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3 years ago

JKLM is a rhombus with an area of 75 square metres. Plz help with mark brainiest !!

Mathematics

1 answer:

Evgesh-ka [11]3 years ago

8 0

Answer:

OJ=5 m

Step-by-step explanation:

Area of rhombus=1/2 d1 *d2

75=1/2*15*d2

75*2/15=d2

10=d2

Therefore, d2=6

d2=10

JL=10

2OJ=10 [ Diagonals of rhombus bisect each other perpendicularly]

OJ=5 m

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Answer:

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4 years ago

Madison has set up a lemonade stand outside her house and sells small cups and large cups of lemonade. Each small cup holds 12 o

Troyanec [42]

Answer:

$12x+22(y+10)=1240, y=x$

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Step-by-step explanation:

let x be the number of small cups and y the number of large cups.

-Given that 10 more cups than small cups of lemonade were sold:

$12x+22(y+10)=1240$

#Before the 10 more were sold, the number of x and y sold were equal>

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7 0

3 years ago

Read 2 more answers

What is the solution to the equation One-third (x minus 2) = one-fifth (x + 4) + 2? x = 12 x = 14 x = 16 x = 26

Elena L [17]

Answer:

x = 26

Step-by-step explanation:

Given

$\frac{1}{3}$ (x - 2) = $\frac{1}{5}$ (x + 4) + 2

Multiply through by 15 to clear the fractions

5(x - 2) =3(x + 4) + 30 ← distribute and simplify both sides

5x - 10 = 3x + 12 + 30, that is

5x - 10 = 3x + 42 ( subtract 3x from both sides )

2x - 10 = 42 ( add 10 to both sides )

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x = 26

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3 years ago

Write a word problem for the equation. <br> 1/2 (x-4) = 1/3x + 2

Ganezh [65]

Answer:

Half the age of Carlos four years ago is 2 years more than one-third his current age.

Step-by-step explanation:

We want to write a word problem for the equation.

$\frac{1}{2} (x - 4) = \frac{1}{3}x + 2$

Let x represent the present age of Carlos.

Four years ago, Carlos was x-4 years.

Half his age four years ago is ½(x-4)

Two years more than one-third his current age is ⅓x+2

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4 years ago

Evaluate the line integral, where C is the given curve. (x + 6y) dx + x2 dy, C C consists of line segments from (0, 0) to (6, 1)

Dima020 [189]

Split $C$ into two component segments, $C_1$ and $C_2$ , parameterized by

$\mathbf r_1(t)=(1-t)(0,0)+t(6,1)=(6t,t)$

$\mathbf r_2(t)=(1-t)(6,1)+t(7,0)=(6+t,1-t)$

respectively, with $0\le t\le1$ , where $\mathbf r_i(t)=(x(t),y(t))$ .

We have

$\mathrm d\mathbf r_1=(6,1)\,\mathrm dt$

$\mathrm d\mathbf r_2=(1,-1)\,\mathrm dt$

where $\mathrm d\mathbf r_i=\left(\dfrac{\mathrm dx}{\mathrm dt},\dfrac{\mathrm dy}{\mathrm dt}\right)\,\mathrm dt$

so the line integral becomes

$\displaystyle\int_C(x+6y)\,\mathrm dx+x^2\,\mathrm dy=\left\{\int_{C_1}+\int_{C_2}\right\}(x+6y,x^2)\cdot(\mathrm dx,\mathrm dy)$

$=\displaystyle\int_0^1(6t+6t,(6t)^2)\cdot(6,1)\,\mathrm dt+\int_0^1((6+t)+6(1-t),(6+t)^2)\cdot(1,-1)\,\mathrm dt$

$=\displaystyle\int_0^1(35t^2+55t-24)\,\mathrm dt=\frac{91}6$

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3 years ago