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2 years ago

The actual length of Wall C is 4 m. To represent Wall C, Noah draws a segment 16 cm long. What scale is he using?

Mathematics

1 answer:

hjlf2 years ago

4 0

Answer:

$\frac{1m}{4cm}$

Step-by-step explanation:

First write the problem out as a fraction.

$\frac{4}{16}$

Next see if there is a GCF

4:1,2,4

16:1,2,4,8,16

Finally divide the numerator and denomenator by the GCF

$\frac{4}{16} /4=\frac{1}{4}$

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Read the following poem:

pentagon [3]

Answer:

7, I believe. I have to fill in the extra character space, so I hope you have a nice day :)

6 0

2 years ago

For the class of 2013's prom Norman's dress shop sold cheaper dresses for $90 each and more expensive dresses for $140 each. The

Tema [17]

Answer:

Number of cheaper dresses sold is 35

Number of expensive dresses sold is 15

Step-by-step explanation:

Given:

Cost of cheaper dresses = $90

Cost of expensive dresses = $140

Total cost of the dresses = $5250

To Find:

Number of cheaper dress = ?

Number of expensive dress = ?

Solution:

Let

The number of cheaper dresses be x

The number of expensive dresses be y

(Number of cheaper dresses X cost of cheap dress) + (Number of Expensive dresses X cost of expensive dress) = $5250

$x \times90 +y \times 140 = 5250$ = $5250

It is given that the 20 more of the cheaper dresses than the expensive dresses is sold

So,

number of cheaper dress = 20 + number of expensive dress

x = 20 + y---------------------------------------(1)

$(20+y) \times90 +y \times 140 = 5250 = 5250$

$(20 \times 90 +y\times 90) +y \times 140= 5250$

$1800 + 90y+ 140y = 5250$

$1800 + 230y = 5250$

$230y =5250 -1800$

$230y = 3450$

$y = \frac{3450}{230}$

y = 15

Substituting y in (1)

x = 20 +15

x= 35

5 0

3 years ago

Evaluate the integral, show all steps please!

Aloiza [94]

Answer:

$\displaystyle \int \dfrac{1}{(9-x^2)^{\frac{3}{2}}}\:\:\text{d}x=\dfrac{x}{9\sqrt{9-x^2}} +\text{C}$

Step-by-step explanation:

Fundamental Theorem of Calculus

$\displaystyle \int \text{f}(x)\:\text{d}x=\text{F}(x)+\text{C} \iff \text{f}(x)=\dfrac{\text{d}}{\text{d}x}(\text{F}(x))$

If differentiating takes you from one function to another, then integrating the second function will take you back to the first with a constant of integration.

Given indefinite integral:

$\displaystyle \int \dfrac{1}{(9-x^2)^{\frac{3}{2}}}\:\:\text{d}x$

Rewrite 9 as 3² and rewrite the 3/2 exponent as square root to the power of 3:

$\implies \displaystyle \int \dfrac{1}{\left(\sqrt{3^2-x^2}\right)^3}\:\:\text{d}x$

Integration by substitution

$\boxed{\textsf{For }\sqrt{a^2-x^2} \textsf{ use the substitution }x=a \sin \theta}$

$\textsf{Let }x=3 \sin \theta$

$\begin{aligned}\implies \sqrt{3^2-x^2} & =\sqrt{3^2-(3 \sin \theta)^2}\\ & = \sqrt{9-9 \sin^2 \theta}\\ & = \sqrt{9(1-\sin^2 \theta)}\\ & = \sqrt{9 \cos^2 \theta}\\ & = 3 \cos \theta\end{aligned}$

Find the derivative of x and rewrite it so that dx is on its own:

$\implies \dfrac{\text{d}x}{\text{d}\theta}=3 \cos \theta$

$\implies \text{d}x=3 \cos \theta\:\:\text{d}\theta$

Substitute everything into the original integral:

$\begin{aligned}\displaystyle \int \dfrac{1}{(9-x^2)^{\frac{3}{2}}}\:\:\text{d}x & = \int \dfrac{1}{\left(\sqrt{3^2-x^2}\right)^3}\:\:\text{d}x\\\\& = \int \dfrac{1}{\left(3 \cos \theta\right)^3}\:\:3 \cos \theta\:\:\text{d}\theta \\\\ & = \int \dfrac{1}{\left(3 \cos \theta\right)^2}\:\:\text{d}\theta \\\\ & = \int \dfrac{1}{9 \cos^2 \theta} \:\: \text{d}\theta\end{aligned}$

Take out the constant:

$\implies \displaystyle \dfrac{1}{9} \int \dfrac{1}{\cos^2 \theta}\:\:\text{d}\theta$

$\textsf{Use the trigonometric identity}: \quad\sec^2 \theta=\dfrac{1}{\cos^2 \theta}$

$\implies \displaystyle \dfrac{1}{9} \int \sec^2 \theta\:\:\text{d}\theta$

$\boxed{\begin{minipage}{5 cm}\underline{Integrating $\sec^2 kx$}\\\\$\displaystyle \int \sec^2 kx\:\text{d}x=\dfrac{1}{k} \tan kx\:\:(+\text{C})$\end{minipage}}$

$\implies \displaystyle \dfrac{1}{9} \int \sec^2 \theta\:\:\text{d}\theta = \dfrac{1}{9} \tan \theta+\text{C}$

$\textsf{Use the trigonometric identity}: \quad \tan \theta=\dfrac{\sin \theta}{\cos \theta}$

$\implies \dfrac{\sin \theta}{9 \cos \theta} +\text{C}$

$\textsf{Substitute back in } \sin \theta=\dfrac{x}{3}:$

$\implies \dfrac{x}{9(3 \cos \theta)} +\text{C}$

$\textsf{Substitute back in }3 \cos \theta=\sqrt{9-x^2}:$

$\implies \dfrac{x}{9\sqrt{9-x^2}} +\text{C}$

Learn more about integration by substitution here:

brainly.com/question/28156101

brainly.com/question/28155016

4 0

1 year ago

A triangle has sides of (2x), (x - 7) and (3x - 20).

LiRa [457]

2x+x-7+3x-20=87cm
6x-27=87
6x=114
x=19
Therefore the sides are:12cm, 38cm, and 37cm

7 0

2 years ago

Evatuate expreeion of the valube 5 (9+d) - 6

sammy [17]

Answer:

39 + 5d

Step-by-step explanation:

First multiply 5 by 9 and 5 by d then subtract 6 from 45 (which you get from multiplying 5 by 9) and you get 39 and multiply 5 by d so you get 39 + 5d

7 0

2 years ago

Read 2 more answers