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Ne4ueva [31]
3 years ago
15

Complete the following proof related to the figure below.

Mathematics
2 answers:
Feliz [49]3 years ago
8 0

Answer:

HL

Step-by-step explanation:

We are given that

\angle A=\angle B=90^{\circ}

AC=BD

We have to find the postulate that is not used in line 3 reason in given proof.

In triangle MAC and BMD

Statement:\angle A=\angle B=90^{\circ}

AC=BD

Reason: Given

Statement:\angle AMC=\angle BMD

Reason: Vertical opposite angle theorem

Statement:\triangle AMC\cong \triangle BMD

Reason: AAS Postulate or LA postulate

AAS postulate: When two angles and one side of one triangle is equal to two corresponding angles and corresponding side of other triangle .Then , two triangles are congruent by AAS postulate.

LA postulate: When one acute angle and  one leg of one triangle is equal to its corresponding angle and corresponding leg of other triangle .Then, two triangles are congruent by LA postulate.

HL Postulate: When hypotenuse and one leg of one triangle is equal to its corresponding hypotenuse and corresponding leg of other triangle .Then, two triangles are congruent by HL postulate.

But we are not given MC=BM

Therefore, we cannot use HL postulate.

Statement:MC=BM

Reason:CPCT

Tju [1.3M]3 years ago
4 0

Answer:

HL

Step-by-step explanation:

Hypotenuse and leg of a right triangle (HL) :Two right triangles are called congruent if the hypotenuses and one corresponding leg are equal in both triangles.

Leg and angle of a right triangle ( LA ) : Two right triangles are called congruent if the corresponding angles and one corresponding leg are equal in both triangles,

AAS ( angle-angle-side) postulate of congruence : Two triangles are called congruent if two pairs of corresponding angles and a pair of opposite sides are equal in both triangles.

Here,

In triangles AMC and BMD,

∠A ≅ ∠B

∠AMC ≅ BMD

AC = BD

Thus, by the above definitions,

We can prove triangles AMC and BMD with help of both AAS and HL postulate,

But, we could not use HL postulate because MC = MD ( where MC and MD are the hypotenuse of triangles AMC and BMD respectively ) is not given.

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The difference of two perfect cubes is 386. If the cube root of the smaller of the two numbers is 7, then the cube root of the l
choli [55]

Answer:

9

Step-by-step explanation:

Let the two perfect cubes be x and y where x > y.

According to the given conditions:

{x}^{3}  -  {y}^{3}  = 386...(1) \\ y = 7...(2) \\ plug \: y = 7 \: in \: equation \: (1) \\  {x}^{3}  -  {7}^{3}  = 386 \\ {x}^{3}  -  343 = 386 \\ {x}^{3}     = 343  +  386 \\ {x}^{3}     = 343  +  386 \\ {x}^{3}   = 729 \\ x =  \sqrt[3]{729}  \\ x = 9

Thus the cube root of the larger number is 9.

8 0
3 years ago
PLZ HELP 70 POINTS!!!!!
marin [14]

5^(x+7)=(1/625)^(2x-13)


We move all terms to the left:


5^(x+7)-((1/625)^(2x-13))=0



Domain of the equation: 625)^(2x-13))!=0


x∈R



We add all the numbers together, and all the variables


5^(x+7)-((+1/625)^(2x-13))=0


We multiply all the terms by the denominator



(5^(x+7))*625)^(2x+1-13))-((=0


We add all the numbers together, and all the variables


(5^(x+7))*625)^(2x-12))-((=0


We add all the numbers together, and all the variables


(5^(x+7))*625)^(2x=0

not sure if this is right :/


8 0
3 years ago
Read 2 more answers
I will give brainliest to people who answer ALL 5 questions.
amid [387]
Here ya go. Hope this helps. Have a great day

8 0
2 years ago
In a right triangle ABC, CD is an altitude, such that AD=BC. Find AC, if AB=3 cm, and CD= 2 cm.
konstantin123 [22]

Consider right triangle ΔABC with legs AC and BC and hypotenuse AB. Draw the altitude CD.

1. Theorem: The length of each leg of a right triangle is the geometric mean of the length of the hypotenuse and the length of the segment of the hypotenuse adjacent to that leg.

According to this theorem,

BC^2=BD\cdot AB.

Let BC=x cm, then AD=BC=x cm and BD=AB-AD=3-x cm. Then

x^2=(3-x)\cdot 3,\\ \\x^2=9-3x,\\ \\x^2+3x-9=0,\\ \\D=3^2-4\cdot (-9)=9+36=45,\\ \\\sqrt{D}=\sqrt{45}=3\sqrt{5},\\ \\x_1=\dfrac{-3-3\sqrt{5} }{2}0.

Take positive value x. You get

AD=BC=\dfrac{-3+3\sqrt{5} }{2}\ cm.

2. According to the previous theorem,

AC^2=AD\cdot AB.

Then

AC^2=\dfrac{-3+3\sqrt{5} }{2}\cdot 3=\dfrac{-9+9\sqrt{5} }{2},\\ \\AC=\sqrt{\dfrac{-9+9\sqrt{5} }{2}}\ cm.

Answer: AC=\sqrt{\dfrac{-9+9\sqrt{5} }{2}}\ cm.

This solution doesn't need CD=2 cm. Note that if AB=3cm and CD=2cm, then

CD^2=AD\cdot DB,\\ \\2^2=AD\cdot (3-AD),\\ \\AD^2-3AD+4=0,\\ \\D

This means that you cannot find solutions of this equation. Then CD≠2 cm.

8 0
3 years ago
Read 2 more answers
Michelle buys a motorcycle for $10,000. The motorcycle loses 20% of its value each year. The depreciation of the motorcycle is s
makkiz [27]
You can use cross product to slove this equation.

3 0
3 years ago
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