Фd —<br> 969 - 9<br> ХОТХЭ: <br> P

3х +5 > – 2x — 5<br> ............

Yanka [14]

Answer:

x>-2

Step-by-step explanation:

6 0

3 years ago

Factor 8q^2 + 3m + 4qm + 6q=

SIZIF [17.4K]

8q² + 4qm + 6q + 3m
4q(2q) + 4q(m) + 3(2q) + 3(m)
4q(2q + m) + 3(2q + m)
(4q + 3)(2q + m)

8 0

4 years ago

Plz help

Angelina_Jolie [31]

Answer:

Part A

The two equations are;

d = 3 m/s × t for Boat A

d = 4.8 m/s × (t - 6 s) for Boat B

Part B

The solution to the system of equation are

t = 16 s and d = 48 m

Part C: The possible values of 't' for Boat A are 0 ≤ t ≤ ∞

The possible values of 't' for Boat B are 6 ≤ t ≤ ∞

Boat B timing starts when Boat B crosses the start line when t = 6 seconds

The solution of the system of equation indicates that Boat B and Boat A arrive at the same point, 'd' 48 meters from the start, 16 seconds from the time Boat A crosses the stat line

Step-by-step explanation:

The given parameters are;

The speed of Boat A = 3 m/s

The speed of Boat B = 4.8 m/s

The time Boat B crosses the start line = 6 seconds after Boat A crosses the start line

Part A

The two equations are;

For Boat A

d = 3 m/s × t...(1)

For Boat B

d = 4.8 m/s × (t - 6 s)...(2)

Where;

t = The time in seconds since Boat A crossed the start line

d = The distance traveled by the boat in meters

Part B

The system of equations from Part A is solved as follows;

d = 3 m/s × t and d = 4.8 m/s × (t - 6 s)

At the (common) solution point to the system of equation, we have;

d = 3 m/s × t, d = 4.8 m/s × (t - 6 s)

∴ 3 m/s × t = 4.8 m/s × (t - 6 s)

3·t m/s = 4.8·t m/s - 28.8 m

3·t m/s + 28.8 m = 4.8·t m/s

4.8·t m/s = 3·t m/s + 28.8 m; By transitive property of equality

4.8·t m/s - 3·t m/s = 28.8 m

1.8·t m/s = 28.8 m

t = 28.8 m/(1.8 m/s) = 16 s

t = 16 s.

d = 3 m/s × t = 3 m/s × 16 s = 48 m

d = 48 m

The solution of the system of equation is t = 16 s, d = 48 m

Part C: The possible values of 't' for Boat A are 0 ≤ t ≤ ∞

The possible values of 't' for Boat B are 6 ≤ t ≤ ∞

The timing variable, 't' for the equations of Boat A and Boat B is with reference to the time which Boat A crosses the start line, which is 6 seconds before Boat B, therefore, Boat B crosses the start line when t = 6 seconds

The solution of the system of equation represents that Boat B reaches Boat A at a point 48 meters from the start, 16 seconds from the time Boat A crosses the stat line.

4 0

3 years ago

The perimeter of a rectangle is equal to 20. If the length is tripled and the width is doubled, the new perimeter is increased b

valkas [14]

16 inches in the length

3 0

3 years ago

please do the steps Solve for d: 1/6d-8=5/8 2. Solve for x: 3x-4+5x=10-2z 3. Solve for c: 7(c-3)=14 4. Solve for m: 11(m/22+3/44

xeze [42]

Answer:

d = 55.5

x = 1

c = 11

m = $\frac{1}{122}$

k = $\frac{a}{(c + 5)}$

Step-by-step explanation:

Sorry, the formatting is slightly hard to understand, but I think this is what you meant.

Q1.

$\frac{1}{6}$ d - 8 = $\frac{5}{8}$ x 2

Step 1. Simplify.

$\frac{5}{8}$ x 2 = $\frac{5}{8}$ x $\frac{2}{1}$ = $\frac{10}{8}$

Step 2. Cancel out the negative 8.

$\frac{1}{6}$ d - 8 = $\frac{10}{8}$

+ 8 to both sides (do the opposite: $\frac{1}{6}$ d is subtracting 8 right now, but to cancel that out, we will do the opposite of subtraction, i.e. addition)

$\frac{1}{6}$ d = $\frac{10}{8}$ + 8

Step 3. Simplify.

$\frac{10}{8}$ + 8 = $\frac{10}{8}$ + $\frac{8}{1}$ = $\frac{10}{8}$ + $\frac{64}{8}$ = $\frac{74}{8}$ = $\frac{37}{4}$

Step 4. Cancel out the $\frac{1}{6}$ .

$\frac{1}{6}$ d = $\frac{37}{4}$

÷ $\frac{1}{6}$ from both sides (do the opposite: d is multiplied by $\frac{1}{6}$ right now, but to cancel that out, we will do the opposite of multiplication, i.e. division)

÷ $\frac{1}{6}$ = x 6

So....

x 6 to both sides

d = $\frac{37}{4}$ x 6 = $\frac{37}{4}$ x $\frac{6}{1}$ = $\frac{222}{4}$ = $\frac{111}{2}$ = 55.5

Step 5. Write down your answer.

d = 55.5

Q2.

3x - 4 + 5x = 10 - 2x × 3

Step 1. Simplify

3x - 4 + 5x = 3x + 5x - 4 = 8x - 4

10 - 2x × 3 = 10 - (2x × 3) = 10 - 6x

Step 2. Cancel out the negative 6x

8x - 4 = 10 - 6x

+ 6x to both sides (do the opposite - you're probably tired of reading this now - right now it's 10 subtract 6x, but the opposite of subtraction is addition)

14x - 4 = 10

Step 3. Cancel out the negative 4

14x - 4 = 10

+ 4 to both sides (right now it's 14x subtract 4, but the opposite of subtraction is addition)

14x = 14

Step 4. Divide by 14

14x = 14

÷ 14 from both sides (out of the [14 × x] we only want the [x], so we cancel out the [× 14])

x = 1

Step 5. Write down your answer.

x = 1

Q3.

7(c - 3) = 14 × 4

Step 1. Expand the brackets

7(c - 3) = (7 x c) - (7 x 3) = 7c - 21

Step 2. Simplify

14 x 4 = 56

Step 3. Cancel out the negative 21

7c - 21 = 56

+ 21

7c = 56 + 21

7c = 77

Step 4. Cancel out the ×7

7c = 77

÷ 7

c = 77 ÷ 7

c = 11

Step 5. Write down your answer.

c = 11

Q4.

11( $\frac{m}{22}$ + $\frac{3}{44}$ ) = 87m + m × 5

Step 1. Expand the brackets

11( $\frac{m}{22}$ + $\frac{3}{44}$ ) = (11 x $\frac{m}{22}$ ) + (11 x $\frac{3}{44}$ ) = ( $\frac{11}{1}$ x $\frac{m}{22}$ ) + ( $\frac{11}{1}$ x $\frac{3}{44}$ ) = $\frac{11m}{22}$ + $\frac{33}{44}$ = $\frac{m}{2}$ + $\frac{3}{4}$