In parallelogram abcd, diagonals ac and bd intersect at point e, be =2x^2-3x, and de=x^2+10. what is bd? enter your answer in the box

70

Step-by-step explanation:

In parallelogram ABCD, diagonals AC and BD intersect at point E

E is the midpoint of BD

So BE = DE

Given: BE =2x^2-3x, and DE=x^2+10

BE = DE

(solve for x)

Subtract x^2 and 10 on both sides

Factor x^2-3x-10

(x-5)(x+2) = 0

x-5 =0  so x=5

x+2=0 so x=-2

Length x cannot be negative so we ignore x=-2

Lets plug in 5 for x  and find out BE  and DE

BE =

DE =

BD = BE + DE= 35+35 = 70

So BD = 70

f(x) is going down by 4

step-by-step explanation:

(its either that or its going left by 4)

x ≥ 18 is your answer

step-by-step explanation:

let "a number" = x

(2/3)x + 17 ≥ 29

isolate the variable (x). do the opposite of pemdas (parenthesis, exponent (& root), multiplication, division, addition, subtraction).

first, subtract 17 from both sides.

(2/3)x + 17 (-17) ≥ 29 (-17)

(2/3)x ≥ 29 - 17

(2/3)x ≥ 12

next, multiply 3/2 to both sides to isolate the x

(3/2)(2/3)x ≥ 12(3/2)

x ≥ 12(3/2)

simplify. multiply, then divide

x ≥ (12 * 3)/2

x ≥ (36)/2

x ≥ 18

x ≥ 18 is your answer

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