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2 years ago

If one line has a slope of -5/4 what would the slope of a line parallel to that line?

Mathematics

2 answers:

artcher [175]2 years ago

7 0

Answer:

$-\frac{5}{4}$

Step-by-step explanation:

Parallel lines always have equal slopes. Therefore any parallel line to a given line will share the same slope.

MAXImum [283]2 years ago

3 0

Answer:

-5/4

Step-by-step explanation:

Parallel lines have the same slope but a different y intercepts, because if you think about parellel lines, they'll never intersect-- if two lines have the same slope, that means they won't be able to intersect. However, if they have the same slope, they would be coinciding (the same). That's why the y intercepts are different.

Since the slope of the first line is -5/4, the slope for the line parallel to it would also be -5/4

hope this helps!

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Round 5.599 to the nearest hundredth

pantera1 [17]

Not entirely sure but its like 5.6 or 5.7

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3 years ago

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The sum of the first two terms of an infinite GP is 6 and each term is 5 times the sum of the succeeding termsthen the second te

kozerog [31]

Answer:

$\dfrac{6}{7}$ .

Step-by-step explanation:

It is given that the sum of first two terms of an infinite GP is 6.

nth term of a GP is

$a_n=ar^{n-1}$

$a+ar=6$ ...(1)

Each term is 5 times the sum of the succeeding terms.

$a_n=5(a_{n+1}+a_{n+2}+...+\infty)$

$ar^{n-1}=5(ar^n+ar^{n+1}+...+\infty)$

$ar^{n-1}=5ar^n(1+r+r^2+...+\infty)$

Divide both sides by a.

$r^{n-1}=5r^n(\dfrac{1}{1-r})$ $[\text{Sum of infinite GP}=\dfrac{a}{1-r}]$

$\dfrac{r^n}{r}=\dfrac{5r^n}{1-r}$

$\dfrac{1}{r}=\dfrac{5}{1-r}$

$1-r=5r$

$1=6r$

$\dfrac{1}{6}=r$

The common ratio is 1/6.

Put r=1/6 in (1).

$a+a(\dfrac{1}{6})=6$

$6a+a=36$

$7a=36$

$a=\dfrac{36}{7}$

Second term $ar=\dfrac{36}{7}\times \dfrac{1}{6}=\dfrac{6}{7}$

Therefore, the second term is $\dfrac{6}{7}$ .

4 0

2 years ago

5-4+7x+1 use the drop down menu

In-s [12.5K]

Answer: 2+7x

Step-by-step explanation:

First you have to do PEMDAS. Then you get this answer.

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2 years ago

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A rocket is launched from the top of a 55-foot cliff with an initial velocity of 138 ft/s.

melisa1 [442]

For this case we have that the equation that describes the height is:

$h = -16t ^ 2 + vt + c$

Where,

v: initial speed

c: initial height

Substituting values we have:

$h = -16t ^ 2 + 138t + 55$

Then, to find the time, we use the quadratic formula:

$t =\frac{-b +/-\sqrt{b^2 - 4ac} }{2a}$

Substituting values we have:

$t =\frac{-138 +/-\sqrt{138^2 - 4(-16)(55)} }{2(-16)}$

Rewriting:

$t =\frac{-138 +/-\sqrt{19044 + 3520} }{-32}$

$t =\frac{-138 +/-\sqrt{22564}}{-32}$

$t =\frac{-138 +/-150.21}{-32}$

We discard the negative root because we want to find the time.

We have then:

$t =\frac{-138 -150.21}{-32}$

$t =\frac{-288.21}{-32} \\t=9$

Answer:

D. $0 = -16t^2 + 138t + 55; 9 s$

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2 years ago

The area of a rectangle is 24 square inches. Determine both dimensions if the length is 6 inches less than three times the width

Rama09 [41]

Answer:

width is 4 inches

length is 6 inches

Step-by-step explanation:

the length is 6 inches less than three times the width

l = 3w - 6

a = l * w

a = (3w - 6)w

a = 3w^2 - 6w

area of the rectangle is 24 square inches

3w^2 - 6w = 24

subtract 24 from both sides

3w^2 - 6w - 24 = 0

Divide both sides by 3

w^2 - 2w - 8 = 0

Factor

(w - 4)(w + 2) = 0

w = {-2, 4}

disregard the negative root since measurement must be positive

w = 4

l = 3w - 6

l = 3(4) - 6

l = 6

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2 years ago