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2 years ago

14

PLEASE HELP!! 20 POINTS! - ✓ 81+ ✓–49 + ✓ 25+ ✓-1

2 answers:

kramer2 years ago

7 0

Answer:

It's indeterminate; because a negative number can't have an rational root when goes under an even radical.

laiz [17]2 years ago

5 0

Answer:

Step-by-step explanation:

=-9+7i+5+i

=-4+8i

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After simplifying, how many terms does the expression 4y-6+y2-9 contain?

Alex73 [517]

Answer:

There are three terms in the simplified expression.

Step-by-step explanation:

We have to simplify the expression and have to count the number of terms that the expression has.

The expression is 4y - 6 + y² - 9

= 4y + y² - 6 - 9

= y² + 4y - 15

Therefore, there are three terms in the simplified expression, one for y² term, another is y term and the constant term. (Answer)

8 0

2 years ago

Read 2 more answers

2+363892917382827363

pentagon [3]

Answer:

363892917382827365

Step-by-step explanation:

thats how one adds

3 0

3 years ago

Read 2 more answers

If there are approximately 1.8 million cars in Vancouver and if each car travels an average of 17,500 kilometres in a year, how

Goryan [66]

Rates are used to measure a quantity over another.

<em>The 1.8 million cars use </em> $1.85 \times 10^{10}$ <em> liters each year</em>

Given

$Cars = 1.8\ million$

$d = 17500km$ --- distance

$r = 40miles/gallon$

First, we calculate the number (n) of gallons used by each car

$r = \frac{d}{n}$

Solve for n

$n = \frac{d}{r}$

So, we have:

$n = \frac{17500km}{40miles/gallon}$

$n = \frac{17500km}{40miles}gallon$

Convert miles to kilometers

$n = \frac{17500km}{40km \times 1.60934}gallon$

$n = \frac{17500km}{64.3738km}gallon$

$n = 271.85\ gallon$

The number of gallons (N), used by all the cars is:

$N = n \times cars$

$N = 271.85 \times 1.8\ million$

$N = 489330000$

Convert to liters

$N = 489330000 \times 3.78541L$

$N = 1852314675.3L$

In scientific notation to 2 decimal places, we have:

$N = 1.85 \times 10^{10}L$

<em>Hence, the number of liters used is </em> $1.85 \times 10^{10}$ <em />

Read more about distance and rates at:

brainly.com/question/24659604

4 0

3 years ago

Need help before 5:30pm i also have to show my work.

SpyIntel [72]

I can help!! But could you tell me what the equations are?

6 0

3 years ago

If a club charges dues of $200 a year, it will have 50 members. For each $5 it raises its dues, it loses a member. USE AN EQUATI

Vikentia [17]

Answer:

The value is $y = \$ 225$

Step-by-step explanation:

From the question we are told that

The amount charge per year is $k = \$ 200$

The number of members it will have at this amount is $n = 50$

The amount amount increase that will lead to the loss of a single member is $z = \$ 5$

Generally the total amount the club would obtain from its members is mathematically represented as

$I = Amount \ due\ paid * Number \ of members$

Now let x denote the number of member lost

Hence

$I = (k + zx) (n-x )$

=> $I = (200 + 5x) (50-x )$

=> $I = 10000+50x-5x^2$

Thus the number of members that be removed to give the maximum income from dues is obtained by differentiating the above equation and equating it to zero

$\frac{dI}{dx} = 50-10x$

=> $x = 5$

So from $I = (200 + 5x) (50-x )$ we have

$I = (200 + 5 (5)) (50-5 )$

$I = \$ 10125$

So the amount the club should charge is

$y = \frac{10125}{50 - 5}$

$y = \$ 225$

7 0

3 years ago