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Ahat [919]
3 years ago
14

If Quadrilateral JKLM is a square, where points J(-6, 0), K(?, ?), L(9, -1) and M(2, 7) are known. Find the coordinates of point

K.
Group of answer choices

(-1, -8)

(5, -15)

(1,8)

(1, -8)
Mathematics
2 answers:
mixer [17]3 years ago
4 0

Answer:

-1 -8

Step-by-step explanation:

drek231 [11]3 years ago
4 0

Answer:

(1,8)

Step-by-step explanation:

only point that doesn't go through other lines

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The mark in a subject for 12 students are as follows 31`, 37 ,35,38 ,42 ,23,17,18 ,35,25, 35,29
just olya [345]

Answer:

Mean (Average): 30.416666666667

Median: 33

Range: 25

Mode: 35, appeared 3 times

Step-by-step explanation:

8 0
2 years ago
Given the area of a triangle is 25x2 + 10x - 8 and the base is 5x + 4, write the algebraic expression for the height.
agasfer [191]

Answer:

the height is 5x-2

Step-by-step explanation:

factor 25x^2+10x-8=(5x-2)(5x+4)

\frac{(5x-2)(5x+4)}{5x+4}

cancel like terms

height = 5x-2

5 0
3 years ago
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frosja888 [35]

Answer:

C. 33 square feet.

Step-by-step explanation:

to get the area of irregular polygon.

divide the area into two parts (see attached image)

total area = A1 + A2

where A = 1/2 base x height

total area = ( 1/2 * 6 * 8) + (1/2 * 6 * 3)

total area = 24 + 9

total area = 33 square feet.

therefore the answer is C. 33 square feet.

3 0
3 years ago
Read 2 more answers
Jeff bought a bottle of water for $2. He also bought some hot dogs for $3 each. Jeff did not spend more than $14 on the hot dogs
lions [1.4K]

Answer:

3H - 2 ≤ 14 I believe.

If i'm wrong forgive me

6 0
3 years ago
Read 2 more answers
Solve the following differential equations or initial value problems. In part (a), leave your answer in implicit form. For parts
shepuryov [24]

Answer:

(a) (y^5)/5 + y^4 = (t^3)/3 + 7t + C

(b) y = arctan(t(lnt - 1) + C)

(c) y = -1/ln|0.09(t + 1)²/t|

Step-by-step explanation:

(a) dy/dt = (t^2 + 7)/(y^4 - 4y^3)

Separate the variables

(y^4 - 4y^3)dy = (t^2 + 7)dt

Integrate both sides

(y^5)/5 + y^4 = (t^3)/3 + 7t + C

(b) dy/dt = (cos²y)lnt

Separate the variables

dy/cos²y = lnt dt

Integrate both sides

tany = t(lnt - 1) + C

y = arctan(t(lnt - 1) + C)

(c) (t² + t) dy/dt + y² = ty², y(1) = -1

(t² + t) dy/dt = ty² - y²

(t² + t) dy/dt = y²(t - 1)

(t² + t)/(t - 1)dy/dt = y²

Separating the variables

(t - 1)dt/(t² + t) = dy/y²

tdt/(t² + t) - dt/(t² + t) = dy/y²

dt/(t + 1) - dt/(t(t + 1)) = dy/y²

dt/(t + 1) - dt/t + dt/(t + 1) = dy/y²

Integrate both sides

ln(t + 1) - lnt + ln(t + 1) + lnC = -1/y

2ln(t + 1) - lnt + lnC = -1/y

ln|C(t + 1)²/t| = -1/y

y = -1/ln|C(t + 1)²/t|

Apply y(1) = -1

-1 = ln|C(1 + 1)²/1|

-1 = ln(4C)

4C = e^(-1)

C = (1/4)e^(-1) ≈ 0.09

y = -1/ln|0.09(t + 1)²/t|

8 0
4 years ago
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