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Alex777 [14]
2 years ago
14

Which number line shows Point A at −2, Point B at 3.5, Point C at −3 and 1 over 2, and point D, which is the opposite of point A

?
A number line from negative 5 to 0 to positive 5 is shown. There are increments of 1 over 2, and only the whole numbers are labeled on either side of 0. A point A is shown on 4 places to the right of 0, A point B is shown on 7 places to the left of 0, and a point C is shown on 7 places to the right of 0. Point D is shown 4 places to the left of 0.
A number line from negative 5 to 0 to positive 5 is shown. There are increments of 1 over 2, and only the whole numbers are labeled on either side of 0. A point A is shown on 4 places to the left of 0, A point B is shown on 7 places to the left of 0, and a point C is shown on 7 places to the right of 0. Point D is shown 4 places to the right of 0.
A number line from negative 5 to 0 to positive 5 is shown. There are increments of 1 over 2, and only the whole numbers are labeled on either side of 0. A point A is shown on 4 places to the left of 0, A point B is shown on 7 places to the right of 0, and a point C is shown on 7 places to the left of 0. Point D is shown 4 places to the right of 0.
A number line from negative 5 to 0 to positive 5 is shown. There are increments of 1 over 2, and only the whole numbers are labeled on either side of 0. A point A is shown on 4 places to the right of 0, A point B is shown on 7 places to the right of 0, and a point C is shown on 7 places to the left of 0. Point D is shown 4 places to the left of 0. (ill give brainly if you awnser correctly) :) 50 Pts!
Mathematics
2 answers:
Leona [35]2 years ago
8 0

Answer:

thanks sa points

Step-by-step explanation:

godbless

Mnenie [13.5K]2 years ago
5 0
Which number line shows Point A at −2, Point B at 3.5, Point C at −3 and 1 over 2, and point D, which is the opposite of point A?

A number line from negative 5 to 0 to positive 5 is shown. There are increments of 1 over 2, and only the whole numbers are labeled on either side of 0. A point A is shown on 4 places to the right of 0, A point B is shown on 7 places to the left of 0, and a point C is shown on 7 places to the right of 0. Point D is shown 4 places to the left of 0.
A number line from negative 5 to 0 to positive 5 is shown. There are increments of 1 over 2, and only the whole numbers are labeled on either side of 0. A point A is shown on 4 places to the left of 0, A point B is shown on 7 places to the left of 0, and a point C is shown on 7 places to the right of 0. Point D is shown 4 places to the right of 0.
A number line from negative 5 to 0 to positive 5 is shown. There are increments of 1 over 2, and only the whole numbers are labeled on either side of 0. A point A is shown on 4 places to the left of 0, A point B is shown on 7 places to the right of 0, and a point C is shown on 7 places to the left of 0. Point D is shown 4 places to the right of 0.
A number line from negative 5 to 0 to positive 5 is shown. There are increments of 1 over 2, and only the whole numbers are labeled on either side of 0. A point A is shown on 4 places to the right of 0, A point B is shown on 7 places to the right of 0, and a point C is shown on 7 places to the left of 0. Point D is shown 4 places to the 50 Pts!nnb
Hope it helps bc d ensnnxbxbcb f dnbwnakxjs
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4y + 4 = 2y - 10 <br> can somebody break this down to solve it ? please
vredina [299]

Answer:

y=-7

Step-by-step explanation:

first, you want variables on one side, numbers on the other side of the equation.

4y+4=2y-10

*plus ten on both side to get rid of -10*

4y+14=2y

*minus 4y on each side to get rid of 4y*

14=-2y

*divide by -2 on each side to get the value of just one y itself*

y=-7

8 0
3 years ago
I don't get this one plz help
ololo11 [35]
Isocoleese means 2 angles are same, normally the base angles

so since it says they measure 42
rememeber that all angles in traingle add to 180
so base angles+vertex angle=180
base angles aer both 42 so
2 times 42+vertex angle=180
42 times 2=84
so
if veretx angle=z then
84+z=180


answer is B
3 0
3 years ago
Read 2 more answers
How can csc^2-cot^2 = 1?
Softa [21]

ANSWER

By simplifying the left hand side using the Pythagorean Identity.

