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Dahasolnce [82]
3 years ago
14

PLEASE HELP! thanks need done asap like oof im tired lol

Mathematics
2 answers:
ohaa [14]3 years ago
7 0

Answer:

first ones 11

Step-by-step explanation:

-22/-2=11

ElenaW [278]3 years ago
6 0

1. 11 because negative/negative is positive

2. 4/6=1/3

3. 11/36

4. 1/12

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What are the steps for using a compass and straightedge to construct the bisector of ZA?
romanna [79]

The order of the steps is:

Step 1. is the fifth statement

Step 2. is the first statement

Step 3. is the second statement

Step 4. is the third statement

Step 5. is the fourth statement

Step-by-step explanation:

Lets revise the steps of construct the bisector of ∠ A

Steps:

1. Place the point of the compass on point A and draw an arc that

intersects the sides of ∠A

Label the points of intersection as points B and C

2. Place the point of the compass on point B and draw an arc in the

interior of the angle

3. Without changing the opening of the compass place the point of

the compass on point C and draw another arc in the interior of the

angle

4. Label the intersection of the arcs in the interior of the angle as

Point D

5. Use the straightedge to draw AD ray

From the steps above

The order of the steps is:

Step 1. is the fifth statement

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Step 5. is the fourth statement

Learn more:

You can learn more about construction in brainly.com/question/3309261

#LearnwithBrainly

6 0
3 years ago
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The sum of two numbers is 19 and their diffrence is 3
Nikolay [14]

Answer:

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Step-by-step explanation:

6 0
3 years ago
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LA plane is trying to travel 250 miles at a bearing of 20° E of S, however, it ends 230 miles away from the
Semmy [17]

Answer:

The wind pushed the plane 65.01 miles in the direction of 45.56 ^{\circ} East of North  with respect to the destination point.

Step-by-step explanation:

Let origin, O, br the starting point and point D be the destination at 250 miles at a bearing of 20° E of S, but due to wind let D' be the actual position of the plane at 230 miles away from the  starting point in the direction of 35° E of South as shown in the figure.

So, we have |OD|=250 miles and |OD'|=230 miles.

Vector \overrightarrow{DD'} is the displacement vector of the plane pushed by the wind.

From figure, the magnitude of the required displacement vector is

|DD'|=\sqrt{|AB|^2+|PQ|^2}\;\cdots(i)

and the direction is \alpha east of north as shown in the figure,

\tan \alpha=\frac{|PQ|}{|AB|}\;\cdots(ii)

From the figure,

|AB|=|OA-OB|

\Rightarrow |AB|=|OD\cos 20 ^{\circ}-OD'\cos 35 ^{\circ}|

\Rightarrow |AB|=|250\cos 20 ^{\circ}-230\cos 35 ^{\circ}|

\Rightarrow |AB|=45.52 miles

Again, |PQ|=|OP-OQ|

\Rightarrow |PQ|=|OD\sin 20 ^{\circ}-OD'\sin 35 ^{\circ}|

\Rightarrow |PQ|=|250\sin 20 ^{\circ}-230\sin 35 ^{\circ}|

\Rightarrow |PQ|=46.42 miles

Now, from equations (i) and (ii), we have

|DD'|=\sqrt{|45.52|^2+|46.42|^2}=65.01 miles, and

\tan \alpha=\frac{|46.42|}{|45.52|}

\alpha=\tan^{-1}\left(\frac{|46.42|}{|45.52|}\right)=45.56 ^{\circ}

Hence, the wind pushed the plane 65.01 miles in the direction of 45.56 ^{\circ} E astof North  with respect to the destination point.

5 0
3 years ago
Enter the y coordinate of the solution to this system of equations.
riadik2000 [5.3K]
-x-y=1
-y=1+x
y=-1-x

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3 0
3 years ago
Describe the transformation of the graph of f into the graph of g as either a horizontal or vertical stretch. f(x)=sqrt(x) and g
SpyIntel [72]
\bf \qquad \qquad \qquad \qquad \textit{function transformations}
\\ \quad \\\\
% left side templates
\begin{array}{llll}
f(x)=&{{  A}}({{  B}}x+{{  C}})+{{  D}}
\\ \quad \\
y=&{{  A}}({{  B}}x+{{  C}})+{{  D}}
\\ \quad \\
f(x)=&{{  A}}\sqrt{{{  B}}x+{{  C}}}+{{  D}}
\\ \quad \\
f(x)=&{{  A}}(\mathbb{R})^{{{  B}}x+{{  C}}}+{{  D}}
\\ \quad \\
f(x)=&{{  A}} sin\left({{ B }}x+{{  C}}  \right)+{{  D}}
\end{array}\\\\
--------------------\\\\

\bf \bullet \textit{ stretches or shrinks horizontally by  } {{  A}}\cdot {{  B}}\\\\
\bullet \textit{ flips it upside-down if }{{  A}}\textit{ is negative}\\
\left. \qquad   \right.  \textit{reflection over the x-axis}
\\\\
\bullet \textit{ flips it sideways if }{{  B}}\textit{ is negative}\\
\left. \qquad   \right.  \textit{reflection over the y-axis}

\bf \bullet \textit{ horizontal shift by }\frac{{{  C}}}{{{  B}}}\\
\left. \qquad  \right. if\ \frac{{{  C}}}{{{  B}}}\textit{ is negative, to the right}\\\\
\left. \qquad  \right.  if\ \frac{{{  C}}}{{{  B}}}\textit{ is positive, to the left}\\\\
\bullet \textit{ vertical shift by }{{  D}}\\
\left. \qquad  \right. if\ {{  D}}\textit{ is negative, downwards}\\\\
\left. \qquad  \right. if\ {{  D}}\textit{ is positive, upwards}\\\\
\bullet \textit{ period of }\frac{2\pi }{{{  B}}}

with that template in mind, let's see    \bf f(x)=\sqrt{x}\qquad 
\begin{array}{llll}
g(x)=&\sqrt{0.5x}\\
&\quad \uparrow \\
&\quad  B
\end{array}

so B went form 1 on f(x), down to 0.5 or 1/2 on g(x)
B = 1/2, thus the graph is stretched by twice as much.
8 0
3 years ago
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