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Arlecino [84]
3 years ago
11

Please hurry! Timed! What is the scale factor in the dilation?

Mathematics
1 answer:
Bess [88]3 years ago
7 0
C. 2

Hope this helps!
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Which of the following statements is always true about similar polygons
Ganezh [65]
The correct answer is D. Since the polygons are /similar/ and not /congruent/, C is not necessarily true.

Hope this helps.
3 0
3 years ago
I need some help with this.
Alenkasestr [34]

Answer:

b is 16 d is 13

Step-by-step explanation:

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7 0
2 years ago
Will give brainliest for answer
sleet_krkn [62]

Answer:

  y > -x -3

Step-by-step explanation:

The graph is shaded <em>above</em> the <em>dashed</em> line, indicating y-values in the solution are greater than those on the line, so your inequality will start with ...

  y >

The y-intercept of the line is (0, -3), so the "b" value in ...

  y > mx +b

will be -3.

The line has a rise of -3 for a run of 3 (between the marked points), so the slope is ...

  m = rise/run = -3/3 = -1

Then the inequality you want is ...

  y > -x -3

7 0
3 years ago
Charles will babysit for up to 4 hours and charges $7 per hour. Write a function in function notation for this situation.
I am Lyosha [343]
Assume Charles charges $7*3=$21 if he babysits for 2,5 hours (that is 2 and a half hours).

That is we are not considering Charles babysitting exactly 1, 2, 3 or 4 hours, but also in between times.


then we can write the following piecewise function to describe the situation:


f is a function from (0, 4] to {$7, $14, $21, $28}

f(x)=\begin{cases} &#10;      \$7 & if \enspace \enspace x\leq 1 \\&#10;      \$14 & if \enspace \enspace 1\ \textless \  x\leq 2 \\&#10;      \$21 & if \enspace\enspace 2\ \textless \ x\leq 3 \\&#10;      \$28 & if \enspace\enspace 3\ \textless \ x\leq 4 \&#10;   \end{cases}
6 0
3 years ago
Plz read and answerrrrrrrrrrr
natulia [17]

Answer:

16.7%.

Step-by-step explanation:

There are initially 24 + 15 + 8 = 47 pencils in the bag.

Take a pencil out of this bag of 47 pencils. 15 out of the 47 pencils blue. Let A represent the event of getting a blue pencil on the first pick. The probability of getting a blue pencil is:

\displaystyle P(A) = \frac{15}{47}.

There are now 47 - 1 = 46 pencils left in the bag. However, given that the first pencil removed from the bag is blue, the number of red pencils in the bag will still be 24. Take another pencil out of this bag of 46 pencils. Let B represent the event of getting a red pencil on the second pick. The possibility that the second pencil is red given that the first pencil is blue will be:

\displaystyle P(B|A) = \frac{24}{46}.

The question is asking for the possibility that the first pencil is blue and the second pencil is red. That is:

\displaystyle P(A\cap B) = P(A) \cdot P(B|A) = \frac{15}{47}\times \frac{24}{46} = 0.166512 \approx 16.7\%.

5 0
3 years ago
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