EXPLANATION

The given identity is

\csc^{2} (x) -  \cot^{2} (x) = 1

Take the left hand side and simplify to get the right hand side.

\csc^{2} (x) -  \cot^{2} (x) =  \frac{1}{\sin^{2} (x)}  - \frac{ \cos^{2} (x)}{\sin^{2} (x)}

Collect LCM for the denominators.

\csc^{2} (x) -  \cot^{2} (x) =  \frac{1 -  \cos^{2} (x)}{\sin^{2} (x)}

Recall the Pythagorean Identity.

\cos^{2} (x) +  \sin^{2} (x) = 1

This implies that:

1 -  \cos^{2} (x) =  \sin^{2} (x)

We substitute this to get,

\csc^{2} (x) -  \cot^{2} (x) =  \frac{\sin^{2} (x)}{\sin^{2} (x)}

\csc^{2} (x) -  \cot^{2} (x) = 1

6 0
3 years ago
The given matrix is the augmented matrix for a linear system. Use technology to perform the row operations needed to transform t
shtirl [24]

Answer:

x_{1} = \frac{176}{127} + \frac{71}{127}x_{4}\\\\ x_{2} = \frac{284}{127} + \frac{131}{254}x_{4}\\\\x_{3} = \frac{845}{127} + \frac{663}{254}x_{4}\\

Step-by-step explanation:

As the given Augmented matrix is

\left[\begin{array}{ccccc}9&-2&0&-4&:8\\0&7&-1&-1&:9\\8&12&-6&5&:-2\end{array}\right]

Step 1 :

r_{1}↔r_{1} - r_{2}

\left[\begin{array}{ccccc}1&-14&6&-9&:10\\0&7&-1&-1&:9\\8&12&-6&5&:-2\end{array}\right]

Step 2 :

r_{3}↔r_{3} - 8r_{1}

\left[\begin{array}{ccccc}1&-14&6&-9&:10\\0&7&-1&-1&:9\\0&124&-54&77&:-82\end{array}\right]

Step 3 :

r_{2}↔\frac{r_{2}}{7}

\left[\begin{array}{ccccc}1&-14&6&-9&:10\\0&1&-\frac{1}{7} &-\frac{1}{7} &:\frac{9}{7} \\0&124&-54&77&:-82\end{array}\right]

Step 4 :

r_{1}↔r_{1} + 14r_{2} , r_{3}↔r_{3} - 124r_{2}

\left[\begin{array}{ccccc}1&0&4&-11&:-8\\0&1&-\frac{1}{7} &-\frac{1}{7} &:\frac{9}{7} \\0&0&- \frac{254}{7} &\frac{663}{7} &:-\frac{1690}{7} \end{array}\right]

Step 5 :

r_{3}↔\frac{r_{3}. 7}{254}

\left[\begin{array}{ccccc}1&0&4&-11&:-8\\0&1&-\frac{1}{7} &-\frac{1}{7} &:\frac{9}{7} \\0&0&1&-\frac{663}{254} &:-\frac{1690}{254} \end{array}\right]

Step 6 :

r_{1}↔r_{1} - 4r_{3} , r_{2}↔r_{2} + \frac{1}{7} r_{3}

\left[\begin{array}{ccccc}1&0&0&-\frac{71}{127} &:\frac{176}{127} \\0&1&0&-\frac{131}{254} &:\frac{284}{127} \\0&0&1&-\frac{663}{254} &:\frac{845}{127} \end{array}\right]

∴ we get

x_{1} = \frac{176}{127} + \frac{71}{127}x_{4}\\\\ x_{2} = \frac{284}{127} + \frac{131}{254}x_{4}\\\\x_{3} = \frac{845}{127} + \frac{663}{254}x_{4}\\

6 0
3 years ago
What is the decimal that is equivalent to the expression 3 * 100 + 7 * 10 + 4 * 1 + 6 * 0.01 + 1 * 0.001
barxatty [35]
I believe it is 0.0318
5 0
2 years ago
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