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3 years ago

What is an equation of the line with slope 3 that goes through point (-1/2, 5/2)?

Mathematics

1 answer:

kkurt [141]3 years ago

4 0

The only equation that has 3 as the coefficient of x (a slope of 3) is ...

... d.) y = 3x+4

_____

Happily, it also goes through the given point.

... 5/2 = 3(-1/2) +4

... 5/2 = -3/2 + 8/2 . . . . true

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PLZ HELP WILL MARK BRANLIEST (No dum answers or I will delete your question - seriously don't be annoying)

Vsevolod [243]

Answer:

A. tan(2π/3) = $-\sqrt{3}$

4 0

2 years ago

Find the midpoint of JK if J(-7. -5) and K(-3,7).

qaws [65]

Answer:

( − 5 , 1 )

Step-by-step explanation:

Use the midpoint formula to find the midpoint of the line segment.

7 0

3 years ago

I know you want to answer this question.

Alik [6]

Answer:

D. x = 3

Step-by-step explanation:

$\frac{1}{2} ^{x-4}$ - 3 = $4^{x-3}$ - 2

First, convert $4^{x-3}$ to base 2:

$4^{x-3}$ = $(2^{2})^{x-3}$

$\frac{1}{2} ^{x-4}$ - 3 = $(2^{2})^{x-3}$ - 2

Next, convert $\frac{1}{2} ^{x-4}$ to base 2:

$\frac{1}{2} ^{x-4}$ = $(2^{-1})^{x-4}$

$(2^{-1})^{x-4}$ - 3 = $(2^{2})^{x-3}$ - 2

Apply exponent rule: $(a^{b})^{c}$ = $a^{bc}$ :

$(2^{-1})^{x-4}$ = $2^{-1*(x-4)}$

$2^{-1*(x-4)}$ - 3 = $(2^{2})^{x-3}$ - 2

Apply exponent rule: $(a^{b})^{c}$ = $a^{bc}$ :

$(2^{2})^{x-3}$ = $2^{2(x-3)}$

$2^{-1*(x-4)}$ - 3 = $2^{2(x-3)}$ - 2

Apply exponent rule: $a^{b+c}$ = $a^{b}$ $a^{c}$ :

$2^{-1(x-4)}$ = $2^{-1x}$ * $2^{4}$ , $2^{2(x-3)}$ = $2^{2x}$ * $2^{-6}$

$2^{-1 * x}$ * $2^{4}$ - 3 = $2^{2x}$ * $2^{-6}$ - 2

Apply exponent rule: $(a^{b})^{c}$ = $a^{bc}$ :

$2^{-1x}$ = $(2^{x})^{-1}$ , $2^{2x}$ = $(2^{x})^{2}$

$(2^{x})^{-1}$ * $2^{4}$ - 3 = $(2^{x})^{2}$ * $2^{-6}$ - 2

Rewrite the equation with $2^{x} = u$ :

$(u)^{-1}$ * $2^{4}$ - 3 = $(u)^{2}$ * $2^{-6}$ - 2

Solve $u^{-1}$ * $2^{4}$ - 3 = $u^{2}$ * $2^{-6}$ - 2:

$u^{-1}$ * $2^{4}$ - 3 = $u^{2}$ * $2^{-6}$ - 2

Refine:

$\frac{16}{u}$ - 3 = $\frac{1}{64}$ $u^{2}$ - 2

Add 3 to both sides:

$\frac{16}{u}$ - 3 + 3 = $\frac{1}{64}$ $u^{2}$ - 2 + 3

Simplify:

$\frac{16}{u}$ = $\frac{1}{64}$ $u^{2}$ + 1

Multiply by the Least Common Multiplier (64u):

$\frac{16}{u}$ * 64u = $\frac{1}{64}$ $u^{2}$ + 1 * 64u

Simplify:

$\frac{16}{u}$ * 64u = $\frac{1}{64}$ $u^{2}$ + 1 * 64u

Simplify $\frac{16}{u}$ * 64u:

1024

Simplify $\frac{1}{64}$ $u^{2}$ * 64u:

$u^{3}$

Substitute:

1024 = $u^{3}$ + 64u

Solve for u:

u = 8

Substitute back u = $2^{x}$ :

8 = $2^{x}$

Solve for x:

x = 3

4 0

3 years ago

Consider the sequence 3, -9, 27, -81, .. Find the 14th term of the sequence.

bagirrra123 [75]

Answer:

let 'a' be the first term, 'd' be the common difference between all the terms of the sequence

Step-by-step explanation:

therefore, a = 3,

and, d = -9 -3

= -12

hence the 14th term would be,

=> a + 13d

=> 3 + 13( -12 )

=> 3 - 156

=> - 153

5 0

3 years ago

Read 2 more answers

Which expression represents the phrase?"4 more than the product of 8 and n."A) 8n – 4B) 8n + 4C) 4(8 + n)D) (8 + n) – 4

Dafna11 [192]

Answer:

B) 8n+4

Step-by-step explanation:

The product of 8 and n means 8n. Four more means +4. These together equal B) 8n+4

6 0

2 years ago

Read 2 more answers