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Rudiy27
2 years ago
12

AD=3, BD=5, AE=4. Find CE

Mathematics
1 answer:
EastWind [94]2 years ago
5 0

Answer:

6.6

Step-by-step explanation:

Triangles ADE and ABC are similar.

Therefore,

AD/AB=AE/AC

3/8=4/AC

3AC=32

AC=10.6(3 s.f)

CE=10.6-4=6.6

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4x2x4 - 2×7×2 + 40

64 - 28 + 40=76

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Find a round to the nearest tenth a 150 12 10cm c a=?cm
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Which expressions can be used to find the volume of the pyramid? Select three choices
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Step-by-step explanation:

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3 years ago
A triangle has base 2x+1 and height 6x-3. What value of x would give an area of 240m2?
GREYUIT [131]

Answer:

The values of x which would give an area of 240m² would be:

x=-\frac{\sqrt{161}}{2},\:x=\frac{\sqrt{161}}{2}

Step-by-step explanation:

Given

The base of triangle b = 2x+1

The height of triangle h = 6x-3

The Area of the triangle A = 240 m²

The Area of the triangle has the formula

A = 1/2 × b × h

substituting b = 2x+1, h = 6x-3 and A = 240

240\:=\:\frac{1}{2}\:\left(2x+1\right)\:\times \left(6x-3\right)

480=\left(2x+1\right)\left(6x-3\right)

480=12x^2-3

Subtract 480 from both sides

12x^2-3-480=480-480

12x^2-483=0

3\left(4x^2-161\right)=0

3\left(2x+\sqrt{161}\right)\left(2x-\sqrt{161}\right)=0

Using the zero factor principle

if ab=0, then a=0 or b=0 (or both a=0 and b=0)

2x+\sqrt{161}=0\quad \mathrm{or}\quad \:2x-\sqrt{161}=0

solving

2x+\sqrt{161}=0

2x=-\sqrt{161}

Divide both sides by 2

\frac{2x}{2}=\frac{-\sqrt{161}}{2}

x=-\frac{\sqrt{161}}{2}

also solving

2x-\sqrt{161}=0

2x=\sqrt{161}

Divide both sides by 2

\frac{2x}{2}=\frac{\sqrt{161}}{2}

x=\frac{\sqrt{161}}{2}

Therefore, the values of x which would give an area of 240m² would be:

x=-\frac{\sqrt{161}}{2},\:x=\frac{\sqrt{161}}{2}

3 0
3 years ago
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