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babymother [125]
3 years ago
6

4. What is the value of x? Show your work! TUV ~ABC

Mathematics
1 answer:
valentinak56 [21]3 years ago
8 0

Answer:

x = 8

Step-by-step explanation:

We can see that in the picture that they are similar triangles and that CA is four times VT, so that means that all the sides in triangle TUV are four times those in triangle ABC. We now know that TU is four times AB, and 8 times 4 is 32, so we can say that 5x - 8 = 32. Now, we can add by 8 on both sides to get that 5x = 40. Now, we divide by five on both sides to get that x = 8.

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4) 90-37 = 53°

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What is the surface area of the cereal box 3 in 8 in 11 in
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Danielle and David each have a job. Danielle earns $15 an hour. She also receives a $50 weekly bonus. Danielle wrote an equation
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For starters, create an equation to show David's earnings. We can do this using Danielle's as a basis, which is set up as y=(# of hours)x+(bonus). This gives us y=12x+80. Now, as we need both their ys to be equal, we just set both equations equal to each other, making 15x+50=12x+80. Now, we solve for x, starting with 15x+50=12x+80, subtracting 12x from both sides to get 3x+50=80, subtracting 50 from both sides to get 3x=30, and dividing three from both sides to get x=10. To check, we just plug in our answer to both equations and see if the ys match up. With Danielle's equation, we get y=15(10)+50=150+50=200 and with David's equation, we get y=12(10)+80=120+80=200, proving that our answer is correct.
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4 years ago
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The diagram shows a logo.
Yakvenalex [24]

Answer:

Step-by-step explanation:

From the picture attached,

Area of ΔECD = \frac{1}{2}(\text{Base})(\text{Height})    

                       = \frac{1}{2}(EF)(CD)

From ΔEFD,

sin(32°) = \frac{\text{Opposite side}}{\text{Hypotenuse}}

sin(32°) = \frac{EF}{ED}

EF = 14 × sin(32°)

    = 7.42 cm

By cosine rule,

EC² = DE² + CD² - 2(DE)(CD)cos(32°)

EC² = 14² + 27² - 2(14)(27)cos(32°)

EC² = 196 + 729 - 641.12

EC² = 283.88

EC = 16.85 cm

Area of ΔECD = Area of ΔAEB = \frac{1}{2}(7.42)(27)

                                                   = 100.17

Area of ΔECD + Area of ΔAEB = 2(100.17)

                                                   = 200.34 cm²

Area of sector BEC = \frac{\theta}{360}(\pi r^{2})

Here, θ = central angle of the sector

Area of sector BEC = \frac{105}{360}(\pi)( EC)^{2}

                                = \frac{105\pi}{360}(16.85)^2

                                = 260.16 cm²

Area of the logo = Area of triangles AEB + Area of triangle ECD + Area of sector BEC

= 200.34 + 260.16

= 460.50

≈ 460 cm²

8 0
3 years ago
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