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3 years ago

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:

Step-by-step explanation:

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

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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)

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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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