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2 years ago

Solve The Following Expression..

Mathematics

2 answers:

d1i1m1o1n [39]2 years ago

8 0

Answer:

$\frac{4 + 3y}{6y - 2} = \frac{ - 3}{5} \\ \\ - 3(6y - 2) = 5(4 + 3y) \\ \\ - 18y + 6 = 20 + 15y \\ \\ 15y + 18y = 6 - 20 \\ \\ 33y = - 14 \\ \\ y = \frac{ - 14}{33} \\ \\ y = - 0.42$

I hope I helped you^_^

ddd [48]2 years ago

8 0

Answer:

$y=-\frac{14}{33}$

Step-by-step explanation:

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6 0

3 years ago

A ferry departs from a port along the Sands River at 10:00 am. The ferry travels at a speed of 15 mph in still water and returns

jek_recluse [69]

Answer:

Speed of river's current = 5 mph

Step-by-step explanation:

Let 'd' be the distance covered

Let x be the speed of the rivers current

Speed upstream = 15-x

speed downstream = 15+x

Time = distance / speed

time upstream = $\frac{d}{15-x}$

time downstream = $\frac{d}{15+x}$

the ferry trip upstream takes twice as long as its return trip downstream

$\frac{d}{15-x}$ = $\frac{2d}{15+x}$

divide both sides by d

$\frac{1}{15-x}$ = $\frac{2}{15+x}$

cross multiply

$15+x= 2(15-x)$

$15+x= 30-2x$

Add 2x on both sideds

$15+3x= 30$

subtract 15 from both sides and divide by 3

$3x=15$

x=5

Speed of river's current = 5 mph

8 0

3 years ago

How do I solve question 6 through 8?<br> Solve for me

rewona [7]

The equations of the functions are y = -4(x + 1)^2 + 2, y = 2(x - 2)^2 + 1 and y = -(x - 1)^2 - 2

<h3>How to determine the functions?</h3>

A quadratic function is represented as:

y = a(x - h)^2 + k

<u>Question #6</u>

The vertex of the graph is

(h, k) = (-1, 2)

So, we have:

y = a(x + 1)^2 + 2

The graph pass through the f(0) = -2

So, we have:

-2 = a(0 + 1)^2 + 2

Evaluate the like terms

a = -4

Substitute a = -4 in y = a(x + 1)^2 + 2

y = -4(x + 1)^2 + 2

<u>Question #7</u>

The vertex of the graph is

(h, k) = (2, 1)

So, we have:

y = a(x - 2)^2 + 1

The graph pass through (1, 3)

So, we have:

3 = a(1 - 2)^2 + 1

Evaluate the like terms

a = 2

Substitute a = 2 in y = a(x - 2)^2 + 1

y = 2(x - 2)^2 + 1

<u>Question #8</u>

The vertex of the graph is

(h, k) = (1, -2)

So, we have:

y = a(x - 1)^2 - 2

The graph pass through (0, -3)

So, we have:

-3 = a(0 - 1)^2 - 2

Evaluate the like terms

a = -1

Substitute a = -1 in y = a(x - 1)^2 - 2

y = -(x - 1)^2 - 2

Hence, the equations of the functions are y = -4(x + 1)^2 + 2, y = 2(x - 2)^2 + 1 and y = -(x - 1)^2 - 2

Solve The Following Expression..​

Solve The Following Expression..