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3 years ago

Just question 1A and 2A with explanation please. Lots of point and brainliest!

Mathematics

2 answers:

Yakvenalex [24]3 years ago

4 0

Answer:

1a. x + 74.5= 90

subtract both sides by 74.5

x= 15.5° the results

2a. Set up the given

x + 180=180

subtract both sides by 180

x = 0 is the results

Butoxors [25]3 years ago

3 0

1/2 = 0.5

74 & 1/2 = 74 + 1/2 = 74 + 0.5 = 74.5

The complement to this angle is 90 minus the angle, so,

complement = 90 - angle

complement = 90 - 74.5

complement = 15.5

Convert this to a mixed number

15.5 = 15 + 0.5 = 15 + 1/2 = 15 & 1/2

If the given angle is 74 & 1/2 degrees, then the complement is 15 & 1/2 degrees

<h3>Answer: 15 & 1/2 degrees</h3>

==========================================

180 is the given angle. The supplement, let's call it x, will add to the given angle to get 180.

x+180 = 180

x+180-180 = 180-180

x = 0

So if we add 0 to the given angle 180, we get the result 180

<h3>Answer: 0 degrees</h3>

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In the △ABC, the height AN = 24 in, BN = 18 in, AC = 40 in. Find AB and BC. Consider all possible cases. Case 1 : N∈BC, AB = __,

aivan3 [116]

Answer:

Case 1:

$AB = 30$

$BC = 50$

Case 2:

$AB = 15.9$

$BC = 36.7$

Case 3: Not possible

Step-by-step explanation:

Given

See attachment for illustration of each case

Required

Find AB and BC

Case 1:

Using Pythagoras theorem in ANB, we have:

$AB^2 = AN^2 + BN^2$

This gives:

$AB^2 = 24^2 + 18^2$

$AB^2 = 576 + 324$

$AB^2 = 900$

Take square roots of both sides

$AB = \sqrt{900$

$AB = 30$

To calculate BC, we consider ANC, where:

$AC^2 = AN^2 + NC^2$

$40^2 = 24^2 + NC^2$

$1600 = 576 + NC^2$

Collect like terms

$NC^2 = 1600 - 576$

$NC^2 = 1024$

Take square roots

$NC = \sqrt{1024$

$NC = 32$

So:

$BC = NC + BN$

$BC = 32 + 18$

$BC = 50$

Case 2:

Using Pythagoras theorem in ANB, we have:

$AN^2 = AB^2 + BN^2$

This gives:

$24^2 = AB^2 + 18^2$

$576 = AB^2 + 324$

Collect like terms

$AB^2 = 576 - 324$

$AB^2 = 252$

Take square roots of both sides

$AB = \sqrt{252$

$AB = 15.9$

To calculate BC, we consider ABC, where:

$AC^2 = AB^2 + BC^2$

$40^2 = 252 + BC^2$

$1600 = 252 + BC^2$

Collect like terms

$BC^2 = 1600 - 252$

$BC^2 = 1348$

Take square roots

$BC = \sqrt{1348$

$BC = 36.7$

Case 3:

This is not possible because in ANC

The hypotenuse AN (24) is less than AC (40)

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3 years ago

I need help with this.

Sveta_85 [38]

Answer:

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2 years ago

564 which statement is false? It is the sum of 56 tens and 4 ones. The number of hundreds is more than the number of tens. To ge

nirvana33 [79]

Answer:

The number of hundreds is more than the number of tens.

Step-by-step explanation:

The sum of 56 tens and 4 ones= 560 + 4 (true)

The number of hundreds is more than the number of tens.

Number of hundreds = 5

Number of tens = 56

5 is not greater than 56.

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3 years ago

Two isosceles triangles share the same base. Prove that the medians to this base are collinear. (There are two cases in this pro

Anna71 [15]

Answer:

Given: Two Isosceles triangle ABC and Δ PBC having same base BC. AD is the median of Δ ABC and PD is the median of Δ PBC.

To prove: Point A,D,P are collinear.

Proof:

→Case 1. When vertices A and P are opposite side of Base BC.

In Δ ABD and Δ ACD

AB= AC [Given]

AD is common.

BD=DC [median of a triangle divides the side in two equal parts]

Δ ABD ≅Δ ACD [SSS]

∠1=∠2 [CPCT].........................(1)

Similarly, Δ PBD ≅ Δ PCD [By SSS]

∠ 3 = ∠4 [CPCT].................(2)

But, ∠1+∠2+ ∠ 3 + ∠4 =360° [At a point angle formed is 360°]

2 ∠2 + 2∠ 4=360° [using (1) and (2)]

∠2 + ∠ 4=180°

But ,∠2 and ∠ 4 forms a linear pair i.e Point D is common point of intersection of median AD and PD of ΔABC and ΔPBC respectively.

So, point A, D, P lies on a line.

CASE 2.

When ΔABC and ΔPBC lie on same side of Base BC.

In ΔPBD and ΔPCD

PB=PC[given]

PD is common.

BD =DC [Median of a triangle divides the side in two equal parts]

ΔPBD ≅ ΔPCD [SSS]

∠PDB=∠PDC [CPCT]

Similarly, By proving ΔADB≅ΔADC we will get, ∠ADB=∠ADC[CPCT]

As PD and AD are medians to same base BC of ΔPBC and ΔABC.

∴ P,A,D lie on a line i.e they are collinear.

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3 years ago

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andrezito [222]

Hello :
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3 years ago