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3 years ago

14

A train drove the distance of 440 feet in 20 seconds. Select ALL the unit rates that are equivalent to the speed of the model ai

rplane. (Use page 26 unit of measure chart to help.)

1 answer:

katrin [286]3 years ago

4 0

Answer:

yo is ya boi Nick you already know what it is yeah big daddy nick

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10) The temperature recorded at 12 noon was 15°C above zero. If it decreases at the rateof 3°C per hour until midnight, at what

aalyn [17]

Answer:

at 8 pm the temperature will be -9 degrees. and at midnight the temperature will be negitive twenty one(-21) degrees

Step-by-step explanation:

the start time is 12 pm and finish is 12 am and every hour your going to subtract 3 from the given degree so 15-3= 12 , 12-3=9 ect.

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

What is the solution to the system of equations? x=4y 6 x-3y=4

ahrayia [7]

If you graph the two equations then you will find that they intersect at (-2,-2).

You then have to check your answer.

x=4y+6
(-2)=4(-2)+6
-2=-8+6
-2=-2

x-3y=4
(-2)-3(-2)=4
-2+6=4
4=4

8 0

3 years ago

Y = -6x + 3 y = 5x - 8 Question 1 options: (-3, -1) (3, -1) (1, -3) (1, 3)

OleMash [197]

Answer:

-3,-1 so a

Step-by-step explanation:

4 0

3 years ago

Read 2 more answers

HelppppI need help this benchmark

serious [3.7K]

Answer:

A: -6.5

Step-by-step explanation:

Take two points that we already know are on the line and find the slope

I used:

(4, -10) and (7, -30)

X1 (first x coordinate): 4

Y1 (first y coordinate): -10

X2 (second x coordinate): 7

Y2 (second y coordinate): 30

These appear to be exactly on the graph at a certain point along the line. We can't choose a coordinate pair that isn't on the line at an exact point.

Use (Y2-Y1)/(X2-X1)

(-30-(-10))/(7-4) = (-30 + 10)/3 = -20/3 = approximately -6.67.

The closest answer to that is A.

3 0

3 years ago

Read 2 more answers