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miss Akunina [59]
3 years ago
10

12.33 to the nearest tenth

Mathematics
2 answers:
ivann1987 [24]3 years ago
7 0
12.3 is the answer i think 
kolbaska11 [484]3 years ago
3 0
The correct answer would be 

12.3 

You would round the hundredth. If the hundredth place is greater than 5, you would round up. If the hundredth place is less than 5, you would round down.
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Which statements can be used to prove that ABC and A’B’C are congruent?
Nadusha1986 [10]

Answer:

O A.

Step-by-step explanation:

<u>Option A</u> identifies two angles (sufficient for similarity) and one side, sufficient (with similarity) for congruence. The applicable congruence theorem is AAS.

<u>Option B</u> identifies two sides and the angle not between them. The two triangles will be congruent in that case only if the angle is opposite the longest side, which is <u>not true</u> in general.

<u>Option C</u> same deal as Option A.

<u>Option D</u> identifies three congruent angles, which will prove the triangles similar, but not necessarily congruent.

7 0
2 years ago
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
PLZ HELP if u know the answer for sure​
krok68 [10]
The answer is b (-2 5/6)
6 0
2 years ago
A dilation has center (0, 0). Find the image of each point for the given scale factor.
pshichka [43]

Answer:

A. (−3,5)

Step-by-step explanation:

The rule for dilation by scale factor k is given by:

(x,y)\to (kx,ky)

The given point is P(-3,5) and the scale factor is k=1

We substitute the point and the scale factor into the rule to get:

P(-3,5)\to P'(-3\times1,5\times1)

This gives us:

P(-3,5)\to P'(-3,5)

The correct answer is A

3 0
3 years ago
Plz help!! Question below!!
Zinaida [17]
A. the discount is 7.5


B. the price is 17.5

C. the tax will be 1.5 on the original price

D. 25+ 1.5=26.5


good luck
8 0
2 years ago
Read 2 more answers
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