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3 years ago

15

Alan set his watch 13 seconds behind, and it falls behind another 1 second every day. How

2 answers:

Vlad1618 [11]3 years ago

5 0

31 additional days, getting behind by 31 seconds; this, added to the original 4 sec behind, gives you a total 35 seconds behind.

Now, that really isn't complicated, right?

Gemiola [76]3 years ago

4 0

It would be 22 days.

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When you flip a coin twice what is the sample space?

Shkiper50 [21]

Sample space is all possible outcome of an event.and denotes in a set form{}

If u flipa coin twice then sample space={HH,TT,HT,TH}

8 0

3 years ago

PLEASE HELP!!!!! What are the values of a and b? -7(x+3)=ax+b

Artist 52 [7]

Step-by-step explanation:

Hope dis helps you......

3 0

3 years ago

PLEASE HELP ASAP

maria [59]

Answer:

First option: $3x^2-5x=-8$

Second option: $2x^2=6x-5$

Fourth option: $-x^2-10x=34$

Step-by-step explanation:

Rewrite each equation in the form $ax^2+bx+c=0$ and then use the Discriminant formula for each equation. This is:

$D=b^2-4ac$

1) For $3x^2-5x=-8$ :

$3x^2-5x+8=0$

Then:

$D=(-5)^2-4(3)(8)=-71$

Since $D$ this equation has no real solutions, but has two complex solutions.

2) For $2x^2=6x-5$ :

$2x^2-6x+5=0$

Then:

$D=(-6)^2-4(2)(5)=-4$

Since $D$ this equation has no real solutions, but has two complex solutions.

3) For $12x=9x^2+4$ :

$9x^2-12x+4=0$

Then:

$D=(-12)^2-4(9)(4)=0$

Since $D=0$ this equation has one real solution.

4) For $-x^2-10x=34$ :

$-x^2-10x-34=0$

Then:

$D=(-10)^2-4(-1)(-34)=-36$

Since $D$ , this equation has no real solutions, but has two complex solutions.

3 0

3 years ago

Solve for the variable c in this equation a(c + b) = d

noname [10]

Multiply everything in the parenthesis by a.

ac + ab = d

Subtract ab from both sides.

ac = d - ab

Divide a on both sides.

c = d - ab / a

Hope this helps!

3 0

3 years ago

Suppose that an insect population’s density, in thousands per acre, during year n, can be modeled by the recursive formula: a1 =

vlabodo [156]

Answer:

The population alternates between increasing and decreasing

Step-by-step explanation:

<u><em>The options of the question are</em></u>

A) The population density decreases each year.

B) The population density increases each year.

C) The population density remains constant.

D) The population alternates between increasing and decreasing

we have

$a_n=2.9(a_n_-_1)-0.2(a_n_-_1)^2$

$a_1=8$

Find the value of $a_2$

For n=2

$a_2=2.9(a_1)-0.2(a_1)^2$

$a_2=2.9(8)-0.2(8)^2=10.4$

Find the value of $a_3$

For n=3

$a_3=2.9(a_2)-0.2(a_2)^2$

$a_3=2.9(10.4)-0.2(10.4)^2=8.528$

For n=4

$a_4=2.9(a_3)-0.2(a_3)^2$

$a_4=2.9(8.528)-0.2(8.528)^2=10.1858$

For n=5

$a_5=2.9(a_4)-0.2(a_4)^2$

$a_5=2.9(10.1858)-0.2(10.1858)^2=8.7887$

therefore

The population alternates between increasing and decreasing

7 0

3 years ago

Read 2 more answers