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2 years ago

PLZ HELP ME PLZ PLZ! Find the product.

Mathematics

2 answers:

cluponka [151]2 years ago

6 0

The answer is -84 because the negative times a positive equals a negative

Pepsi [2]2 years ago

3 0

$= 7(-12) \\ \\ (7)(-12) \\ \\ Answer = - 84$

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The equation A = x(x-4) describes the area, A, of a rectangular flower garden, where x is the width in feet.

BabaBlast [244]

Answer:

The flower garden width would be 6 ft

Step-by-step explanation:

A = 12 ft^2 = x(x + 4) becomes:

x^2 + 4x - 12 = 0, or (by completing the square) x^2 + 4x + 4 - 4 - 12 = 0

Thus, we have (x + 2)^2 - 16 = 0, or (x + 2)^2 = 16

Taking the square root of both sides yields x + 2 = ±4.

In this context only a positive x value makes sense: x = 2 + 4 => x = 6.

The flower garden width would be 6 ft. The other dimension would be 6 - 4, or 2, ft.

4 0

3 years ago

Simplify x0 y-3/x2 y-1

Natali5045456 [20]

Answer:

x^0 y^-3 / x^2 y^-1

= 1 / x^2 y^-1 (y^3) ...because x^0 = 1 and [(y^-1) (y^3)] = y^2

= 1/(x^2 y^2)

6 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

A photography studio charges $50 that includes a sitting fee and 6 prints. Luigi increased his order to 11 prints and paid $65.

Effectus [21]

Answer:

Sitting fee - 32$

Step-by-step explanation:

This is a system of equations(let x represent the sitting fee)

x+6y=50

x+11y=65

You want to isolate the x variable - x+6y-6y=50-6y ; x = 50-6y

Input this into the 2nd equation: 50-6y+11y=65 ; 50+5y=65

Subtract 50 from both sides. 5y=15 (Divided 5y on both sides) ; y=3

Now that y = 3 input this into any equation I choose the 1st one.

x+6(3) = 50 ; x + 18 = 50 (Subtract 18 on both sides to get x)

x = 32

Prove: 32 + 6(3) = 50 ; 32+18 = 50 ; 50 = 50 True

7 0

3 years ago

Can someone help me.

marshall27 [118]

Answer:

Step-by-step explanation:

yeah just pick B man lol

7 0

2 years ago