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3 years ago

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How do we show that two lines are parallel when using slope? How do we show they are perpendicular when using slope?

1 answer:

Amanda [17]3 years ago

3 0

Answer:

QUESTION:

How do we show that two lines are parallel when using slope? How do we show they are perpendicular when using slope?

ANSWER:

<u><em>How to determine if a line is a parallel perpendicular or neither?</em></u>

The slopes of parallel lines are equal, and the slopes of perpendicular lines are opposite reciprocals. This is a worked example of determining whether given lines are parallel or perpendicular.

Step-by-step explanation:

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B - 1 > -17 what does b equal?

Nastasia [14]

Answer:

b > -16

Step-by-step explanation:

The inequality is given as:

b - 1 > -17

To solve this problem, add +1 to both sides of the expression:

b - 1 + 1 > -17 + 1

Now;

b > -16

So, all values for which b is more than -16 satisfy the solution for this problem.

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2 years ago

X^3+13x^2+32x+20<br> please help me factorizing above equation

Mazyrski [523]

$x^3+13x^2+32x+20=x^3+x^2+12x^2+ 12x+20x+20 =\\\\=x^2(x+1)+12x(x + 1 )+20(x+1 )=(x+1)(x^2+12x+20)=\\\\=(x+1)[(x^2+10x+2x+20)]=(x+1)[x(x +10)+2(x+10))]=\\\\=(x+1) (x +2) (x+10)$

7 0

3 years ago

The angle looking up at the sun is 70°. A flagpole casts a shadow of 15 ft. Draw a diagram and find the height of the flagpole

deff fn [24]

Answer:

The length of the flagpole be 41.2 ft .

Option (A) is correct .

Step-by-step explanation:

As given

The angle looking up at the sun is 70°.

A flagpole casts a shadow of 15 ft .

Now by using the trignometric identity .

$tan \theta = \frac{Perpendicular}{Base}$

$\ theta = 70^{\circ}$

$tan\ 70^{\circ} = \frac{AB}{BC}$

BC = 15 ft

tan 70° = 2.75 (Approx)

$2.75 = \frac{AB}{15}$

AB = 2.75 × 15

= 41.2 ft

Therefore the length of the flagpole be 41.2 ft .

Option (A) is correct .

4 0

3 years ago

Read 2 more answers

Solve each equation.<br> 3) 4= 2x + 3.x

Andrej [43]

Answer:

Step-by-step explanation:

2x+3x=4

5x=4

x=4/5

6 0

2 years ago

Read 2 more answers

Find the oth term of the geometric sequence 7, 14, 28, ...

yaroslaw [1]

Answer:

The nth term of the geometric sequence 7, 14, 28, ... is:

$a_n=7\cdot \:2^{n-1}$

Step-by-step explanation:

Given the geometric sequence

7, 14, 28, ...

We know that a geometric sequence has a constant ratio 'r' and is defined by

$a_n=a_1\cdot r^{n-1}$

where a₁ is the first term and r is the common ratio

Computing the ratios of all the adjacent terms

$\frac{14}{7}=2,\:\quad \frac{28}{14}=2$

The ratio of all the adjacent terms is the same and equal to

$r=2$

now substituting r = 2 and a₁ = 7 in the nth term

$a_n=a_1\cdot r^{n-1}$

$a_n=7\cdot \:2^{n-1}$

Therefore, the nth term of the geometric sequence 7, 14, 28, ... is:

$a_n=7\cdot \:2^{n-1}$

6 0

2 years ago