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BabaBlast [244]

2 years ago

7

Mr. George has a bank balance of -42 dollars at the start of the month. After he deposits 6 dollars, the new balance is (with fu

ll method pls)

1 answer:

OLga [1]2 years ago

5 0

Answer:

$-36

Step-by-step explanation:

-42+6

-36

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Find the value or measure. Round answers to the nearest tenth, as needed.

Leni [432]

Answer:

its 34

Step-by-step explanation:

its half of the center angles

3 0

2 years ago

The Walking Club is one of the after-school activities at Jefferson Middle School.

saveliy_v [14]

You first need to know how much miles it takes on the trail. Then you look 20 x 3 is one hour. And if the 3 group said it takes 2 hours waking for them its like this, the first group takes one hour to go thru the trail, but the second group takes and hour and half the 3 group takes 2 hours.

I wish this helps. :)

4 0

3 years ago

He counted 206 wheels and 170 pedals. How many bicycles and tricycles does he have.

noname [10]

Bicycle has 2 wheels and 2 pedals
tricycle has 3 wheels and 2 pedals

EQ 1 : 2b + 2t = 170 pedals

EQ 2: 2b +3t = 206 wheels

subtract EQ 1 from EQ 2

2b +3t = 206 - 2b + 2t = 170 = t=36

there were 36 tricycles

36 x 2 = 72 pedals

170 pedals - 72 = 98 pedals left
98/2 = 49 bicycles

36 Tricycles and 49 bicycles

8 0

3 years ago

PLEASE HELP ME ASAP ILL GIVE 100 POINTS

Airida [17]

Answer:

8(+4)

Step-by-step explanation:

6 0

2 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