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spayn [35]
2 years ago
5

Transversal Problems with Equations (Level 1)

Mathematics
1 answer:
elena-14-01-66 [18.8K]2 years ago
8 0

Answer:

x is 21°

Step-by-step explanation:

The angles formed between the transversal <em>t</em> and lines <em>m</em> and <em>n</em> are; (8·x - 7)°, and (9·x - 28)°

Based on the similar location the angles are formed by the transversal, <em>t</em>, and lines <em>m</em> and <em>n</em>, the angles are corresponding angles

Given that lines, <em>m</em> and <em>n</em> are parallel, we have the corresponding angles formed by the transversal, <em>t</em>, and the lines are equal, therefore;

(8·x - 7)° = (9·x - 28)°

Simplifying the above equation to make <em>x</em> the subject, we get

(28 - 7)° = 9·x - 8·x = x

∴ 21° = x

x = 21°.

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Which graph represents exponential decay?
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Answer:

  (d) -- see attached

Step-by-step explanation:

A graph that shows exponential decay is one that tends toward a horizontal asymptote as x gets large.

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A basic (parent) exponential decay curve is concave upward and tends toward zero as x gets large. The fractional change in any interval is the same as for any other interval of equal size. The curve attached decreases by a factor of 2 when x increases by 1.

6 0
2 years ago
What else would need to be congruent to show that JKL= MNO by AAS?
taurus [48]
Answer: Choice B) Angle L = Angle O

---------------------------------------------------
---------------------------------------------------

If we know that Angle L is congruent to Angle O, then we can use the AAS (angle angle side) congruence property. We have one pair of angles marked by the square marker (angle J and angle M). So they are congruent angles. We have a pair of congruent sides JK = MN = 3. So we're just missing a pair of angles. 

Note: The answer is NOT angle K = angle N because this would mean ASA would be used instead of AAS. The order of the letters is important as it establishes how the sides and angles relate. With ASA, the side is between the angles. With AAS, the side is not between the angles. 

3 0
3 years ago
Read 2 more answers
What is the length of the hypotenuse of a right triangle with legs of length 6 and 4 ?
liberstina [14]

Answer:

6²+4²=h²

36+16=h²

\sqrt{52}  = h \:  \:  \:  \:  \:  \\ hyphotenuse \: is \: 2 \sqrt{13  \:} \\ or \: 7.2111

6 0
3 years ago
Please please help!!!! According to the probability distribution below, what is the expected value of X? X 0 1 2 3 4 5 P(X) 0.3
Novay_Z [31]

Answer:

D.) The expected value is 1.78

Step-by-step explanation:

The expected value of a probability distribution is evaluated using the formula.

Expected Value, E(X)=\sum^{n}_{i=1}x_iP(x_i)

Therefore, from the given probability distribution, we have:

E(X)=(0*0.3)+(1*0.2)+(2*0.16)+(3*0.2)+(4*0.04)+(5*0.1)

E(X)=1.78

The Expected value of X is 1.78.

The correct option is D.

4 0
3 years ago
Help! Algebra!!!
balandron [24]
Thet contradict each other, that's why both of them are incorrect.
<span>Suppose that a polynomial has four roots: s, t, u, and v. If the polynomial were evaluated at any of these values, it would have to be zero. Therefore, the polynomial can be written in this form. 

p(x)(x - s)(x - t)(x - u)(x - v), where p(x) is some non-zero polynomial 

This polynomial has a degree of at least 4. It therefore cannot be cubic. 

Now prove Kelsey correct. We have already proved that there can be no more than three roots. To prove that a cubic polynomial with three roots is possible, all we have to do is offer a single example of that. This one will do. 

(x - 1)(x - 2)(x - 3) 

This is a cubic polynomial with three roots, and four or more roots are not possible for a cubic polynomial. Kelsey is correct. 

Incidentally, if this is a roller coaster we are discussing, then a cubic polynomial is not such a good idea, either for a vertical curve or a horizontal curve. I hope this helps</span><span>
</span>
4 0
3 years ago
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