All categories

2 years ago

PLEASE HELP!! ASAP! Thank you!

Mathematics

2 answers:

ASHA 777 [7]2 years ago

6 0

X^2+2x=19
x^2+2x+1=20
(x+1)^2=20

nexus9112 [7]2 years ago

3 0

$x^2+2x=19\\
x^2+2x+1=20\\
(x+1)^2=20$

You might be interested in

X^2/3 can be written as

Nonamiya [84]

X to the power of two thirds

5 0

3 years ago

Read 2 more answers

Pls help I can't do anything I will give the brain thingy

Sidana [21]

Step-by-step explanation:

Blank C :

Given total = 68

No. of TV news = 13

Therefore, in Internet it would be = Total - TV news

= 68 - 13 = 55 (ans)

Blank D :

No. of Internet news in 7th grade = 55

No. of Internet news in 8th grade = 64

Hence, total Internet base news = 64 + 55 = 119 (ans)

Blank B :

All total = 138

Internet news total = 119

Hence, TV news total = All total - Internet news

= 138 - 119 = 19 (ans)

Blank A :

No. of TV news in 7th grade = 13

Total news in TV = 19

Therefore, no. of TV news in 8th grade = 19 - 13 = 6(ans)

Blank E :

All total = 138

Total of 7th grade news = 68

Hence, total of 8th grade news = 138 - 68

= 70 (ans)

6 0

3 years ago

Pleaseee helppp meeeeee

Alinara [238K]

I think its either A or C. Leaning more to A. Do you think you could explain the question just a bit more?

3 0

2 years ago

Jackson bought 3 shirts costing $15.98, $19, and $24.50.How much money did he spend altogether?

marusya05 [52]

Answer:

Jackson spent $59.48 altogether

Step-by-step explanation:

15.98+19= 34.98

34.98+24.50= 59.48

6 0

2 years ago

In a class of 40 students, everyone has either a pierced nose or a pierced ear. The professor asks everyone with a pierced nose

ladessa [460]

Answer:3

Step-by-step explanation:

Given

There are total 40 students in a class

No of students with Pierced Nose n(N)=7

No of students with Pierced ear n(E)=36

and we know using sets

$n\left ( N\cup E\right )=n\left ( E\right )+n\left ( N\right )-n\left ( N\cap E\right )$

where $n\left ( N\cup E\right )=$ Students either nose or ear Pierced

$n\left ( N\cap E\right )=$ no of students having both nose and ear pierced

$40=7+36-n\left ( N\cap E\right )$

$n\left ( N\cap E\right )=43-40=3$

3 0

3 years ago