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vaieri [72.5K]
3 years ago
9

Help! Based on the given information, what can you conclude and why?

Mathematics
1 answer:
DochEvi [55]3 years ago
6 0

Answer: \triangle HIJ \cong \triangle LKJ

Step-by-step explanation:

Given: ∠H ≅∠L and line HJ congruent to Line JL.

Since, In triangles HIJ and LKJ,

∠H ≅ ∠L

HJ ≅ JL

And, ∠HJI ≅ ∠ LJK

Therefore, By ASA postulate,

Δ HIJ ≅ Δ LKJ

This is why Option first is correct.

Note: Option third ΔHIJ≅ΔJKL is not correct,

Because,  HI≠JK



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The graph of 15x + 7y = 15 is shown on the grid. Which ordered pair is in the solution set of 15x + 7y ≥ 15?
elena55 [62]

Answer:

The correct option is c.

Step-by-step explanation:

The given equation is

15x+7y=15

The given inequality is

15x+7y\geq 15

A point will contained in solution set if the above inequality satisfied by that point.

Check the point (0,-3),

15(0)+7(-3)\geq 15

-21\geq 15

This statement is false, therefore option a is incorrect.

Check the point (-3,0),

15(-3)+7(0)\geq 15

-45\geq 15

This statement is false, therefore option b is incorrect.

Check the point (3,3),

15(3)+7(3)\geq 15

66\geq 15

This statement is true, therefore option c is correct.

Check the point (-3,-3),

15(-3)+7(-3)\geq 15

-66\geq 15

This statement is false, therefore option d is incorrect.

7 0
3 years ago
In the graph above, the coordinates of the vertices of QPR are Q(3, 0), P(5, 6), and R(7, 0). If AQPR is reflected across
ozzi

Answer:

D. (7, 0)

Step-by-step explanation:

The rule for a reflection over the y-axis is (x, y) → (x, -y)

This means that the x-values stay the same while the y-values change.

Q(x, y) → (x, -y)

Q(3, 0) → (3, 0)

Q'(3, 0)

P(x, y) → (x, -y)

P(5, 6) → (5, -6)

P'(5, -6)

R(x, y) → (x, -y)

R(7, 0) → (7, 0)

R'(7, 0)

Therefore, the correct answer is D.

Hope this helps!

7 0
2 years ago
Pls help me i need this to get a 100
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Answer:

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3 0
3 years ago
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Anika [276]
Sure I can help to the best of my ability if possible . Only on mathematics
6 0
3 years ago
How to factor a tri, quad, or polynomial.
Akimi4 [234]

Explanation:

Factoring to linear factors generally involves finding the roots of the polynomial.

The two rules that are taught in Algebra courses for finding real roots of polynomials are ...

  • Descartes' rule of signs: the number of positive real roots is equal to the number of coefficient sign changes when the polynomial is written in standard form.
  • Rational root theorem: possible rational roots will have a numerator magnitude that is a divisor of the constant, and a denominator magnitude that is a divisor of the leading coefficient when the coefficients of the polynomial are rational. (Trial and error will narrow the selection.)

In general, it is a difficult problem to find irrational real factors, and even more difficult to find complex factors. The methods for finding complex factors are not generally taught in beginning Algebra courses, but may be taught in some numerical analysis courses.

Formulas exist for finding the roots of quadratic, cubic, and quartic polynomials. Above 2nd degree, they tend to be difficult to use, and may produce results that are less than easy to use. (The real roots of a cubic may be expressed in terms of cube roots of a complex number, for example.)

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Personally, I find a graphing calculator to be exceptionally useful for finding real roots. A suitable calculator can find irrational roots to calculator precision, and can use that capability to find a pair of complex roots if there is only one such pair.

There are web apps that will find all roots of virtually any polynomial of interest.

_____

<em>Additional comment</em>

Some algebra courses teach iterative methods for finding real zeros. These can include secant methods, bisection, and Newton's method iteration. There are anomalous cases that make use of these methods somewhat difficult, but they generally can work well if an approximate root value can be found.

6 0
3 years ago
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