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2 years ago

Can someone help me with this?

Mathematics

1 answer:

Bas_tet [7]2 years ago

8 0

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A total of 817 tickets were sold for the play. They were either adult or student tickets. There were 67 more student tickets sol

hammer [34]

Answer:

375

Step-by-step explanation:

"817 tickets were sold"

a and s are the number of adults and students, respectively.

a+s = 817

s = 817-a

"67 more student tickets sold than adults"

s = a+67

817-a = a+67

750 = 2a

a = 375

4 0

3 years ago

Read 2 more answers

Find the missing number 2 ?/4 = 11/4 ?/10 =4/5

fgiga [73]

Answer:

Step-by-step explanation:

?/10 =4/5

?/5×2 =4/5

4×2/5×2 =8/10 ... the number is : 8

?/4 = 11/4....the number is : 11

4 0

3 years ago

Daniel has $650 in his checking account. He would like to spend some of his money on a new Playstation 4 and a game. He wants to

d1i1m1o1n [39]

Answer:

The answer would be D t is less than or equal to 375

Step-by-step explanation:

Because he wants to save 275 and 275+375 is 650. so if he wants to save at least 275 he can only spend a max of 375.

3 0

3 years ago

Biologists tagged 196 fish in a lake on January 1. On February 1, they returned and collected a random sample of 42 fish, 14 o

erastovalidia [21]

Answer:

approximentley 588 fish live in the lake

Step-by-step explanation:

42 divided by 14 =3

196 multiplied by 3=588

3 0

3 years ago

For what value(s) of k will the relation not be a function

skelet666 [1.2K]

We are given two relations

(a)

Relation (R)

$R=[((k-8.3+2.4k),-5),(-\frac{3}{4}k,4)]$

We know that

any relation can not be function when their inputs are same

so, we can set both x-values equal

and then we can solve for k

$k-8.3+2.4k=-\frac{3}{4} k$

$3.4k-8.3=-\frac{3}{4}k$

$3.4k\cdot \:10-8.3\cdot \:10=-\frac{3}{4}k\cdot \:10$

$4k-83=-\frac{15}{2}k$

$34k=-\frac{15}{2}k+83$

$\frac{83}{2}k=83$

$83k=166$

$k=2$ ............Answer

(b)

S = {(2−|k+1| , 4), (−6, 7)}

We know that

any relation can not be function when their inputs are same

so, we can set both x-values equal

and then we can solve for k

$2-|k+1|=-6$

$2-\left|k+1\right|-2=-6-2$

$-\left|k+1\right|=-8$

$\left|k+1\right|=8$

Since, this is absolute function

so, we can break it into two parts

$|f\left(k\right)|=a\quad \Rightarrow \:f\left(k\right)=-a\quad \mathrm{or}\quad \:f\left(k\right)=a$

$k+1=-8\quad \quad \mathrm{or}\quad \:\quad \:k+1=8$

we get

$k+1=-8\quad$

$k=-9$

$k+1=8\quad$

$k=7$

so,

$k=-9\quad \mathrm{or}\quad \:k=7$ ...............Answer

5 0

3 years ago