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Eddi Din [679]
3 years ago
11

What are the slope and the y-intercept of the linear function that is represented by the graph?

Mathematics
2 answers:
andreev551 [17]3 years ago
7 0
The answer is -3/4 and y-intercept is 3 because another way is of slope is rise over run. Since it is negative 3, you go down three and since the run is 4, you go to the right 4 times to finally hit a perfect point in the graph.
MrRissso [65]3 years ago
3 0
For the y-intercept, look at the graph, and find the point of intersection of the line and the y-axis. You will see that the line and the y-axis intersect at 3.
The y-intercept is 3.

To find the slope, you need to find two points that are easy to read on the graph. You must go from one point to the other, but you can only move vertically and horizontally along grid lines. Look at (0, 3) and (4, 0). Slope is rise over run. Rise is up and down. Run is left and right. Start at (0, 3), the y-intercept. To go from that point to (4, 0), you go 3 units down. Down is negative, so the run is -3. Then you are at the origin, (0, 0). To go to (4, 0), now you go 4 units to the right. That is a run of 4. The rise is -3, and the run is 4.

slope = rise/run = -3/4
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Help ASAP please! Problem is on pictures below.
Elis [28]

Answer: figures C and D.


Explanation:


The question is which two figures have the same volume. Hence, you have to calculate the volumes of each figure until you find the two with the same volume.


1) Figure A. It is a slant cone.

Dimensions:

  • slant height, l = 6 cm
  • height, h: 5 cm
  • base area, b: 20 cm²

The volume of a slant cone is the same as the volume of a regular cone if the height and radius of both cones are the same.

Formula: V = (1/3)(base area)(height) = (1/3)b·h

Calculations:

  • V = (1/3)×20cm²×5cm = 100/3 cm³

2. Figure B. It is a right cylinder

Dimensions:

  • base area, b: 20 cm²
  • height, h: 6 cm

Formula: V = (base area)(height) = b·h

Calculations:

  • V = 20 cm²· 6cm = 120 cm³

3. Figure C. It is a slant cylinder.

Dimensions:

  • base area, b: 20 cm²
  • slant height, l: 6 cm
  • height, h: 5 cm

The volume of a slant cylinder is the same as the volume of a regular cylinder if the height and radius of both cylinders are the same.

Formula: V = (base area)(height) = b·h

Calculations:

  • V = 20cm² · 5cm = 100 cm³

4. Fiigure D. It is a rectangular pyramid.

Dimensions:

  • length, l: 6cm
  • base area, b: 20 cm²
  • height, h: 5 cm

Formula: V = (base area) (height) = b·h

Calculations:

  • V = 20 cm² · 5 cm = 100 cm³

→ Now,  you have found the two figures with the same volume: figure C and figure D. ←

7 0
4 years ago
What is 9 over 10 as an decimal
iragen [17]

Answer: 0.9

Step-by-step explanation:

3 0
3 years ago
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The degree measure of angle a is 135°. which expression below is equivalent to the radian measure of angle a?
Kazeer [188]

It is given that \angle a=135^{\circ}.

Now, know that in 180 degrees there are \pi radians. This can be written as:

180^{\circ}=\pi radians

\therefore 1^{\circ}=\frac{\pi}{180} radians (dividing both sides by 180)

Thus, to find the measure of the given angle of 135^{\circ} in radians, we will have to multiply the above equation by 135. Thus, we get:

135^{\circ}=\frac{\pi}{180}\times 135 radians

135^{\circ}\approx2.356 radians

Thus, equivalent to the radian measure of angle a is 2.356


7 0
4 years ago
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Find the longer leg of the triangle.
Paha777 [63]

Answer:

Choice A. 3.

Step-by-step explanation:

The triangle in question is a right triangle.

  • The length of the hypotenuse (the side opposite to the right angle) is given.
  • The measure of one of the acute angle is also given.

As a result, the length of both legs can be found directly using the sine function and the cosine function.

Let \text{Opposite} denotes the length of the side opposite to the 30^{\circ} acute angle, and \text{Adjacent} be the length of the side next to this 30^{\circ} acute angle.

\displaystyle \begin{aligned}\text{Opposite} &= \text{Hypotenuse} \times \sin{30^{\circ}}\\ &=2\sqrt{3}\times \frac{1}{2} \\&= \sqrt{3}\end{aligned}.

Similarly,

\displaystyle \begin{aligned}\text{Adjacent} &= \text{Hypotenuse} \times \cos{30^{\circ}}\\ &=2\sqrt{3}\times \frac{\sqrt{3}}{2} \\&= 3\end{aligned}.

The longer leg in this case is the one adjacent to the 30^{\circ} acute angle. The answer will be 3.

There's a shortcut to the answer. Notice that \sin{30^{\circ}} < \cos{30^{\circ}}. The cosine of an acute angle is directly related to the adjacent leg. In other words, the leg adjacent to the 30^{\circ} angle will be the longer leg. There will be no need to find the length of the opposite leg.

Does this relationship \sin{\theta} < \cos{\theta} holds for all acute angles? (That is, 0^{\circ} < \theta?) It turns out that:

  • \sin{\theta} < \cos{\theta} if 0^{\circ} < \theta;
  • \sin{\theta} > \cos{\theta} if 45^{\circ} < \theta;
  • \sin{\theta} = \cos{\theta} if \theta = 45^{\circ}.

4 0
3 years ago
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