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Llana [10]
3 years ago
9

Help Image attached Please

Mathematics
2 answers:
mojhsa [17]3 years ago
5 0
Answer: A

Steps:
-5x + 10 > -15
-5x > -15 - 10
-5x > -25
x < -25 / -5 (sign flipped bc we divided by a negative)
x < 5


Stels [109]3 years ago
4 0

Answer:

A

Step-by-step explanation:

-5x + 10 > -15

-5x > -25

x < 5

(switch direction of inequality sign when multiplying or dividing by a negative number)

Because x < 5, the arrow would go left (more negative) starting at 5, so the answer would be A.

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Figure RST is congruent is R´S´T´

Step-by-step explanation:

Given that R´S´T is a reflection of RST, we know that is a rigid transformation, so it is congruent (the same size & shape).

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Write the equation of the line perpendicular to + = that passes through (−,−). Write your answer in slope-intercept form.
Ad libitum [116K]

Answer:

y = 0.25x - 5

Step-by-step explanation:

Given line is 4x + y = 3

or y = -4x + 3

Comparing this with slope-intercept form y = mx + c :

slope of this line is -4

Product of slopes of perpendicular lines = -1

⇒ slope of a line perpendicular to this is \frac{-1}{-4} = \frac{1}{4} = 0.25

This line also passes through the point (-4,-6)

The equation of a line having slope m and passing through a point (h,k) is

y - k = m(x - h)

⇒ equation of line perpendicular to given line is y - (-6) = 0.25×{x - (-4)}

⇒ y + 6 = 0.25×(x + 4)

⇒ y + 6 = 0.25x + 1

⇒ <u>y = 0.25x - 5</u>

This is in the slope-intercept form y = mx + c with slope m = 0.25 and y-intercept (0,-5)

6 0
4 years ago
Find the nature of the root
GREYUIT [131]

Answer:

1) Real and same.

2) Real and distinct

3) Real and distinct

4) Real and distinct

5) Real and distinct

6) Real and distinct

7) Real and distinct

8) Real and distinct

Step-by-step explanation:

If a quadratic equation is in the form of ax² + bx + c = 0, then Discriminant of the equation D = b² - 4ac, which governs the nature of roots the equation has.

If D > 0, then there will be two different real roots.

If D = 0, then two same and real roots.

If D < 0, then two distinct but imaginary roots.

Now, 1) x² + 6x + 9 = 0 has D = 6² - 4 × 9 × 1 = 0. So, there will be two same and real roots.

2) 5x² - x = 4x² + 2

⇒ x² - x - 2 = 0

It has D = (-1)² - 4 × 1 × (-2) = 9. Therefore, the roots will be two distinct and real.

3) \frac{2}{x} + \frac{3}{x} = x - 4

⇒ 5 = x² - 4x

⇒ x² - 4x - 5 = 0.

So, D = (-4)² - 4 × (1) × (- 5) = 36. So, the equation has two real and distinct rools.

4) x(x - 5) = 4(5 - x)

⇒ x² - x - 20 = 0

Hence, D = (-1)² - 4 × 1 × (-20) = 81

So, the roots will be real and distinct.

5) x² + 7x + 1 = 0 has D = 7² - 4 × 1 × 1 = 45

So, the roots will be real and distinct.

6) 2x² + 9x + 3 = 0, has D = 9² - 4 × 2 × 3 = 57.

Hence, the roots will be real and distinct.

7) 5x² - 6 = 13x

⇒ 5x² - 13x - 6 = 0

So, D = (-13)² - 4 × 5 × (-6) = 289

So, the roots will be real and distinct.

8) x² - x = 3(x + 7)

⇒ x² - 4x - 21 = 0

It has D = (-4)² - 4 × 1 × (-21) = 100

So, the roots will be real and distinct. (Answer)

6 0
3 years ago
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