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PSYCHO15rus [73]
3 years ago
10

Which dimensions can create more than one triangle? A. Three angles measuring 75°,45°, and 60°. B. 3 sides measuring 7, 10, 12?

C. Three angles measuring 40,50,and60°? D. 3 sides measuring 3,4,and 5
Mathematics
2 answers:
NeTakaya3 years ago
7 0

This is vague.  Any dimensions that make a triangle can make more than one, just draw another right next to it.  What's really being asked is which dimensions can make more than one non-congruent triangle.

<span>A. Three angles measuring 75°,45°, and 60°.

That's three angles, and 75+45+60 = 180, so it's a legit triangle. The angles don't determine the sides, so we have whole family of similar triangles with these dimensions.  TRUE

<span>B. 3 sides measuring 7, 10, 12?

</span>Three sides determine the triangles size and shape uniquely; FALSE

<em>C. Three angles measuring 40</em></span><span><em>°</em></span><em>, 50°</em><span><em>, and 60°? </em>

40+50+60=150, no such triangle exists.  FALSE

<em>D. 3 sides measuring 3,4,and 5</em>

Again, three sides uniquely determine a triangle's size and shape;  FALSE


</span>
Elina [12.6K]3 years ago
5 0

Answer: A. Three angles measuring 75°,45°, and 60°.

Step-by-step explanation: thats the answer i know

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Maths question on algebraic equations ​= due in 10 minutes
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Answer:

108 degrees.

Step-by-step explanation:

2x + 16 = 3x - 12  (corresponding angles; AB parallel to CD).

16 + 12 = 3x - 2x

x = 28.

So the angle marked 3x - 12 = 3(28) - 12

= 72 degrees.

Angle y is adjacent to the angle so

y = 180 - 72

= 108 degrees.

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3 years ago
List all possible rational roots. Then use synthetic division to confirm which rational roots exist:
Kisachek [45]

Answer:

\boxed{(1) \, x = \, \pm \dfrac{1}{2}, \pm 1, \pm2, \pm \dfrac{5}{2}, \pm 5, \pm 10; (2) \, x = -2}

Step-by-step explanation:

2x³+ 6x² - x - 10 = 0

(1) Possible roots

The Rational Roots Theorem states that, if a polynomial has any rational roots, they will have the form p/q, where p is a factor of the constant term  and q is a factor of the leading coefficient.

\text{Possible rational root} = \dfrac{ p }{ q } = \dfrac{\text{factor of constant term}}{\text{factor of leading coefficient}}

In your function, the constant term is -10 and the leading coefficient is 2, so

\text{Possible root} = \dfrac{\text{factor of 10}}{\text{factor of 2}}

Factors of 10 = ±1, ±2, ±5, ±10

Factors of 2 = ±1, ±2

\text{Possible roots are } \large \boxed{\mathbf{x = \pm \dfrac{1}{2}, \pm 1, \pm2, \pm \dfrac{5}{2}, \pm 5, \pm 10}}

(2) Synthetic division

Rather than work through all 12 possibilities, I will do one that works.

\begin{array}{r|rrrr}-2 & 2 & 6 & -1 & -10\\& & -4& -4 & 10\\& 2 & 2& -5 & 0\\\end{array}

So, x = -2 is a root, and the quotient is 2x² + 2x - 5.

(3) Check for other rational roots

2x² + 2x - 5 = 0

D = b² - 4ac =2²- 4(2)(-5) = 4 + 40 = 44

√44 = 2√11, which is irrational.

Since irrational roots come in pairs, the cubic equation has two real, irrational roots and one rational root at x = -2.

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3 years ago
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