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Bad White [126]
3 years ago
11

The perimeter of an isosceles trapezoid is 28 in and the ratio of lengths of its bases is 5:3. Find the lengths of the sides of

the trapezoid if its diagonal bisects the angle at the longer base. Answer:
Mathematics
1 answer:
Natalka [10]3 years ago
8 0
The measurements are:
b₁ = 8.4 inches; b₂ = 14 inches; leg₁ = 2.8 inches; leg₂ = 2.8 inches

Let x = length of base and leg

The formula for perimeter of isosceles trapezoid is P = b₁ + b₂ + 2(leg)
Where: b = base

       P = b₁ + b₂ + 2(leg)
     28 = 3x + 5x + 2(x)
     28 = 8x + 2x
28/10 = 10x/10
    2.8 = x

Now, substitute the values:
       P = b₁ + b₂ + 2(leg)
     28 = 3(2.8) + 5(2.8) + 2(2.8)
     28 = 8.4 + 14 + 5.6
     28 = 28

Hence the measurements are:
b₁ = 8.4 inches; b₂ = 14 inches; leg₁ = 2.8 inches; leg₂ = 2.8 inches
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Alexus [3.1K]

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the answer is 95

Step-by-step explanation:

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4 0
3 years ago
Can you solve 5a please greatly appreciated.
timurjin [86]
Part I)
 
The module of vector AB is given by:
 lABl = root ((- 3) ^ 2 + (4) ^ 2)
 lABl = root (9 + 16)
 lABl = root (25)
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 Part (ii)
 The module of the EF vector is given by:
 lEFl = root ((5) ^ 2 + (e) ^ 2)
 We have to:
 lEFl = 3lABl
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3 0
3 years ago
What expressions is equivalent to -6(-2/3 + 2x)
yuradex [85]

Answer:

=−12x+4

Step-by-step explanation:

−6( −2 /3 +2x)

=(−6)( −2 /3 +2x)

=(−6)( −2 /3 )+(−6)(2x)

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6 0
3 years ago
I need help Please help.
asambeis [7]

Answer:

Step-by-step explanation:

Number of students 10

Problem 1. $625 for the bus hire per friday, So 625*4=$2500

Problem 2. 2500/25=$100 each for the whole 4 weeks

Problem 3.

10 students tickets 220= 2200 for all tickets. The bus, 625/10 = $62.5*4= $250 dollars for the whole 4 weeks for the bus so in all each student pays $470 each

20 students, tickets 220=4400 for all tickets. The bus, 625/20=$31.25*4=$125 for the whole 4 weeks for the bus, so in all each student must pay $345 each

30 students, tickets 220 = 6600 for all tickets. The bus, 625/30 =$20.83*4=$83.32 for the whole 4 weeks for the bus, so in all each student must pay $303.32 each

41 students, tickets $160=$6560 for all tickets. The bus, because you need 2 buses at 625 each so $1250 for both buses 1250/41= 30.49*4=$121.96 for the whole 4 weeks for the bus. So in al each student must pay $281.96 each

Hope this is correct

8 0
3 years ago
<img src="https://tex.z-dn.net/?f=f%28x%29%20%3D%20%E2%88%923x2%E2%88%9211x%2B7" id="TexFormula1" title="f(x) = &minus;3x2&minus
bixtya [17]

Answer:

x=   \frac{ - 11 +  \sqrt{205} }{6}  \:  \:  \:  \:  \: or \:  \:  \:  \:  \\x  =  \frac{ - 11 -  \sqrt{205} }{6}

Step-by-step explanation:

- 3x {}^{2}  - 11x + 7 = 0 \\ 3x {}^{2}  + 11x - 7 = 0

here,

a = 3 \\ b = 11 \\ c =  - 7

now,

D = b {}^{2}  - 4ac \\  = (11) {}^{2}  - 4(3)( - 7) \\  = 121 + 84 \\  = 205  > 0 \\

so,

x =  \frac{ - b +  -  \sqrt{b {}^{2}  - 4ac} }{2a}  \\  \\  =   \frac{ - 11 +  -  \sqrt{205} }{2(3)}  \\  =   \frac{ - 11 +  -   \sqrt{205}  }{6}  \\  \\  =   \frac{ - 11 +  \sqrt{205} }{6}  \:  \:  \:  \:  \: or \:  \:  \:  \:  =  \frac{ - 11 -  \sqrt{205} }{6}

4 0
3 years ago
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