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3 years ago

A square photograph has a perimeter of 20 inches what is the area of the photograph

1 answer:

7 0

Since it's a square it must have 4 sides.
20÷4=5

area=l×w or b×h
5×5=25 (the sides of a square have the same length)

the area of the photograph is 25in²

hope this helped

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An object of mass weighing 5.24 kilograms is raised to a height of 1.63 meters. What is the potential energy of the object at th

Trava [24]

Find force,
9.8 m/s2 x 5.24 kg = 51.352 N 

Find work,
51.352 N x 1.63 m= 83.70 N*m

Example:
To solve this given word problem we can first identify the given and the apt formula to use in this phenomenon: Given: Force = 4, 500 N = 4, 500 kg-m/s^2 Acceleration = 5 m/s^2 
Formula: f=ma 
Derivation: m = f/a 
Solution: 
1. M = f/a 
2. M = 4,500 kg-m/s^2 / 5 m/s^2 
3. M = 900 kg 

Hence, the object’s mass is 900 kg.

6 0

3 years ago

Read 2 more answers

Help Please! Solve for d. 5+d>5−d Solve for p. 2p+3>2(p−3)

Keith_Richards [23]

Solve for d.
5+d>5−d Subtract 5 from both sides of this inequality:

d>d There is no value for d that satisfies this inequality.
No value can be greater than itself.

Solve for p.
2p+3>2(p−3) Multiply this out: 2p+3>2p-6
 Subtr 3 from both sides: 2p> 2p-9

This is equivalent to 2p+9>2p.
We could subtr. 2p from both sides: 0>-9.

0> -9 is always true. Thus, the given inequality has infinitely many solutions.

3 0

3 years ago

Find the length of each segment with endpoints A and B. Select the exact answer. (This means that it is either an integer or in

kondaur [170]

It’s b bucko I did that

4 0

3 years ago

In the circle below, AD is a diameter and AB is tangent at A. suppose mADC=228. Find the measures of mCAB and mCAD. Type your nu

Tju [1.3M]

Answer:

m∠CAB = 66°

m∠CAD = 24°

Step-by-step explanation:

m∠CAB

The given parameters are;

The measure of arc m $\widehat{ADC}$ = 228°

The diameter of the given circle = $\overline{AD}$

The tangent to the circle = $\underset{AB}{\leftrightarrow}$

The measure of m∠CAB and m∠CAD = Required

By the tangent and chord circle theorem, we have;

m∠CAB = (1/2) × m $\widehat{AC}$

However, we have;

m $\widehat{AC}$ + m $\widehat{ADC}$ = 360° the sum of angles at the center of a circle is 360°

∴ m $\widehat{AC}$ = 360° - m $\widehat{ADC}$

Which gives;

m $\widehat{AC}$ = 360° - 228° = 132°

m $\widehat{AC}$ = 132°

Therefore;

m∠CAB = (1/2) × 132° = 66°

m∠CAB = 66°

m∠CAD

Given that $\overline{AD}$ is the diameter of the given circle, we have

The tangent, $\underset{AB}{\leftrightarrow}$ , is perpendicular to the radius of the circle, and therefore $\underset{AB}{\leftrightarrow}$ is also perpendicular to the diameter of the circle

∴ m∠DAB = 90° which is the measure of the angle formed by two perpendicular lines

By angle addition property, we have;

m∠DAB = m∠CAB + m∠CAD

∴ m∠CAD = m∠DAB - m∠CAB

By substitution, we have;

m∠CAD = 90° - 66° = 24°

m∠CAD = 24°

7 0

3 years ago

Let PQR be a triangle such that angle P = angle Q + angle R. Among all three exterior angles of triangle PQR, what is the smalle

Phantasy [73]

Answer:

The smaller exterior angle is the exterior angle in the vertex P, and it measures 90°

Step-by-step explanation:

The sum of the internal angles of a triangle needs to be 180 degrees, so we have two equations:

$mP = mQ + mR$

$mP + mQ + mR = 180\°$

Substituting $mQ + mR$ by $mP$ in the second equation, we have:

$mP + mP = 180\°$

$mP = 90\°$

The other two angles need to be lesser than mP, and the exterior angle is the supplement of the internal angle, so the bigger the internal angle, the smaller the exterior angle.

So if the bigger internal angle in this triangle is mP, the smaller exterior angle is also the angle in the vertex P:

$exteriorP = 180\° - mP$

$exteriorP = 90\°$

7 0

3 years ago