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sweet-ann [11.9K]
3 years ago
6

Order of Operations with and without variables

Mathematics
1 answer:
lyudmila [28]3 years ago
6 0
I think the answer is -4
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Number 1d please help me analytical geometry
lesantik [10]
For a) is just the distance formula

\bf \textit{distance between 2 points}\\ \quad \\
\begin{array}{lllll}
&x_1&y_1&x_2&y_2\\
%  (a,b)
A&({{ x}}\quad ,&{{ 1}})\quad 
%  (c,d)
B&({{ -4}}\quad ,&{{ 1}})
\end{array}\qquad 
%  distance value
\begin{array}{llll}

d = \sqrt{({{ x_2}}-{{ x_1}})^2 + ({{ y_2}}-{{ y_1}})^2}
\\\\\\
\sqrt{8} = \sqrt{({{ -4}}-{{ x}})^2 + (1-1)^2}
\end{array}
-----------------------------------------------------------------------------------------
for b)  is also the distance formula, just different coordinates and distance

\bf \textit{distance between 2 points}\\ \quad \\
\begin{array}{lllll}
&x_1&y_1&x_2&y_2\\
%  (a,b)
A&({{ -7}}\quad ,&{{ y}})\quad 
%  (c,d)
B&({{ -3}}\quad ,&{{ 4}})
\end{array}\ \ 
\begin{array}{llll}

d = \sqrt{({{ x_2}}-{{ x_1}})^2 + ({{ y_2}}-{{ y_1}})^2}
\\\\\\
4\sqrt{2} = \sqrt{(-3-(-7))^2+(4-y)^2}
\end{array}
--------------------------------------------------------------------------
for c)  well... we know AB = BC.... we do have the coordinates for A and B
so... find the distance for AB, that is \bf \textit{distance between 2 points}\\ \quad \\
\begin{array}{lllll}
&x_1&y_1&x_2&y_2\\
%  (a,b)
A&({{ -3}}\quad ,&{{ 0}})\quad 
%  (c,d)
B&({{ 5}}\quad ,&{{ -2}})
\end{array}\qquad 
%  distance value
\begin{array}{llll}

d = \sqrt{({{ x_2}}-{{ x_1}})^2 + ({{ y_2}}-{{ y_1}})^2}\\\\
d=\boxed{?}

\end{array}

now.. whatever that is, is  = BC, so  the distance for BC is

\bf \textit{distance between 2 points}\\ \quad \\
\begin{array}{lllll}
&x_1&y_1&x_2&y_2\\
%  (a,b)
B&({{ 5}}\quad ,&{{ -2}})\quad 
%  (c,d)
C&({{ -13}}\quad ,&{{ y}})
\end{array}\qquad 
%  distance value
\begin{array}{llll}

d = \sqrt{({{ x_2}}-{{ x_1}})^2 + ({{ y_2}}-{{ y_1}})^2}\\\\
d=BC\\\\
BC=\boxed{?}

\end{array}

so... whatever distance you get for AB, set it equals to BC, BC will be in "y-terms" since the C point has a variable in its ordered points

so.. .solve AB = BC for "y"
------------------------------------------------------------------------------------

now d)   we know M and N are equidistant to P, that simply means that P is the midpoint of the segment MN

so use the midpoint formula

\bf \textit{middle point of 2 points }\\ \quad \\
\begin{array}{lllll}
&x_1&y_1&x_2&y_2\\
%  (a,b)
M&({{-2}}\quad ,&{{ 1}})\quad 
%  (c,d)
N&({{ x}}\quad ,&{{ 1}})
\end{array}\qquad
%   coordinates of midpoint 
\left(\cfrac{{{ x_2}} + {{ x_1}}}{2}\quad ,\quad \cfrac{{{ y_2}} + {{ y_1}}}{2} \right)=P
\\\\\\


\bf \left(\cfrac{{{ x_2}} + {{ x_1}}}{2}\quad ,\quad \cfrac{{{ y_2}} + {{ y_1}}}{2} \right)=(1,4)\implies 
\begin{cases}
\cfrac{{{ x_2}} + {{ x_1}}}{2}=1\leftarrow \textit{solve for "x"}\\\\
\cfrac{{{ y_2}} + {{ y_1}}}{2}=4
\end{cases}

now, for d), you can also just use the distance formula, find the distance for MP, then since MP = PN, find the distance for PN in x-terms and then set it to equal to MP and solve for "x"


7 0
3 years ago
Two airplanes leave the airport. Plane A departs at a 44° angle from the runway, and plane B departs at a 40° angle from the run
lawyer [7]

Answer: Answer:

Plane B was farthest away from the airport

Step-by-step explanation:

"This question requires you to visualize the run way as the horizontal distance to be covered, the height from the ground as the height gained by the plane after take of and the distance from the airport as the displacement due to the angle of take off.

In plane A

The take-off angle is 44° and the height gained is 22 ft.

Apply the relationship for sine of an angle;

Sine Ф°= opposite side length÷hypotenuse side length

The opposite side length is the height gained by plane which is 22 ft

The angle is 44° and the distance the plane will be away from the airport after take-off will be represented by the value of hypotenuse

Applying the formula

sin Ф=O/H where O=length of the side opposite to angle 44° and H is the hypotenuse

In plane B

Angle of take-off =40°, height of plane=22miles finding the hypotenuse

Solution

After take-off and reaching a height of 22 ft from the ground, plane A will be 31.67 miles from the airport

After take-off and reaching a height of 22 ft from the ground, plane B will be 34.23 miles away from the airport."

Excerpt from Brainly

3 0
2 years ago
Read 2 more answers
Find the outlier of 22,26,31,37,3126,35,28,24
olya-2409 [2.1K]

Answer:

3126

Step-by-step explanation:

Because it is so much bigger than the other numbers.

5 0
2 years ago
Read 2 more answers
Hep with this please
pishuonlain [190]

Answer:

x = 23

Step-by-step explanation:

6x+19+x = 180

7x = 180 - 19

7x =  161

x = 161/7

x = 23

6 0
2 years ago
How do you solve (3x+)(x+3), I dont want the answer, I just want to know how to solve it in complete sentences.
artcher [175]

Answer:

???????????//?///

Step-by-step explanation:

7 0
2 years ago
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