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Ulleksa [173]
3 years ago
11

PL HELP ME GIVING 100

Mathematics
2 answers:
il63 [147K]3 years ago
8 0

Answer:

C ≤ 11

Step-by-step explanation:

c − 5.5 + 5.5 ≤ 5.5 + 5.5

add 5.5 to both sides

Mkey [24]3 years ago
6 0
YOU COULD DO c=6.....6 - 5.5= 0.5 AND THATS LESS THAN 5.5
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SKETCHPAD
Irina-Kira [14]

The attached figure represents the image of A"B"C" after the transformation

<h3>How to transform the triangle?</h3>

The transformation rule is given as:

A"B"C" = Ro90° (T(-4,3)(ABC))

This means that we rotate the triangle 90 degrees clockwise, and then translate the triangle

From the figure, the coordinates of ABC are

A = (-1, 2)

B = (1, 4)

C = (3, -1)

The rule of 90 degrees clockwise rotation is

(x,y) ⇒ (y,-x)

So, we have

A' = (2, 1)

B' = (4, -1)

C' = (-1, -3)

The translation of the triangle by T(-4,3) is

(x,y) ⇒ (x - 4, y + 3)

So, we have

A'' = (-2, 4)

B'' = (0, 2)

C'' = (-5, 0)

See attachment for the image of the transformation

Read more about transformation at:

brainly.com/question/11709244

#SPJ1

5 0
2 years ago
Drag one or more expressions into each box to
ASHA 777 [7]

Answer:

the right answer is x(x+2)

3 0
3 years ago
5.349 round to the nearest tenth
soldi70 [24.7K]
5.4 because the 9 bumps it to 5.35.

And the 5 bumps it to 5.4.                                                                                                    
6 0
3 years ago
Read 2 more answers
Each year for 4 years, a farmer increased the number of trees in a certain orchard by of the number of trees in the orchard the
Neko [114]

Answer:

The number of trees at the begging of the 4-year period was 2560.

Step-by-step explanation:

Let’s say that x is number of trees at the begging of the first year, we know that for four years the number of trees were incised by 1/4 of the number of trees of the preceding year, so at the end of the first year the number of trees wasx+\frac{1}{4} x=\frac{5}{4} x, and for the next three years we have that

                             Start                                          End

Second year     \frac{5}{4}x --------------   \frac{5}{4}x+\frac{1}{4}(\frac{5}{4}x) =\frac{5}{4}x+ \frac{5}{16}x=\frac{25}{16}x=(\frac{5}{4} )^{2}x

Third year    (\frac{5}{4} )^{2}x-------------(\frac{5}{4})^{2}x+\frac{1}{4}((\frac{5}{4})^{2}x) =(\frac{5}{4})^{2}x+\frac{5^{2} }{4^{3} } x=(\frac{5}{4})^{3}x

Fourth year (\frac{5}{4})^{3}x--------------(\frac{5}{4})^{3}x+\frac{1}{4}((\frac{5}{4})^{3}x) =(\frac{5}{4})^{3}x+\frac{5^{3} }{4^{4} } x=(\frac{5}{4})^{4}x.

So  the formula to calculate the number of trees in the fourth year  is  

(\frac{5}{4} )^{4} x, we know that all of the trees thrived and there were 6250 at the end of 4 year period, then  

6250=(\frac{5}{4} )^{4}x⇒x=\frac{6250*4^{4} }{5^{4} }= \frac{10*5^{4}*4^{4} }{5^{4} }=2560.

Therefore the number of trees at the begging of the 4-year period was 2560.  

7 0
3 years ago
What is the improper fraction 17/10 as a decimal
Rina8888 [55]
17 / 10 = 1 7/10 = 1.7
4 0
3 years ago
Read 2 more answers
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