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2 years ago

2x + 5x = -14 What is the answer

Mathematics

1 answer:

Mekhanik [1.2K]2 years ago

6 0

Answer:

x= -2

Step-by-step explanation:

2(-2)+5(-2)= -4+-10= -14

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Suppose a ball is thrown upward to a height of h 0 meters. After each bounce, the ball rebounds to a fraction r of its previous

gladu [14]

Answer:

A)1st term:45

2nd term:48.75

3rd term:49.6875

4th term:49.921875

B) Sₙ = h₀ + 2h₀((∞, n=1) Σrⁿ)

Step-by-step explanation:

We are given;

h₀ = Initial height of the ball = 30

r = Rebound fraction = 0.25

a) The arithmetic sequence of bouncing balls is given by the following;

Sₙ=h₀+2h₀(r¹+r²+r³+r⁴.........rⁿ)

The first term of the sequence is;

S₁ = h₀ + 2h₀r¹

S₁ = 30 + (2 × 30 × 0.25)

S₁ = 45

The second term of the sequence is;

S₂ = h₀ + 2h₀(r¹+r²)

S₂ = 30 + (2 × 30 × (0.25 + 0.25²)) = 48.75

The third term of the sequence is;

S₃ = h₀ + 2h₀(r¹ + r² + r³) = 30 + (2 × 30 × (0.25 + 0.25² + 0.25³)) = 49.6875

S₄ = h₀ + 2h₀(r¹ + r² + r³ + r⁴)

S₄ = 30 + (2 × 30 × (0.25 + 0.25² + 0.25³ + 0.25⁴)) = 49.921875

B) The general expression for the nth term of the sequence is;

Sₙ = h₀ + 2h₀((∞, n=1) Σrⁿ)

4 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

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3 years ago

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