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3 years ago

The slope of a line is -1/2 what is the slope of a line that is parallel to it?

1 answer:

5 0

Parallel lines have the exact same slope ( or else they would eventually run into each other). So the answer is -1/2 is the slope.

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im stupid so, if my grade was at a 90% (A) and my tests were worth 50% and i get a 40% on the test, what's my current grade now?

Butoxors [25]

I think it would be a B

6 0

2 years ago

If a tree 28 meters tall casts a shadow 32 meters long, what is the angle of elevation of the sun to the nearest degree?

avanturin [10]

Since we need an angle measure we have to use an inverse trig ratio, sin^-1, cos^-1, or tan^-1.
Setting up the right triangle we have the opposite and adjacent so we use tan.
tan angle = 28/32
tan^-1 (28/32) = 41 degrees

4 0

3 years ago

Please help me with domain and range of graphs Part 1

dybincka [34]

Answer:

see the attached for domain
Range: (-∞, 4], [3, ∞), [-3, 1]

Step-by-step explanation:

The domain is the horizontal extent of the graph. When the graph has arrows on the end, it extends to infinity in the indicated direction.

The range is the vertical extent of the graph.

<u>Left Graph</u>

We assume the curve is intended to touch, not cross, the line y=4, so the range is ...

range: (-∞, 4]

<u>Middle Graph</u>

We assume the graph continues along the line y=3 to the right indefinitely. Then this value is the minimum of the range.

range: [3, ∞)

<u>Right Graph</u>

The low point on the graph is at y=-3, and it looks like the graph continues to the right along the line y=1. Then the range is ...

range: [-3, 1]

8 0

2 years ago

For ΔABC, which two relationships are true?

Phoenix [80]

The two correct relationships are:

$cos(\theta)=\frac{AB}{AC} = sin(90-\theta)\\\\sin(\theta)=\frac{BC}{AC} = cos(90-\theta)$

<h3>What is a right-angled triangle?</h3>

The triangle shown is a right-angled triangle

Considering < $(90-\theta)^o$

$cos(90-\theta)=\frac{BC}{AC} \\\\sin(90-\theta)=\frac{AB}{AC}\\\\tan(90-\theta)=\frac{AB}{BC}$

Considering < $\theta$

$cos(\theta)=\frac{AB}{AC} \\\\sin\theta)=\frac{BC}{AC}\\\\tan(\theta)=\frac{BC}{AB}$

Comparing < $\theta$ and < $(90-\theta)^o$ , the true statements are:

$cos(\theta)=\frac{AB}{AC} = sin(90-\theta)\\\\sin(\theta)=\frac{BC}{AC} = cos(90-\theta)$

Learn more on trigonometry here: brainly.com/question/20519838

6 0

2 years ago

How to find angles in tranversal

xeze [42]

We have some couples of angles that are equal

5 0

3 years ago