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2 years ago

A rectangle has an area of 8/9 square yard. Which could be the dimensions of the rectangle?

Mathematics

1 answer:

katen-ka-za [31]2 years ago

6 0

Answer:

There could be any number of dimensions as long as the two dimensions multiply to 8/9

Step-by-step explanation:

One example of this would be 2 by 4/9

2 * 4/9 = 8/9

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IceJOKER [234]

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The tick mark on the lines means they are equal, so that means we have two lines that are equal to 10

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3 years ago

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3 years ago

$1111 at 11% for 11 years

maksim [4K]

Answer:

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Step-by-step explanation:

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3 years ago

$100 is invested at 12% per year. If the amount is compounded annually, write the total amount after 2 years in exponential func

Sladkaya [172]

Answer:

A = $100(1.12)^2

Step-by-step explanation:

The standard formula for compound interest is given as;

A = P(1+r/n)^(nt) .....1

Where;

A = final amount/value

P = initial amount/value (principal)

r = rate yearly

n = number of times compounded yearly.

t = time of investment in years

For this case;

P = $100

t = 2years

n = 1

r = 12% = 0.12

Substituting the values, we have;

A = $100(1+0.12)^(2)

A = $100(1.12)^2

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3 years ago

Es 18 de junio de 1815, las tropas napoleónicas se encuentran justo adelante, tu eres el ingeniero en balística y el general Wel

Airida [17]

Answer:

89.1° or -1.4°

Step-by-step explanation:

1. Location:

You are on the Mont-Saint-Jean escarpment, near the Belgian town of Waterloo.

The French troops are about 50 m below you and 1.2 km distant.

2. Finding the firing angle

Data:

R = 1200 m

u = 600 m/s

h = -50 m (the height of the target)

a = 9.8 m/s²

We have two conditions.

Horizontal distance

(1) 1200 = 600t cosθ

Vertical distance

(2) -50 = 600t sinθ - 4.9t²

Divide each side of (1) by 600cosθ.

$(3) \, t =\dfrac{2}{\cos \theta}$

Substitute (3) into (2)

$-50 = 600t \sin \theta - 4.9t^{2} = 600 \left( \dfrac{2}{\cos \theta} \right ) \sin \theta - 4.9 \left( \dfrac{2}{\cos \theta} \right )^{2}\\\\(4) \, -50 = 1200 \tan \theta - \dfrac{19.6}{\cos^{2} \theta}$

Recall that

(5) sec²θ = 1/cos²θ = tan²θ + 1

Substitute (5) into (4)

$-50 = 1200 \tan \theta - 19.6 \left(\tan^{2} \theta}+ 1\right )$

Set up a quadratic equation

$\begin{array}{rcl}-50 & = & 1200 \tan \theta - 19.6\tan^{2} \theta -19.6 \\0 & = & 1200 \tan \theta - 19.6\tan^{2} \theta + 30.4\\0 & =&19.6\tan^{2} \theta - 1200 \tan \theta - 30.4\\0 & =&\tan^{2} \theta - 61.224 \tan \theta - 1.551\\\end{array}$

Solve for θ

Use the quadratic formula.

tanθ = 61.249 or -0.025

θ = arctan(61.249) = 89.1° or

θ = arctan(-0.025) = -1.4°

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3 years ago