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iogann1982 [59]
3 years ago
13

Helppp how do u do 12

Mathematics
2 answers:
Zina [86]3 years ago
6 0
\begin{array}{ccc}\$53.00&-&106\%\\\$x&-&100\%\end{array}\\\\cross\ multiply\\\\106x=53\cdot100\\\\106x=5,300\ \ \ \ |:106\\\\\boxed{x=50}\\\\Answer:\ \boxed{\$50}
Delicious77 [7]3 years ago
4 0
B is the correct answer



\frac{x}{100}  =  \frac{53}{106}  \\ x =  \frac{53 \times 100}{106}  =  \frac{5300}{106}  = 50




good luck
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Answer all correctly
zalisa [80]

Answer:

2. y = -9x + 7

2. y = 3x - 2

3. y - 6 = 10(2x - 1)

1. 5x + 2y = 10

4. 5x = 3y + 9

3. y + 2 = 6(x - 4)

Step-by-step explanation:

7 0
2 years ago
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Use Simpson's Rule with n = 10 to estimate the arc length of the curve. Compare your answer with the value of the integral produ
SOVA2 [1]

y=\ln(6+x^3)\implies y'=\dfrac{3x^2}{6+x^3}

The arc length of the curve is

\displaystyle\int_0^5\sqrt{1+\frac{9x^4}{(6+x^3)^2}}\,\mathrm dx

which has a value of about 5.99086.

Let f(x)=\sqrt{1+\frac{9x^4}{(6+x^3)^2}}. Split up the interval of integration into 10 subintervals,

[0, 1/2], [1/2, 1], [1, 3/2], ..., [9/2, 5]

The left and right endpoints are given respectively by the sequences,

\ell_i=\dfrac{i-1}2

r_i=\dfrac i2

with 1\le i\le10.

These subintervals have midpoints given by

m_i=\dfrac{\ell_i+r_i}2=\dfrac{2i-1}4

Over each subinterval, we approximate f(x) with the quadratic polynomial

p_i(x)=f(\ell_i)\dfrac{(x-m_i)(x-r_i)}{(\ell_i-m_i)(\ell_i-r_i)}+f(m_i)\dfrac{(x-\ell_i)(x-r_i)}{(m_i-\ell_i)(m_i-r_i)}+f(r_i)\dfrac{(x-\ell_i)(x-m_i)}{(r_i-\ell_i)(r_i-m_i)}

so that the integral we want to find can be estimated as

\displaystyle\sum_{i=1}^{10}\int_{\ell_i}^{r_i}p_i(x)\,\mathrm dx

It turns out that

\displaystyle\int_{\ell_i}^{r_i}p_i(x)\,\mathrm dx=\frac{f(\ell_i)+4f(m_i)+f(r_i)}6

so that the arc length is approximately

\displaystyle\sum_{i=1}^{10}\frac{f(\ell_i)+4f(m_i)+f(r_i)}6\approx5.99086

5 0
3 years ago
Examples of geometric transformations can be found throughout the real world. Think about some places where you might use or see
lina2011 [118]

Answer:

We know that there are four types of geometric transformations namely translation, reflection, rotation and dilation.

1. Translation: It is a rigid transformation that moves the objects in any direction. Few examples of translation we see in real life are:

A. When we move furniture in the house from one place to another.

B. The position of a car on the road is an example of translation.

C. When we changes the places of pieces on the chessboard.

2. Reflection: is a rigid transformation that flips the object. It can be easily observed in real life in the following ways:

A. The reflection of light from any object is the most easily viewed example of reflection.

B. Looking at yourself in the mirror is because of reflection.

C. When we can see our faces through clear water is also because of reflection.

3. Rotation: It is a rigid transformation that turns the object to a certain degree. The few examples of rotation are:

A. Opening of the door to a certain angle.

B. The spinning of Earth, Sun and other planets.

C. The moving of the hands of the clock is an example of rotation.

4. Dilation: It is a rigid transformation that changes the shape of the object. It's few examples are:

A. Enlarging the image on a computer.

B. Blowing air in the balloon to make it large.

C. Miniature version of buildings that are in proportion to the actual size of the buildings.

Hence, the transformations can be easily observed in real life.

5 0
3 years ago
What is the corresponding growth decay factor of the given annual rate of +200%
Alina [70]
2000 should be it I'm pretty sure
4 0
3 years ago
How do you say 750 in words
Agata [3.3K]

750 in word form:

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8 0
3 years ago
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