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2 years ago

Graph y=-2x+5 please help fast

Mathematics

1 answer:

Vlad [161]2 years ago

4 0

Answer:

y=−2x+5

Use the slope-intercept form to find the slope and y-intercept.

The slope-intercept form is

y=mx+b , where m is the slope and b is the y-intercept.

y=mx+b

Find the values of

m and b

using the form

y=mx+b.

m=−2b=5

The slope of the line is the value of m , and the y-intercept is the value of

Slope: −2

y-intercept:

(0,5)

Any line can be graphed using two points. Select two x values, and plug them into the equation to find the corresponding y values.

Graph the line using the slope and the y-intercept, or the points.

Slope: −2

y-intercept: (0,5)

X|Y

0 5

1 3

Step-by-step explanation:

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Read 2 more answers

Teresa graphs the following 33 equations: y=2xy=2x, y=x2+2y=x2+2, and y=2x2y=2x2. She says that the graph of y=2xy=2x will event

Neko [114]

The correct question is
Teresa graphs the following 3 equations: y=2x, y=x2+2, and y=2x2. She says that the graph of y=2x will eventually surpass both of the other graphs. Is Teresa correct? Why or why not?

we have that
y=2x
y=x²+2
y=2x²

using a graph tool
see the attached figure

We can affirm the following
the three graphs present the same domain-----> the interval (-∞,∞)

The range of the graph y=2x is the interval (-∞,∞)

The range of the graphs y=x²+2 and y=2x² is the interval [0,∞)

therefore
Teresa is not correct because the graph of y = 2x will not surpass the other two graphs since in the interval of [0, infinite) the three graphs present the same range

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3 years ago

The rental rate for a power washer has a one time fee of $60 plus $20 for each day of the rental. The graph shows the points rep

Kamila [148]

I think it’s

Y=25x+50

Hope this helps

7 0

2 years ago

Select all the conditions for which it is possible to construct a triangle. (7.G.1.2) Group of answer choices a. A triangle with

saw5 [17]

Answer:

b, d, e, f

Step-by-step explanation:

Here are the applicable restrictions:

The sum of angles in a triangle is 180°, no more, no less.
The sum of the lengths of the two shortest sides exceeds the longest side.
When two sides and the angle opposite the shortest is given, the sine of the given angle must be at most the ratio of the shortest to longest sides.

a. A triangle with angle measures 60°, 80°, and 80° (angle sum ≠ 180°, not OK)

b. A triangle with side lengths 4 cm, 5 cm, and 6 cm (4+5 > 6, OK)

c. A triangle with side lengths 4 cm, 5 cm, and 15 cm (4+5 < 15, not OK)

d. A triangle with side lengths 4 cm, 5 cm, and a 50° angle across from the 4 cm side (sin(50°) ≈ 0.766 < 4/5, OK)

e. A triangle with angle measures 30° and 60°, and an included 3 cm side length (OK)

f. A triangle with angle measures 60°, 20°, and 100° (angle sum = 180°, OK)

_____

Additional comment

In choice "e", two angles and the side between them are specified. As long as the sum of the two angles is less than 180°, a triangle can be formed. The length of the side is immaterial with respect to whether a triangle can be made.

The congruence postulates for triangles are ...

SSS, SAS, ASA, AAS, and HL

These essentially tell you the side and angle specifications necessary to define a singular triangle. As we discussed above, the triangle inequality puts limits on the side lengths specified in SSS. The angle sum theorem puts limits on the angles when only two are specified (ASA, AAS).

In terms of sides and angles, the HL postulate is equivalent to an SSA theorem, where the angle is 90°. In that case, the angle is opposite the longest side (H). In general, SSA will specify a singular triangle when the angle is opposite the longest specified side, regardless of that angle's measure. However, when the angle is opposite the shortest specified side, the above-described ratio restriction holds. If the sine of the angle is less than the ratio of sides, then two possible triangles are specified.

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2 years ago