he quickly capitulated. Not only were the halos of her breasts noticeable, her. Breast tenderness was always her strong weakness. The endings of the hemispheres in moments of intense excitement in Ira stretched out and became dark, like small plums.

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