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2 years ago

Miranda bought a square frame that has an area of 30 square inches what is the approximate side lenght of the frame

Mathematics

2 answers:

Whitepunk [10]2 years ago

4 0

Answer:

The side length of the square frame is 5.5 inches approximately.

Step-by-step explanation:

Given:

Area of the square frame bought by Miranda= 30 square inches

To find:

Side length of the frame=?

Solution:

we know that,

$\text{{Area of a square}} = \text{( side length )}}^2$

Here, area of square = 30 square inches .

$\text{( side length )}}^2 = 30$ square inches

Side length = $\sqrt30$ inches

Side length = 5.4772 inches

Alexxandr [17]2 years ago

3 0

Area of a square equals side squared

(A = s²)

30 = s²

√30 = s

You can use a calculator to find out that √30 ≈ 5.48

or you can use estimation (√25 < √30 < √36) which means that √30 is between 5 and 6.

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schepotkina [342]

Answer:

zero (0)

Step-by-step explanation:

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where as males have 50% chance of being affected with hemophilia.

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2 years ago

The top of a ladder is 4m up a vertical wall.The bottom is 2m from the wall.The coordinate axes are the wall and the horizontal

mash [69]

Answer:

Part a) $y=-2x+4$

Part b) The coordinates of the point are $(\frac{4}{3},\frac{4}{3})$

Step-by-step explanation:

Part a) Find the equation representing the ladder

we have the ordered pairs

(0,4) and (2,0)

Find the slope

$m=(0-4)/(2-0)=-2$

Find the equation of the line in slope intercept form

$y=mx+b$

we have

$m=-2\\b=4$

substitute

$y=-2x+4$

Part b) A square box just fits under the ladder.Find the coordinates of the point where the box touches the ladder.

If the box is a square

the x-coordinate of the point where the box touches the ladder must be equal to the y-coordinate

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$y=-2x+4$

substitute

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

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1 year ago

Solve the given initial value problem and determine how the interval in which the solution exists depends on the initial value y

zalisa [80]

Answer:

y has a finite solution for any value y_0 ≠ 0.

Step-by-step explanation:

Given the differential equation

y' + y³ = 0

We can rewrite this as

dy/dx + y³ = 0

Multiplying through by dx

dy + y³dx = 0

Divide through by y³, we have

dy/y³ + dx = 0

dy/y³ = -dx

Integrating both sides

-1/(2y²) = - x + c

Multiplying through by -1, we have

1/(2y²) = x + C (Where C = -c)

Applying the initial condition y(0) = y_0, put x = 0, and y = y_0

1/(2y_0²) = 0 + C

C = 1/(2y_0²)

1/(2y²) = x + 1/(2y_0²)

2y² = 1/[x + 1/(2y_0²)]

y² = 1/[2x + 1/(y_0²)]

y = 1/[2x + 1/(y_0²)]½

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For y_0 = 0, the solution has an expression 1/0, which makes the solution infinite.

With this, y has a finite solution for any value y_0 ≠ 0.

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