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aliina [53]
2 years ago
9

Which of the following best describes the slope of the line below? PLSSS HELP

Mathematics
1 answer:
12345 [234]2 years ago
4 0

The slope is zero. Slope formula is Y=mx+b and since B is 1.5 and it is a straight line, Y=mx+1.5. What plus 1.5 is 1.5? 0. Hope this helps.

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Which relation below represents a one to one function
musickatia [10]

Answer:

<h2>No table shows one-to-one function</h2>

Step-by-step explanation:

<em>One-to-one function is a function that preserves distinctness: it never maps distinct elements of its domain to the same element of its codomain. Every element of the function's domain is the image of at most one element of its domain.</em>

First table:

No. Because for x = 12 and x = 14 we have the same value of y = 197

Second table:

No. Because for x = -2 and x = 2 we have the same value of y = 5

Third table:

No. Because for x = 7.25 and x = 8.5 we have the same value of y = 11

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2 years ago
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Use Pythagorean identities to prove whether ΔLMN is a right, acute, or obtuse triangle. Show all work for full credit.
diamong [38]

Answer:  The given triangle LMN is an obtuse-angled triangle.

Step-by-step explanation:  We are given to use Pythagorean identities to prove whether ΔLMN is a right, acute, or obtuse triangle.

From the figure, we note that

in ΔLMN, LM = 5 units, MN = 13 units  and  LN = 14 units.

We know that a triangle with sides a units, b units and c units (a  > b, c) is said to be

(i) Right-angled triangle if b^2+c^2=a^2,

(ii) Acute-angled triangle if b^2+c^2>a^2,

(iii) Obtuse-angled triangle if b^2+c^2

For the given triangle LMN, we have

a = 14, b = 13 and c = 5.

So,

b^2+c^2=13^2+5^2=169+25=194,\\\\a^2=14^2=196.

Therefore,  b^2+c^2

Thus, the given triangle LMN is an obtuse-angled triangle.

5 0
2 years ago
The length, l cm, of a simple pendulum is directly proportional to the square of its period (time taken to complete one oscillat
Greeley [361]

Answer:

1) L \propto T^2

Using the condition given:

2.205 m = K (3)^2

K = 0.245 \approx \frac{g}{4\pi^2}

So then if we want to create an equation we need to do this:

L = K T^2

With K a constant. For this case the period of a pendulumn is given by this general expression:

T = 2\pi \sqrt{\frac{L}{g}}

Where L is the length in m and g the gravity g = 9.8 \frac{m}{s^2}.

2) T = 2\pi \sqrt{\frac{L}{g}}

If we square both sides of the equation we got:

T^2 = 4 \pi^2 \frac{L}{g}

And solving for L we got:

L = \frac{g T^2}{4 \pi^2}

Replacing we got:

L =\frac{9.8 \frac{m}{s^2} (5s)^2}{4 \pi^2} = 6.206m

3) T = 2\pi \sqrt{\frac{0.98m}{9.8\frac{m}{s^2}}}= 1.987 s

Step-by-step explanation:

Part 1

For this case we know the following info: The length, l cm, of a simple pendulum is directly proportional to the square of its period (time taken to complete one oscillation), T seconds.

L \propto T^2

Using the condition given:

2.205 m = K (3)^2

K = 0.245 \approx \frac{g}{4\pi^2}

So then if we want to create an equation we need to do this:

L = K T^2

With K a constant. For this case the period of a pendulumn is given by this general expression:

T = 2\pi \sqrt{\frac{L}{g}}

Where L is the length in m and g the gravity g = 9.8 \frac{m}{s^2}.

Part 2

For this case using the function in part a we got:

T = 2\pi \sqrt{\frac{L}{g}}

If we square both sides of the equation we got:

T^2 = 4 \pi^2 \frac{L}{g}

And solving for L we got:

L = \frac{g T^2}{4 \pi^2}

Replacing we got:

L =\frac{9.8 \frac{m}{s^2} (5s)^2}{4 \pi^2} = 6.206m

Part 3

For this case using the function in part a we got:

T = 2\pi \sqrt{\frac{L}{g}}

Replacing we got:

T = 2\pi \sqrt{\frac{0.98m}{9.8\frac{m}{s^2}}}= 1.987 s

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