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Andrej [43]
3 years ago
11

Suppose a hexagonal traffic sign is dilated by a scale factor of 2.5 about its center. Which statement is true? A. The dilated s

ign will have corresponding line segments that are congruent to those of the pre-image and corresponding angles that are proportional to that of the pre-image. B. The dilated sign will have corresponding line segments that are proportional to those of the pre-image and corresponding angles that are congruent to that of the pre-image. C. The dilated sign will have corresponding line segments and angles that are congruent to those of the pre-image. D. The dilated sign will have corresponding line segments and angles that are proportional to the pre-image but not congruent
Mathematics
1 answer:
xeze [42]3 years ago
4 0

<em>Hexagonal</em> implies that the <em>traffic sign</em> has six straight sides. But <u>dilating</u> it changes its <em>length</em> of sides, but not its angles. So that the <em>statement</em> that is <u>true</u> is option B.

i.e B. The <u>dilated</u> sign will have corresponding line segments that are <em>proportional</em> to those of the pre-image and <u>corresponding</u> angles that are congruent to that of the pre-image.

Dilation involves <u>increasing</u> or <u>decreasing</u> the line segments of a given shape by a <em>scale factor</em>. This process do not affect the measure of the internal <u>angles</u> of the shape.

Given a <u>hexagonal</u> traffic sign in the question, this implies that the traffic sign has six sides. <u>Dilating</u> it by a scale factor of 2.5 about its center would affect its <em>length</em> of sides. So that the <em>statement</em> that is <u>true</u> is option B.

i.e B. The dilated sign will have <u>corresponding</u> line segments that are proportional to those of the pre-image and <em>corresponding</em> angles that are congruent to that of the pre-image.

Visit: brainly.com/question/8285347

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A recent study asked U. S. Adults to name 10 historic events that occurred during their lifetime that have had the greatest impa
Nadusha1986 [10]

Using the z-distribution, as we are working with a proportion, it is found that the 99% confidence interval for the proportion of all U. S. Adults who would include the 9/11 attacks on their list of 10 historic events is (0.7458, 0.7942). It means that we are 99% sure that the true proportion for all U.S. adults is between these two bounds.

<h3>What is a confidence interval of proportions?</h3>

A confidence interval of proportions is given by:

\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}

In which:

  • \pi is the sample proportion.
  • z is the critical value.
  • n is the sample size.

In this problem, the parameters are:

z = 2.575, n = 2000, \pi = 0.77

The lower limit of this interval is:

\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.77 - 2.575\sqrt{\frac{0.77(0.23)}{2000}} = 0.7458

The upper limit of this interval is:

\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.77 + 2.575\sqrt{\frac{0.77(0.23)}{2000}} = 0.7942

The 99% confidence interval for the proportion of all U. S. Adults who would include the 9/11 attacks on their list of 10 historic events is (0.7458, 0.7942). It means that we are 99% sure that the true proportion for all U.S. adults is between these two bounds.

More can be learned about the z-distribution at brainly.com/question/25890103

7 0
2 years ago
Jacob wants to know the favorite sport among 6th grade students. Which types of survey below would give Jacob valid results?
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Because they will have people that don’t like sports and do like sports where as if you survey them in a sports shop there gonna like sports otherwise they wouldn’t be there
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What is the y-intercept, axis of symmetry and vertex of the following function.
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Complete the square:

F(x) = -3x² - 6x - 5

F(x) = -3 (x² + 2x) - 5

F(x) = -3 (x² + 2x + 1 - 1) - 5

F(x) = -3 ((x + 1)² - 1) - 5

F(x) = -3 (x + 1)² + 3 - 5

F(x) = -3 (x + 1)² - 2

The y-intercept has x-coordinate equal to 0, so it corresponds to the value of F(0) :

F(0) = -3 (0 + 1)² - 2 = -3 - 2 = -5

The axis of symmetry is the vertical line running through the vertex of this parabola, so we'll come back to this.

The vertex of the parabola is (-1, -2). This represents the maximum value of F(x), which follows from

(x + 1)² ≥ 0   ⇒   -3 (x + 1)² ≤ 0   ⇒   -3 (x + 1)² - 2 ≤ -2

This is to say, every point on the parabola has a y-coordinate no greater than -2.

As mentioned earlier, the axis of symmetry is the vertical line through the vertex, and its equation is determined by the x-coordinate of the vertex. Hence the AoS is the line x = -1.

