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3 years ago

A sidewalk tile measures 4 feet by 5 feet. How many square inches are in its area?

Mathematics

2 answers:

Bas_tet [7]3 years ago

7 0

Answer: Area = 2880 square inches.

Step-by-step explanation: The tile is measured in feet, but it requests the measurement of area in inches. So:

1 ft = 12 in

4 ft = 12*4 = 48 in

5 ft = 12*5 = 60 in

A tile has a rectangular shape. Its are is calculated as:

A = width * length

A = 48 * 60

A = 2880

The tiel's area is 2880 square inches.

Mashcka [7]3 years ago

6 0

Convert 4ft. and 5ft. to inches (48 and 60). then multiply your length by your width to find the area of the tile. 48*60=2,880sq in.

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Don't let fractions fool you.

$\frac{1}{12}$ of a tank in $\frac{1}{3}$ an hour is equal to:
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Thus we know in an hour, an oil company can fill 3/12 of a tank (60 minutes in an hour).

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3 years ago

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3 years ago

Read 2 more answers

I need help on 17 and 23 pls

ludmilkaskok [199]

Question # 17 Solution

Answer:

$x_{1}= \frac{mx_{2}-y_{2}+y_{1}}{m}$

Step-by-step Explanation:

The given expression

$m = \frac{y_{2}-y_{1}}{x_{2}-x_{1}}$

And we have to solve for x₁

So,

Lets solve for x₁.

$m = \frac{y_{2}-y_{1}}{x_{2}-x_{1}}$

Multiply both sides by x₂ - x₁

$m(x_{2}-x_{1})= (x_{2}-x_{1})\frac{y_{2}-y_{1}}{x_{2}-x_{1}}$

$m(x_{2}-x_{1})= (y_{2}-y_{1})$

$mx_{2}-mx_{1}= (y_{2}-y_{1})$

$-mx_{1}= y_{2}-y_{1}-mx_{2}$

Divide both sides by -m

$\frac{-mx_{1}}{-m}= \frac{y_{2}-y_{1}-mx_{2}}{-m}$

$x_{1}= \frac{mx_{2}-y_{2}+y_{1}}{m}$

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Question # 23 Solution

Answer:

$n= \frac{bx}{-b + x}$

Step-by-step Explanation:

The given expression

$\frac{nx}{b}-x=x$

And we have to solve for n

So,

Let's solve for n.

$\frac{nx}{b}-x=x$

Multiply both sides by b.

$-bn + nx = bx$

Factor out n.

$n(-b + x)= bx$

Divide both sides by -b + x.

$\frac{n(-b + x)}{-b + x}= \frac{bx}{-b + x}$

$n= \frac{bx}{-b + x}$

Therefore, $n= \frac{bx}{-b + x}$

Keywords: solution, equation

Learn more about the solution of equations from brainly.com/question/12864981

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3 years ago

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Grace [21]

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Step-by-step explanation:

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3 years ago

Find the exact values of sin2 θ for cos θ = 3/18 on the interval 0° ≤ θ ≤ 90°

mote1985 [20]

Answer:

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Step-by-step explanation:

Recall the formula for the sine of the double angle:

$sin(2\theta)=2*sin(\theta)*cos(\theta)$

we know that $cos(\theta)=\frac{3}{18}$ , and that $\theta$ is in the interval between 0 and 90 degrees, where both the functions sine and cosine are non-negative numbers. Based on such, we can find using the Pythagorean trigonometric property that relates sine and cosine of the same angle, what $sin(\theta)$ is:

$cos^2(\theta)+sin^2(\theta)=1\\sin^2(\theta)=1-cos^2(\theta)\\sin(\theta)=\sqrt{1-cos^2(\theta)} \\sin(\theta)=\sqrt{1-(\frac{3}{18} )^2}\\sin(\theta)=\sqrt{1-\frac{9}{324} }\\sin(\theta)=\sqrt{\frac{324-9}{324} }\\sin(\theta)=\sqrt{\frac{315}{324} }\\\\sin(\theta)=\frac{3}{18}\sqrt{35 }$

With this information, we can now complete the value of the sine of the double angle requested:

$sin(2\theta)=2*sin(\theta)*cos(\theta)\\sin(2\theta)=2*\frac{3}{18} \,\sqrt{35} \,\frac{3}{18}\\sin(2\theta)=\frac{2*3*3}{18*18}\,\sqrt{35} \\sin(2\theta)=\frac{\sqrt{35} }{18}$

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3 years ago