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olganol [36]
3 years ago
14

Please help me with this is due right now thamk

Mathematics
1 answer:
almond37 [142]3 years ago
5 0

Answer:

1/2

Step-by-step explanation:

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What is the solution to this inequality?
AlladinOne [14]

Answer:

D

Step-by-step explanation:

The answer is D, because when you divide x by a negative the sign changes

7 0
2 years ago
Label this graph as “linear” or “nonlinear.”
butalik [34]
That graph would be non-linear
7 0
2 years ago
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please help in the picture below (top left is A) (bottom left is B) top right is C ) and bottom right is D) (will report for non
Phoenix [80]
For this you count rise over run aka how many up divided by how many over starting from 0. here it’s 7 up and 3.5 side, 7 divided by 3.5 is 2 and the line is going down so it’s negative 2 (D)
3 0
3 years ago
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Can you help me please??
dusya [7]

Answer:

y = \frac{4}{5}x+\frac{54}{5}

Step-by-step explanation:

Equation of a line has been given as,

y=\frac{4}{5}x+\frac{3}{5}

Here, slope of the line = \frac{4}{5}

y-intercept = \frac{3}{5}

"If the two lines are parallel, there slopes will be equal"

By this property slope of the parallel line to the given line will be equal.

Therefore, slope 'm' = \frac{4}{5}

Since, slope intercept form of a line is,

y = mx + b

Therefore, equation of the parallel line will be,

y = \frac{4}{5}x+b

Since, this line passes through a point (-6, 6),

6 = \frac{4}{5}(-6)+b

6 = -\frac{24}{5}+b

b = 6+\frac{24}{5}

b = \frac{30+24}{5}

b = \frac{54}{5}

Equation of the parallel line will be,

y = \frac{4}{5}x+\frac{54}{5}

4 0
3 years ago
Which sum or difference identity would you use to verify that cos (180° - q) = -cos q?
Phantasy [73]

Answer:

\cos (a-b)=\cos a \cos b+\sin a \sin b

Step-by-step explanation:

 Given : \cos (180^{\circ}-q)=-\cos q

We have to write which identity we will use to prove the given statement.

Consider \cos (180^{\circ}-q)=-\cos q

Take left hand side of given expression \cos (180^{\circ}-q)

We know

\cos (a-b)=\cos a \cos b+\sin a \sin b

Comparing , we get, a= 180° and b = q

Substitute , we get,

\cos (180^{\circ}-q)=\cos 180^{\circ}  \cos (q)+\sin q \sin 180^{\circ}

Also, we know \sin 180^{\circ}=0 and \cos 180^{\circ}=-1

Substitute, we get,

\cos (180^{\circ}-q)=-1\cdot \cos (q)+\sin q \cdot 0

Simplify , we get,

\cos (180^{\circ}-q)=-\cos (q)

Hence, use difference identity to  prove the given result.

7 0
3 years ago
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