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i need help with some pre calc. plz?

6. a triangle with an area of 16 has a base of (x+1) and a height of (x-13) fine the value of x along with the base and the height of the triangle.

1. between what two consecutive intergers is the value of log 66

2

7. express (6-3i)^2 in a+bi form.

2. which equation represents a circle with a center of (2,-3) and a radius of 6

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  • 1 decade ago
    Favorite Answer

    6.

    A = 1/2bh

    16 = 1/2(x + 1)(x - 13) = 1/2(x² -12x -13) = 1/2x² - 6x - 13/2

    32 = x² - 12x - 13

    0 = x² - 12x - 45

    0 = (x - 15)(x + 3)

    Throw out -3, so x = 15, the base is 16 and the height is 2.

    2)

    log 66 is between 1 and 2, because it's bigger than 10 (log 10 = 1) and less than 100 (log 100 = 2).

    3)

    (6 - 3i)² = 36 - 36i + 9i² = 36 - 36i - 9 = 27 - 36i

    4)

    (x - h)² + (y - k)² = r²

    (x - 2)² + (y + 3)² = 6²

    or in "standard form":

    x² - 2x + 4 + y² + 6x + 9 = 36

    x² - 2x + y² + 6x - 23 = 0

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  • 1 decade ago

    6. a triangle with an area of 16 has a base of (x+1) and a height of (x-13) fine the value of x along with the base and the height of the triangle

    A=16=.5bh=.5(x+1)(x-13)

    32=x^2-12x-13

    x^2-12x-45=0

    (x-15)(x+3)

    only x-15=0

    x=15 has physical meaning

    so x=15

    b=x+1=16

    h=x-13=2

    1. between what two consecutive intergers is the value of log 66

    log 10=1

    log 100=2

    66 is between 10 & 100, so log 66 is between 1 & 2.

    7. express (6-3i)^2 in a+bi form.

    36-2*3*6i+9i^2

    36-9-36i

    25-36i

    2. which equation represents a circle with a center of (2,-3) and a radius of 6

    (x-2)^2+(y+3)^2=36

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  • 1 decade ago

    6. you just need to set up an equation and solve for x. area of a triangle = 1/2 base * height so...

    16 = ((x+1)/2) * (x-13)

    simplify this function by multiplying each side by 2.

    32 = (x + 1) * 2(x-13)

    you should be able to solve from here

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  • aeiou
    Lv 7
    1 decade ago

    Area Triangle = a base x height

    Area triangle = (x + 1)(x -13)

    X' = -1

    x" = 13

    <>

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  • 1 decade ago

    6.)

    A = (1/2)bh

    16 = (1/2)(x + 1)(x - 13)

    32 = x^2 - 13x + x - 13

    x^2 - 12x - 45 = 0

    (x - 15)(x + 3) = 0

    x = 15 or -3

    Since you can't have a negative length

    ANS : 16in base and 2 in height.

    ----------------------------------------------------------

    1.) log(66) = 1.82

    ANS : 1 and 2

    -------------------------------------

    7.)

    (6 - 3i)^2

    (6 - 3i)(6 - 3i)

    36 - 18i - 18i + 9i^2

    36 - 36i + 9(-1)

    36 - 36i - 9

    27 - 36i

    -----------------------------------

    2.)

    (x - 2)^2 + (y - (-3))^2 = 6^2

    (x - 2)^2 + (y + 3)^2 = 36

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