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Veseljchak [2.6K]
3 years ago
13

Triangle ABC is translated 4 units down and 6 units right, resulting in triangle A'B'C'.

Mathematics
2 answers:
gogolik [260]3 years ago
7 0

Answer:

The new coordinate of B' will be ( -1, 6)

kotykmax [81]3 years ago
6 0

Answer:

( -1, 6)

Step-by-step explanation:

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Which of the following is not a way to represent the solution of the inequality 3(2x − 1) greater than or equal to 4(2x − 3) − 3
vfiekz [6]
3(2x - 1) > = 4(2x - 3) - 3
6x - 3 > = 8x - 12 - 3
6x - 3 > = 8x - 15
6x - 8x > = -15 + 3
-2x > = -12
x < = -12/-2
x < = 6


the incorrect one is : A number line with a closed circle on 6, shading to the right


3 0
3 years ago
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A function of the form fx) = ab' is modified so that the b value remains the same but the a value is increased by 2.
ehidna [41]

Answer:

please dont delete ill give points back

Step-by-step explanation:

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4 0
3 years ago
HELP, it’s due in 30 minutes when my class ends. Will give brainliest
ziro4ka [17]

Answer:

Because elena must have went higher above ground

Step-by-step explanation:

4 0
3 years ago
Which expressions are equivalent to the one below? Check all that apply.
dem82 [27]

Answer:

Recall from Logarithm law Log a - Log b = Log (a/b)

so, Log 2 - Log 4 = Log ( 2/4)

                            = Log ( 1/2).

Also from the law

Log (a/b) = Log a - Log b

i.e Log (1/2) = Log 1 - Log 2

Recall: Log 1 = 0

which implies : Log 1 - Log 2 = 0 - Log 2

                                               = - Log 2

Step-by-step explanation:

3 0
3 years ago
Read 2 more answers
Use the upper and lower sums to approximate the area of the region using the given number of subintervals (of equal width)
Mashutka [201]
From the figure shown, the interval is divided into 5 equal parts making each subinterval to be 0.2.

Part A:

y= \sqrt{1-x^2}

The approximate the area of the region shown in the figure using the lower sums is given by:

 Area= [y(0.2)\times0.2]+[y(0.4)\times0.2]+[y(0.6)\times0.2]+[y(0.8)\times0.2] \\ +[y(1)\times0.2] \\  \\ =[\sqrt{1-(0.2)^2}\times0.2]+[\sqrt{1-(0.4)^2}\times0.2]+[\sqrt{1-(0.6)^2}\times0.2] \\ +[\sqrt{1-(0.8)^2}\times0.2]+[\sqrt{1-(1)^2}\times0.2] \\  \\ =(0.9798\times0.2)+(0.9165\times0.2)+(0.8\times0.2)+(0.6\times0.2)+(0\times0.2) \\  \\ =0.196+0.183+0.16+0.12=0.659



Part B:

The approximate the area of the region shown in the figure using the lower sums is given by:

 Area= [y(0)\times0.2]+[y(0.2)\times0.2]+[y(0.4)\times0.2]+[y(0.6)\times0.2] \\ +[y(0.8)\times0.2] \\ \\ =[\sqrt{1-(0)^2}\times0.2]+[\sqrt{1-(0.2)^2}\times0.2]+[\sqrt{1-(0.4)^2}\times0.2] \\ +[\sqrt{1-(0.6)^2}\times0.2] +[\sqrt{1-(0.8)^2}\times0.2] \\ \\ =(1\times0.2)+(0.9798\times0.2)+(0.9165\times0.2)+(0.8\times0.2)+(0.6\times0.2) \\ \\ =0.2+0.196+0.183+0.16+0.12=0.859



Part C:

The approximate area of the given region is given by

Area= \frac{0.659+0.859}{2} = \frac{1.518}{2} =0.759
7 0
4 years ago
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