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4 years ago

12

Need help with this answer !!!

2 answers:

KengaRu [80]4 years ago

8 0

To find the area of a rectangle, multiple the width by the length.

(And simply the fractions for a simpler equation)

For piece A:
The length 1 and 3/5 can be turned into an improper fraction by multiplying 1 by the denominator (5) and adding it to the numerator (3). 1 and 3/5 = 8/5

(3/4) • (8/5) = area

Multiple the numerators with each other and the denominators with each other (3 times 8 = 24) (4 times 5 = 20)

The area of piece A is 24/20

If you do the same for piece B:

(2/5) • (21/8) = area

The answer is 42/40

vivado [14]4 years ago

8 0

It’s should be I think42/5

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The equation below represents how much money Addy makes delivering papers. In the equation p represents the number of papers del

Fittoniya [83]

Step-by-step explanation:

Step one:

we are given the equation for Addy And Rachel's earning as

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Step two:

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3 years ago

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Montano1993 [528]

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3 years ago

Find the center, vertices, and foci of the ellipse with equation 4x^2 + 9y^2 =36

Anvisha [2.4K]

<u>Answer: </u>

The center, vertices and foci of the ellipse with equation $4 x^{2}+9 y^{2}=36 is (0,0),(\pm 3,0),(\pm \sqrt{5}, 0)$ respectively

<u> Solution: </u>

The equation of ellipse with centre (0, 0) in the form of $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$ --- eqn 1

Where,

x is the major axis

Centre (0, 0)

Vertices is $(\pm \mathrm{a}, 0)$

Foci is $(\pm \mathrm{c}, 0)$ where $c=\sqrt{a^{2}-b^{2}}$

Now given that the equation of ellipse is

$4 x^{2}+9 y^{2}=36$ --- eqn 2

On dividing equation (2) by 36,

$\frac{x^{2}}{9}+\frac{y^{2}}{4}=1$

On comparing equations (1) and (2),

We get a = 3, b= 2

$c=\sqrt{a^{2}-b^{2}}=\sqrt{9-4}=\sqrt{5}$

So centre of $\frac{x^{2}}{9}+\frac{y^{2}}{4}=1 is (0, 0)$

Vertices of $\frac{x^{2}}{9}+\frac{y^{2}}{4}=1 is (\pm 3,0)$

Foci of $\frac{x^{2}}{9}+\frac{y^{2}}{4}=1 is (\pm \sqrt{5}, 0)$

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4 years ago