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Order these numbers from least to greatest: -3.8, 3<br> 3.-4, 0.
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The answer would be,
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8 0
3 years ago
Please help!!<br> Write a matrix representing the system of equations
frozen [14]

Answer:

(4, -1, 3)

Step-by-step explanation:

We have the system of equations:

\left\{        \begin{array}{ll}            x+2y+z =5 \\    2x-y+2z=15\\3x+y-z=8        \end{array}    \right.

We can convert this to a matrix. In order to convert a triple system of equations to matrix, we can use the following format:

\begin{bmatrix}x_1& y_1& z_1&c_1\\x_2 & y_2 & z_2&c_2\\x_3&y_2&z_3&c_3 \end{bmatrix}

Importantly, make sure the coefficients of each variable align vertically, and that each equation aligns horizontally.

In order to solve this matrix and the system, we will have to convert this to the reduced row-echelon form, namely:

\begin{bmatrix}1 & 0& 0&x\\0 & 1 & 0&y\\0&0&1&z \end{bmatrix}

Where the (x, y, z) is our solution set.

Reducing:

With our system, we will have the following matrix:

\begin{bmatrix}1 & 2& 1&5\\2 & -1 & 2&15\\3&1&-1&8 \end{bmatrix}

What we should begin by doing is too see how we can change each row to the reduced-form.

Notice that R₁ and R₂ are rather similar. In fact, we can cancel out the 1s in R₂. To do so, we can add R₂ to -2(R₁). This gives us:

\begin{bmatrix}1 & 2& 1&5\\2+(-2) & -1+(-4) & 2+(-2)&15+(-10) \\3&1&-1&8 \end{bmatrix}\\\Rightarrow\begin{bmatrix}1 & 2& 1&5\\0 & -5 & 0&5 \\3&1&-1&8 \end{bmatrix}

Now, we can multiply R₂ by -1/5. This yields:

\begin{bmatrix}1 & 2& 1&5\\ -\frac{1}{5}(0) & -\frac{1}{5}(-5) & -\frac{1}{5}(0)& -\frac{1}{5}(5) \\3&1&-1&8 \end{bmatrix}\\\Rightarrow\begin{bmatrix}1 & 2& 1&5\\ 0 & 1 & 0& -1 \\3&1&-1&8 \end{bmatrix}

From here, we can eliminate the 3 in R₃ by adding it to -3(R₁). This yields:

\begin{bmatrix}1 & 2& 1&5\\ 0 & 1 & 0& -1 \\3+(-3)&1+(-6)&-1+(-3)&8+(-15) \end{bmatrix}\\\Rightarrow\begin{bmatrix}1 & 2& 1&5\\ 0 & 1 & 0& -1 \\0&-5&-4&-7 \end{bmatrix}

We can eliminate the -5 in R₃ by adding 5(R₂). This yields:

\begin{bmatrix}1 & 2& 1&5\\ 0 & 1 & 0& -1 \\0+(0)&-5+(5)&-4+(0)&-7+(-5) \end{bmatrix}\\\Rightarrow\begin{bmatrix}1 & 2& 1&5\\ 0 & 1 & 0& -1 \\0&0&-4&-12 \end{bmatrix}

We can now reduce R₃ by multiply it by -1/4:

\begin{bmatrix}1 & 2& 1&5\\ 0 & 1 & 0& -1 \\ -\frac{1}{4}(0)&-\frac{1}{4}(0)&-\frac{1}{4}(-4)&-\frac{1}{4}(-12) \end{bmatrix}\\\Rightarrow\begin{bmatrix}1 & 2& 1&5\\ 0 & 1 & 0& -1 \\0&0&1&3 \end{bmatrix}

Finally, we just have to reduce R₁. Let's eliminate the 2 first. We can do that by adding -2(R₂). So:

\begin{bmatrix}1+(0) & 2+(-2)& 1+(0)&5+(-(-2))\\ 0 & 1 & 0& -1 \\0&0&1&3 \end{bmatrix}\\\Rightarrow\begin{bmatrix}1 & 0& 1&7\\ 0 & 1 & 0& -1 \\0&0&1&3 \end{bmatrix}

And finally, we can eliminate the second 1 by adding -(R₃):

\begin{bmatrix}1 +(0)& 0+(0)& 1+(-1)&7+(-3)\\ 0 & 1 & 0& -1 \\0&0&1&3 \end{bmatrix}\\\Rightarrow\begin{bmatrix}1 & 0& 0&4\\ 0 & 1 & 0& -1 \\0&0&1&3 \end{bmatrix}

Therefore, our solution set is (4, -1, 3)

And we're done!

3 0
3 years ago
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