All categories

3 years ago

Are my answer correct

Mathematics

2 answers:

soldi70 [24.7K]3 years ago

5 0

Yep, this looks good!

Elena-2011 [213]3 years ago

3 0

Isnt the answer for qn 3 c?

You might be interested in

when professional football player dan marino retired in 1999, he held the record for most career passing yards of 61,361. what w

ollegr [7]

61,361 rounded to the nearest hundred = 61,400

3 0

3 years ago

community gym charges a $50 membership fee and a $60 monthly fee. Workout gym charges a $225 fee and a $5 monthly fee. After how

DIA [1.3K]

Answer:

3.1818... months

Step-by-step explanation:

x = months

50 + 60x = 225 + 5x

55x = 175

x = 3.1818...

7 0

3 years ago

PLEEEEEEEAAAAAAASSSSSSSSSSEEEEEEEEEEE help me!!!!!!!!!!

defon

Answer:

$\boxed{\sf \ \ \ \dfrac{18}{(x-7)(x+12)(x+30)} \ \ \ }$

Step-by-step explanation:

hello,

first of all we will study the quadratic expressions

we can write that, (the different answers provide good clues :-) )

$x^2+5x-84=(x-7)(x+12)$

and

$x^2+23x-210=(x-7)(x+30)$

so first of all as we cannot divide by 0 we need to take x different from 7, -12 and -30 and then we can write

$\dfrac{1}{x^2+5x-48}-\dfrac{1}{x^2+23x-210}=\dfrac{1}{(x-7)(x+12)}-\dfrac{1}{(x-7)(x+30)}\\\\=\dfrac{(x+30) -(x+12)}{(x-7)(x+12)(x+30)}=\dfrac{x+30-x-12}{(x-7)(x+12)(x+30)}\\\\=\dfrac{18}{(x-7)(x+12)(x+30)}$

hope this helps

7 0

3 years ago

A uranium mining town reported population declines of 3.2%, 5.2%, and 4.7% for the three successive five-year periods 1985–89, 1

KATRIN_1 [288]

Answer:

Step-by-step explanation:

Heres the complete question:

A uranium mining town reported population declines of 3.2%, 5.2%, and 4.7% for the three successive five-year periods 1985–89, 1990–94, and 1995–99. If the population at the end of 1999 was 9,320:

How many people lived in the town at the beginning of 1985? (Round your answer to the nearest whole number.)

solution:

Let the population of the town at the beginning of 1985 be P. Then, given that in the first five-year period the population declined by 3.2%, i.e., 0.032, the population of the town at the end of 1989 would be

(1 – 0.032)P = 0.968P.

Again, given that in the second five-year period the population declined by 5.2%, i.e., 0.052, the population of the town at the end of 1994 would be

(1 – 0.052)(0.968P) = 0.948 x 0.968P = 0.917664P.

Finally, given that in the third five-year period the population declined by 4.7%, i.e., 0.047, the population of the town at the end of 1999 would be

(1 – 0.047)(0.917664P) = 0.874533792P.

We are given, 0.874533792P = 9320 or

P = 9320/0.874533792 = 10657.11.

Thus, 10657 people lived in the town at the beginning of 1985

3 0

3 years ago

Please Show Work.

s344n2d4d5 [400]

Answer:

$\huge\boxed{x=17}$