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valina [46]
3 years ago
6

B. what is a real-world example or use of surface area?

Mathematics
1 answer:
Anuta_ua [19.1K]3 years ago
5 0

Answer:

Surface area : It is the total area of a surface

The surface area is used in many places.

1) If we want to paint a wall, we have to find the total surface area of the wall.  

2). If we want to place tiles in our floor we want to find surface area.

3)If we want to roof our houses we have to find the surface area.



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Answer:

a to the top left hope this help

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2 years ago
Which of the following statements is true if m∠E = m∠Y and m∠F = m∠X?
mars1129 [50]

Answer:

segment FE over segment XY equals segment EG over segment YZ equals segment GF over segment ZX

Step-by-step explanation:

see the attached figure to better understand the problem  

we know that  

If m∠E = m∠Y and m∠F = m∠X  

then  

Triangles EFG and YXZ are similar by AA Similarity Theorem  

Remember that  

If two figures are similar, then the ratio of its corresponding sides is proportional and its corresponding angles are congruent  

In this problem  

The corresponding sides are  

FE and XY  

EG and YZ  

GF and ZX  

so therefore  

segment FE over segment XY equals segment EG over segment YZ equals segment GF over segment ZX Firstly , let us learn about trigonometry in mathematics.  

Suppose the ΔABC is a right triangle and ∠A is 90°.  

sin ∠A = opposite / hypotenuse

cos ∠A = adjacent / hypotenuse

tan ∠A = opposite / adjacent

Let us now tackle the problem!  

A similar triangle has the same angle, in other words the triangle has the same shape but different sizes.  

From the figure in the attachment , we can conclude that:  

m∠E = m∠Y  

m∠F= m∠X  

m∠G = m∠Z  

∴ ΔEFG ~ ΔYXZ  ( ΔEFG is similar to ΔYXZ )    

Because of the similarity , then:  

FE : XY = EG : YZ = GF : ZX  

Conclusion:

ΔFG is similar to ΔYXZ.  

Segment FE over segment XY equals segment EG over segment YZ equals segment GF over segment ZX , i.e:    

Keywords: Sine , Cosine , Tangent , Opposite , Adjacent , Hypotenuse , Triangle , Fraction , Lowest , Function , Angle

3 0
2 years ago
Read 2 more answers
Please help with geometry. 20 points, and another question worth 98 on my profile!
vovikov84 [41]

Answer:

Part 1) m∠1 =(1/2)[arc SP+arc QR]

Part 2) PR^{2} =PS*PT

Part 3) PQ=PR

Part 4) m∠QPT=(1/2)[arc QT-arc QS]

Step-by-step explanation:

Part 1)

we know that

The measure of the inner angle is the semi-sum of the arcs comprising it and its opposite.

we have

m∠1 -----> is the inner angle

The arcs that comprise it and its opposite are arc SP and arc QR

so

m∠1 =(1/2)[arc SP+arc QR]

Part 2)

we know that

The <u>Intersecting Secant-Tangent Theorem,</u> states that the square of the measure of the tangent segment is equal to the product of the measures of the secant segment and its external secant segment.

so

In this problem we have that

PR^{2} =PS*PT

Part 3)

we know that

The <u>Tangent-Tangent Theorem</u>  states that if from one external point, two tangents are drawn to a circle then they have equal tangent segments

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PQ=PR

Part 4)

we know that

The measurement of the outer angle is the semi-difference of the arcs it encompasses.

In this problem

m∠QPT -----> is the outer angle

The arcs that it encompasses are arc QT and arc QS

therefore

m∠QPT=(1/2)[arc QT-arc QS]

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You take (8x - 6) then add 80 =90 then solve for x
8x-6+80 =90
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The ratio of the number of text messages sent by Lucas to the number of text messages sent by his sister is 3 to 4. Lucas sent 1
vitfil [10]

Answer:

the answer is 24

Step-by-step explanation:

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7 0
3 years ago
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