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3 years ago

2.6 more than the quotient of h and 3 is w

Mathematics

2 answers:

dangina [55]3 years ago

7 0

<span>So, "2.5 more than the quotient of h and 3 is w" becomes 2.5 + h/3 = w. </span>

iris [78.8K]3 years ago

5 0

Answer:

$2.6 + \frac{h}{3} =w$

Step-by-step explanation:

In this problem, we have to express givens into an equation. So, 2.6 more than refers to $2.6 +$ , quotient of h and 3, refers to a fraction or division between these two $\frac{h}{3}$ , this sum results in <em>w. </em>Therefore the expression is:

$2.6 + \frac{h}{3} =w$

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A hot air balloon descends at a rate of 320 feet per minute for 3 minutes. Use integers to calculate and

Amanda [17]

Given:
descends at a rate of 320 feet per minute for 3 minutes.

320 feet per minute * 3 minutes = 960 feet

Total change in altitude is 960 feet. Since it is descending, I believe it would be a negative 960 feet.

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3 years ago

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There were 65 people waiting in two lines to buy a snowball ice drink they were joined in the line by 11 more people a total of

Pavlova-9 [17]

Answer:

67 people were still left waiting in line.

Step-by-step explanation:

If there are 65 plus 11 more, there'd be 76 people in line.

I guess the fact that it says "waiting in two lines" can be completely discarded because bottom line is 9 people got their drinks- which means that "waiting in two lines" means nothing.

76 - 9 = 67 people still left waiting in line.

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3 years ago

The average value of two positive numbers is 30% less than one of the two numbers. By which percentage is average value bigger t

Viefleur [7K]

Answer:

75% bigger

Step-by-step explanation:

Let the first number be x and the second be y.

So:

$Average = \frac{x + y}{2}$

And:

$Average = (1 - 30\%) * x$ -----30% less than x

Substitute $Average = (1 - 30\%) * x$ in $Average = \frac{x + y}{2}$

$\frac{x + y}{2} = (1 - 30\%) * x$

$\frac{x + y}{2} = (1 - 0.30) * x$

$\frac{x + y}{2} = 0.70 * x$

$\frac{x + y}{2} = 0.70x$

Cross multiply

$x + y = 0.70x * 2$

$x + y = 1.40x$

Collect like terms

$1.40x - x = y$

$0.40x = y$

Make x the subject

$x = \frac{y}{0.40}$

$x = 2.50y$

So, the average is:

$Average = (1 - 30\%) * x$

$Average = (1 - 30\%) * 2.50y$

$Average =0.70 * 2.50y$

$Average =1.75y$

Rewrite as:

$Average =(1 + 0.75)y$

Express as percentage

$Average =(1 + 75\%)y$

This means that the average is 75% bigger than the second number

6 0

3 years ago

Can someone please help me solve this?

aleksklad [387]

Answer:

centre = (-5, -4)

radius = 7.42

Step-by-step explanation:

The general form for center-radius is

(x - h)² + (y - k)²= r²

Center = (h, k)

Radius = r

Rearrange the equation.

(x² + 10x) + (y² + 8y) - 14 = 0

Add 14 on both sides.

(x² + 10x) + (y² + 8y) = 14

(10/2)² = 25

(8/2)² = 16

Add 25 and 16 to both sides.

(x² + 10x + 25) + (y² + 8y + 16) = 14 + 25 + 16

Factor left side of the equation.

(x+5)² + (x+4)² = 55

(55) = (7.416198)²

(x+5)² + (x+4)² = 7.42²

The radius is 7.42.

The centre is at (-5, -4).

8 0

3 years ago

Read 2 more answers

Greg swam 4 kilometers against the current in the same amount of time it took him to swim 16 kilometers with the current. The ra

ad-work [718]

Answer:

5 kmph

Step-by-step explanation:

Given: Greg swam 4 kilometers against the current.

Greg swam 16 kilometers with the current.

Rate of the current was 3kmph.

Lets assume the speed of greg swimming with no current be "x".

∴ Speed of swimming against current= $(x-3)\ kmph$

Speed of swimming with current= $(x+3)\ kmph$

As given, it took same amount of time to swim both with current and against current.

∴ Forming an equation to find the value of x, which is speed of greg swimming with no current.

We know, $Time= \frac{Distance}{Speed}$

⇒ $\frac{4}{(x-3)} = \frac{16}{(x+3)}$

Multiplying both side by (x+3) and (x-3)

⇒ $4\times (x+3)= 16\times (x-3)$

Using distributive property of multiplication.

⇒ $4x+12= 16x-48$

Subtracting both side by 4x and adding by 48.

⇒ $60= 12x$

Dividing both side by 12

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Hence, Greg can swim at the rate of 5kmph with no current.

7 0

4 years ago