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3 years ago

For triangle TRI the following facts are given:

1 answer:

5 0

For the answer to the question above,b.Triangle TAN being an image of triangle TRI is smaller than triangle TRI, and so is a contraction.while,
c.) TA/AN = TR/RI3.3/6 = (3.3 + AR)/86(3.3 + AR) = 8 * 3.3 = 26.43.3 + AR = 26.4/6 = 4.4AR = 4.4 - 3.3 = 1.1AR = 1.1
d.) TA/TN = TR/TI3.3/TN = (3.3 + 1.1)/(TN + 1.6)3.3(TN + 1.6) = 4.4TN3.3TN + 5.28 = 4.4TN4.4TN - 3.3TN = 5.281.1TN = 5.28TN = 5.28/1.1 = 4.8TN = 4.8I hope my answer helped you.

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Bad White [126]

Answer:

Should be $160

<h3>Step-by-step explanation:</h3>

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3 years ago

Read 2 more answers

Two angles are supplementary. If 5 times of one angle is 10 times of the other angle, find the two angles.

Luba_88 [7]

5x + 10x = 180
15x = 180
X = 12
5 x 12 = 60
10 x 12 = 120

The angles are 60° and 120°

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3 years ago

1 The prism-shaped roof has equilateral triangular bases. Use the model you created in question #1 to calculate the height of th

Diano4ka-milaya [45]

Answer:

Step-by-step explanation:

Step-by-step explanation: As shown in the attached figure, the prism-shaped roof has equilateral triangular bases, one of which is ΔABC. We need to create an equation that models the height of one of the roof's triangular bases in terms of its sides. Let ii be AD.

See the figure attached herewith, ΔABC forms an equilateral triangle, in which AD is the height. So, D will be the mid-point of BC and ∠ADB = ∠ADC = 90°.

Now, in ΔADB, we have

AD^2=AB^2-BD^2

AD^2=AB^2-(1/2AB^2)^2

AD=√3/4AB^2

we can find the height of any one of the roof's triangular bases.

2.1. Check picture 1. Let the one side of the triangle be a, drop one perpendicular, CD. Then triangle ADB is a right triangle, with hypothenuse a and one side equal to 1/2a. By the Pythagorean theorem, as shown in the picture, the height is √3/2a

2. if a=25 ft, then the height is √3/2a=√3/2*25=1.732/2*25=21.7(ft)

3. consider picture 2. Let the length of the roof be l feet.

one side of the prism (the roof) is a rectangle with dimensions a and l, so the area of one side is a*l

the lateral Area of the roof is 3a*l

the area of the equilateral surfaces is 2*(1/2*a*√3/2a)=√3/2a^2

so the total area of the roof is

4. The total area was the 2 triangular surfaces + the 3 equal lateral rectangular surfaces. Now instead of 3 lateral triangular surfaces, we have 2.

So the total area found previously will be decreased by al

5. so the area now is √3/2a^2 + 2al

6. now a=25 and l=2a=50

Area= √3/2a^2+2al=√3/2*25^2+2*25*50=25^2(√3/2+4)=625*4.866

=3041.3 (ft squared)

6 0

3 years ago

HeLP!! will mark brainlest!

shtirl [24]

Perimeter=54 a.2ft b.9 missing length=4 a.light rectangle=15 dark rectangle=10 b. sum=25 5*5=25

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3 years ago

Triangle RST has vertices R(2, 0), S(4, 0), and T(1, –3). The image of triangle RST after a rotation has vertices R'(0, –2), S'(

sleet_krkn [62]

Answer with explanation:

Pre -image= Vertices of Δ R ST=R(2, 0), S(4, 0), and T(1, –3)

Image of Δ R ST after rotation= R'(0, –2), S'(0, –4), and T'(–3, –1)

Pre-Image lies in Fourth Quadrant and Image lies in Third Quadrant.

If triangle is rotated by different angles in anticlockwise direction,then

$(a,b)_{90^{\circ}}=(-b, a)\\\\(a,b)_{180^{\circ}}=(-b, -a)\\\\(a,b)_{270^{\circ}}=(b, -a)$

If triangle is rotated by different angles in Clockwise direction,then

$(a,b)_{90^{\circ}}=(b, -a)\\\\(a,b)_{180^{\circ}}=(-b, -a)\\\\(a,b)_{270^{\circ}}=(-b, a)$

⇒→So, Pre image that is ,Δ R ST having vertices ,R(2, 0), S(4, 0), and T(1, –3) when rotated by either 90° in clockwise direction or by 270°, in anticlockwise Direction to get Image Δ R' S'T' having vertices R'(0, –2), S'(0, –4), and T'(–3, –1) .

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3 years ago