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exis [7]
4 years ago
15

ASAP PLEASE HELP WILL MARK BRAINLIEST

Mathematics
1 answer:
Lubov Fominskaja [6]4 years ago
6 0

Answer:

Clockwise 270° around the origin: (-5,-4)

Counterclockwise 270° around the origin: (5,4)

Step-by-step explanation:

Clockwise

A(x,y) -> A'(-y,x)

Counterclockwise 270° rotation rule:

A(x,y) -> A'(y,-x)

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Determine whether a triangle can be formed with the given side lengths. If so, use Heron's formula to find the area of the trian
kipiarov [429]
Triangle inequality rule states that the sum of eachpair of sides must be greater than the third sides.
 240 + 123 > 207 true
240 + 207 > 123 true
123 + 207 > 240 true

A = √(s(s-a)(s-b)(s-c))
s is the semi-perimeter (a + b + c)/2
s = (240 + 123 + 207)/2
s = 285
A = √(285*(285-249)(285-123)(285-207))
A = 12,730 units^2
4 0
3 years ago
Pls help me on this question.<br>Do NOT answer 13 and 15. Only answer 14.
Ilia_Sergeevich [38]
1,700 good luck hope it helps you 100*17=1,700
4 0
3 years ago
Read 2 more answers
Determine which situations would be debits from a bank account. Select all that apply. paying an electric bill
siniylev [52]

Answer:

im not sure but i think its c

Step-by-step explanation:

3 0
3 years ago
Read 2 more answers
Marely walked around 2 rectangular parks. One measures 54 feet by 38 feet and the other measures 32 feet by 22 feet.
galben [10]

Answer:

130 feet or 39.6m

Step-by-step explanation:

for Marely's first rectangular park, L=54 and B= 38

Total perimeter of distance walked in first rectangular park is 2(L+B), which is 2(54+38) =184 feet

For Marely's second rectangular park, L= 32 and B=22

Total perimeter of second distance 2(L+B), which is 2(32+22) = 108 feet

Total perimeter covered by Marely = (P1+P2) 184+108 = 292 feet

Marely's brother = 2(L+B)

2(48+33) = 162 feet

Hence, difference is 292-162 = 130 feet or 39.6m

3 0
3 years ago
3. Solve the system using elimination (not substitution or matrices). negative 2 x plus y minus 2 z equals negative 8A N D7 x pl
riadik2000 [5.3K]

Elimination Method

\begin{gathered} -2X+Y-2Z=-8 \\ 7X+Y+Z=-1 \\ 5X+2Y-Z=-9 \end{gathered}

If we multiply the equation 3 by (-1) we obtain this:

\begin{gathered} -2X+Y-2Z=-8 \\ 7X+Y+Z=-1 \\ -5X-2Y+Z=9 \end{gathered}

If we add them we obtain 0, therefore there are infinite solutions. So, let's write it in terms of Z

1. Using the 3rd equation we can obtain X(Y,Z)

\begin{gathered} 5X=-9-2Y+Z \\ X=\frac{-9-2Y+Z}{5} \\  \end{gathered}

2. We can replace this value of X in the 1st and 2nd equations

\begin{gathered} -2\cdot(\frac{-9-2Y+Z}{5})+Y-2Z=-8 \\ 7\cdot(\frac{-9-2Y+Z}{5})+Y+Z=-1 \end{gathered}

3. If we simplify:

\begin{gathered} \frac{-9Y+12Z-63}{5}=-1 \\ \frac{9Y-12Z+18}{5}=-8 \end{gathered}

4. We can obtain Y from this two equations:

\begin{gathered} Y=-\frac{-12Z+58}{9} \\  \end{gathered}

5. Now, we need to obtain X(Z). We can replace Y in X(Y,Z)

\begin{gathered} X=\frac{-9-2Y+Z}{5} \\ X=\frac{-9-2(-\frac{-12Z+58}{9})+Z}{5} \end{gathered}

6. If we simplify, we obtain:

X=\frac{-3Z+7}{9}

7. In conclusion, we obtain that

(X,Y,Z) =

(\frac{-3Z+7}{9},-\frac{-12Z+58}{9},Z)

8 0
1 year ago
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