What is 20 - 16 ÷ 8 + 7

Margarita [4]

70 is the answer I am pretty sure

7 0

3 years ago

Mikes movie Rentals Charge $3.00 for each rental plus $0.50 for each additional day. Reed's Rentals Charge $1.50 each rental plu

iris [78.8K]

Answer:

2

Step-by-step explanation:

What we need here is to express both charges trough equations, and then find out for which value these equations are equal.

Let X be the amount of additional days of rent.

So, Mikes Movie Rentals cost equations is:

CostMike = 3 + 0.5X

This implies that if there is no additional day (X=0), the cost is $3, if there is one additional day the cost is $3.5, for 2 days $4, and so on ...

Now lets find Reed's Rental cost.:

CostReed = 1.5 + 1.25X

And the interpretation is analogous to the one above. If there is no additional day, Reed's charge is $1.5, for 1 days $2.75, for 2 days $4

(Notice that with the examples I made about interpreting the equations I found out that for 2 days the cost of both is the same. Now lets find it analytically)

For finding the additional days that give the same costs we need to equal both equations:

3 + 0.5 X = 1.5 + 1.25 X

Lets subtract 1.5 from both sides:

3 - 1.5 + 0.5 X = 1.5 - 1.5 + 1.25 X

1.5 + 0.5 X = 1.25 X

Now subtract 0.5 X in both sides:

1.5 + 0.5 X - 0.5 X = 1.25 X - 0.5 X

1.5 = 0.75 X

Dividing both sides by 0.75:

1.5 / 0.75 = 0.75 X / 0.75

2 = X

So, for 2 additional days bith charges are equal.

6 0

3 years ago

Does anyone know this and if it is extraneous or not?

Anna11 [10]

Addd 4 to both sides
$\sqrt{x+9}=5$
sqare both sides
x+9=25
minus 9 both sides
x=16

plug it in for x and see
$\sqrt{16+9}-4=1$
$\sqrt{25}-4=1$
$5-4=1$
1=1
true

not extraneous
an extraneous root would be x=-34

5 0

3 years ago

20 points pls help no links

aniked [119]

Answer:

(-15, 13)

Step-by-step explanation:

-3(8x + 9y = -3)

4( 6x + 7y = 1)

-24x - 27y = 9

24x + 28y = 4 (add both equations)

y = 13

6x + 7*13= 1

6x + 91= 1

6x = 1 -91

x = -91/6

x = -15

3 0

3 years ago

If f (n)(0) = (n + 1)! for n = 0, 1, 2, , find the taylor series at a=0 for f.

Pie

Given that $f^{(n)}(0)=(n+1)!$ , we have for $f(x)$ the Taylor series expansion about 0 as

$f(x)=\displaystyle\sum_{n=0}^\infty\frac{(n+1)!}{n!}x^n=\sum_{n=0}^\infty(n+1)x^n$

Replace $n+1$ with $n$ , so that the series is equivalent to

$f(x)=\displaystyle\sum_{n=1}^\infty nx^{n-1}$

and notice that

$\displaystyle\frac{\mathrm d}{\mathrm dx}\sum_{n=0}^\infty x^n=\sum_{n=1}^\infty nx^{n-1}$

Recall that for $|x|$ , we have

$\displaystyle\sum_{n=0}^\infty x^n=\frac1{1-x}$

which means

$f(x)=\displaystyle\sum_{n=1}^\infty nx^{n-1}=\frac{\mathrm d}{\mathrm dx}\frac1{1-x}$
$\implies f(x)=\dfrac1{(1-x)^2}$

5 0

3 years ago