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never [62]
3 years ago
10

Sherri made the trellis shown for her climbing rose. The trellis is made so that rhombus ABCD is ≅ rhombus MNOP.

Mathematics
1 answer:
motikmotik3 years ago
5 0

Answer:

It should be 12!!

Step-by-step explanation:

I took the assignment on Compass Learning!!

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M is between L and C, LM= 11, MC= 2x+5, and LC= 6x-6, what is MC?
Margarita [4]
MC=16... Use this equation to solve for X:  LM+MC=LC... Then sub the value of X into the equation for MC to find your answer.
5 0
3 years ago
In ΔLMN, MN = 15, NL = 20, and LM = 6. Which list has the angles of ΔLMN in order from smallest to largest?
Gnom [1K]

Answer:

m<N, m<L, m<M

Step-by-step explanation:

The angle that sees the longest side length has the bigger angle measurement

so the angles would be listed as following :

from smallest to largest :

m<N, m<L, m<M

6 0
3 years ago
What fraction of 2 1/8 is 6?
Dima020 [189]
1/8th is the fraction that is equal to six
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Equation of the line that is perpendicular to y= 1/8x -2 and goes through the point <br> (-2,-3)
larisa [96]
The answer to this equation is y=-8x-19



I hope this helps!!!!!!!!
7 0
3 years ago
Read 2 more answers
Which of the following graphs shows the solution set for the inequality below? 3|x + 1| &lt; 9
Bas_tet [7]

Step-by-step explanation:

The absolute value function is a well known piecewise function (a function defined by multiple subfunctions) that is described mathematically as

                                 f(x) \ = \ |x| \ = \ \left\{\left\begin{array}{ccc}x, \ \text{if} \ x \ \geq \ 0 \\ \\ -x, \ \text{if} \ x \ < \ 0\end{array}\right\}.

This definition of the absolute function can be explained geometrically to be similar to the straight line   \textbf{\textit{y}} \ = \ \textbf{\textit{x}}  , however, when the value of x is negative, the range of the function remains positive. In other words, the segment of the line  \textbf{\textit{y}} \ = \ \textbf{\textit{x}}  where \textbf{\textit{x}} \ < \ 0 (shown as the orange dotted line), the segment of the line is reflected across the <em>x</em>-axis.

First, we simplify the expression.

                                             3\left|x \ + \ 1 \right| \ < \ 9 \\ \\ \\\-\hspace{0.2cm} \left|x \ + \ 1 \right| \ < \ 3.

We, now, can simply visualise the straight line,  y \ = \ x \ + \ 1 , as a line having its y-intercept at the point  (0, \ 1) and its <em>x</em>-intercept at the point (-1, \ 0). Then, imagine that the segment of the line where x \ < \ 0 to be reflected along the <em>x</em>-axis, and you get the graph of the absolute function y \ = \ \left|x \ + \ 1 \right|.

Consider the inequality

                                                    \left|x \ + \ 1 \right| \ < \ 3,

this statement can actually be conceptualise as the question

            ``\text{For what \textbf{values of \textit{x}} will the absolute function \textbf{be less than 3}}".

Algebraically, we can solve this inequality by breaking the function into two different subfunctions (according to the definition above).

  • Case 1 (when x \ \geq \ 0)

                                                x \ + \ 1 \ < \ 3 \\ \\ \\ \-\hspace{0.9cm} x \ < \ 3 \ - \ 1 \\ \\ \\ \-\hspace{0.9cm} x \ < \ 2

  • Case 2 (when x \ < \ 0)

                                            -(x \ + \ 1) \ < \ 3 \\ \\ \\ \-\hspace{0.15cm} -x \ - \ 1 \ < \ 3 \\ \\ \\ \-\hspace{1cm} -x \ < \ 3 \ + \ 1 \\ \\ \\ \-\hspace{1cm} -x \ < \ 4 \\ \\ \\ \-\hspace{1.5cm} x \ > \ -4

           *remember to flip the inequality sign when multiplying or dividing by

            negative numbers on both sides of the statement.

Therefore, the values of <em>x</em> that satisfy this inequality lie within the interval

                                                     -4 \ < \ x \ < \ 2.

Similarly, on the real number line, the interval is shown below.

The use of open circles (as in the graph) indicates that the interval highlighted on the number line does not include its boundary value (-4 and 2) since the inequality is expressed as "less than", but not "less than or equal to". Contrastingly, close circles (circles that are coloured) show the inclusivity of the boundary values of the inequality.

3 0
2 years ago
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