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3 years ago

6

A garden has an area of 286ft. It’s length is 9ft more than it’s width. What are the dimensions of the garden?

1 answer:

Gelneren [198K]3 years ago

3 0

Answer:

13

Step-by-step explanation:

x × (x + 9) = x² + 9x = 286
x² + 9x - 286 = 0
(x - 13)(x + 22)
x = 13 or -22
Since you cannot have negative length, it’s 13

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The distance from the centroid of a triangle to its vertices are 16cm, 17cm, and 18cm. What is the length of the shortest median

Marizza181 [45]

Answer:

$24$ $\text{cm}$

Step-by-step explanation:

Given: The distance from the centroid of a triangle to its vertices are $16\text{cm}$ , $17\text{cm}$ , and $18\text{cm}$ .

To Find: Length of shortest median.

Solution:

Consider the figure attached

A centroid is an intersection point of medians of a triangle.

Also,

A centroid divides a median in a ratio of 2:1.

Let G be the centroid, and vertices are A,B and C.

length of $\text{AG}$ $=16\text{cm}$

length of $\text{BG}$ $=17\text{cm}$

length of $\text{CG}$ $=18\text{cm}$

as centrod divides median in ratio of $2:1$

length of $\text{AD}$ $=\frac{3}{2}\text{AG}$

$=\frac{3}{2}\times16$

$=24\text{cm}$

length of $\text{BE}$ $=\frac{3}{2}\text{BG}$

$=\frac{3}{2}\times17$

$=\frac{51}{2}\text{cm}$

length of $\text{CF}$ $=\frac{3}{2}\text{CG}$

$=\frac{3}{2}\times18$

$=27\text{cm}$

Hence the shortest median is $\text{AD}$ of length $24\text{cm}$

7 0

3 years ago

I have a math problem:<br> I also attached the possible values I was given. Please Help!

vekshin1

Answer:

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8 0

3 years ago

How can i do this equation 8y-(5y+2)=16

AlladinOne [14]

Answer:

y=6

Step-by-step explanation:

8y-5y-2=16

8y-5y=16+2

3y=18

y=18/3

y=6

4 0

4 years ago

Read 2 more answers

Find Mx, My, and (x, y) for the lamina of uniform density rho bounded by the graphs of the equations. y = x2/3, y = 0, x = 1

erik [133]

Answer:

$\mathbf{(\overline x , \overline y ) = (\dfrac{5}{8}, \dfrac{5}{14})}$

Step-by-step explanation:

Given that:

$y = x^{2/3}$ at y = 0 , x = 1

Then:

Area = $\int^{1}_{0} x^{2/3} \ dx$

Area = $\begin {bmatrix} \dfrac{3}{5}x^{5/3} \end {bmatrix} ^1_0$

Area = $\dfrac{3}{5}$

Then:

$\overline x = \dfrac{1}{A} \int^b_a x (f(x) -g(x) ) \ dx$

$\overline x = \dfrac{5}{3} \int^1_0 x (x^{2/3} -0 ) \ dx$

$\overline x = \dfrac{5}{3} \int^1_0 x^{5/3} \ dx$

$\overline x = \dfrac{5}{3} \ [\dfrac{3}{8}x^{8/3}]^1_0$

$\overline x = \dfrac{5}{3} \times \dfrac{3}{8}$

$\overline x = \dfrac{5}{8}$

Similarly;

$\overline y = \dfrac{1}{A} \int^b_a \dfrac{1}{2} \begin{bmatrix} (f(x)^)2 - (g(x))^2 \end {bmatrix} \ dx$

$\overline y = \dfrac{5}{3} \int^1_0 \dfrac{1}{2} \begin{bmatrix} (f(x^{2/3})^2 -0 \end {bmatrix} \ dx$

$\overline y = \dfrac{5}{3} \int^1_0 \dfrac{1}{2} \begin{bmatrix} (x^{4/3} ) \end {bmatrix} \ dx$

$\overline y = \dfrac{5}{3} \begin{bmatrix} \dfrac{1}{2} (x^{7/3} ) \times \dfrac{3}{7} \end {bmatrix} ^1_0$

$\overline y = \dfrac{5}{3} \begin{bmatrix} \dfrac{3}{14} (x^{7/3} ) \end {bmatrix} ^1_0$

$\overline y = (\dfrac{5}{3} \times \dfrac{3}{14} )$

$\overline y = \dfrac{5}{14}$

Thus; $\mathbf{(\overline x , \overline y ) = (\dfrac{5}{8}, \dfrac{5}{14})}$

4 0

3 years ago

Find the value of each variable so that the quadrilateral is a parallelogram.

Dmitry [639]

Answer:

<em>x = -2</em>

<em>y = -5</em>

Step-by-step explanation:

<u>Properties of a Parallelogram</u>

Two pairs of opposite sides are parallel.
Two pairs of opposite sides are equal in length.
Two pairs of opposite angles are equal in measure
The diagonals bisect each other.
Adjacent angles are supplementary.
Each diagonal divides the quadrilateral into two congruent triangles.

Each diagonal is divided into two measures expressed with variables. Since diagonals bisect each other:

y + 23 = -4y - 2. And

-2x + 6 = x + 12

Solve the first equation. Adding 4y:

5y + 23 = - 2

Subtracting 23:

5y = -25

Dividing by 5:

y = -5

Solve the second equation. Subtracting x and 6:

-3x = 6

Dividing by -3:

x = -2

Solution:

x = -2

y = -5

4 0

3 years ago