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2 years ago

MATH HELP ASAP!!! PLSSS

Mathematics

2 answers:

Alex2 years ago

6 0

First one is 13. Second is 9. Third is 5. Fourth is 1.

MAXImum [283]2 years ago

6 0

Step-by-step explanation:

i think the answers are 1)13 2)9 3)5 4)1

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An adult and 5 children rent a skates. it costs $4 per adult for skates and a total of $16

Elenna [48]

Answer:

$2.20 per child

Step-by-step explanation:

16-4=11

11/5=2.2

3 0

3 years ago

A waitress earned $9 per hour at her job plus an additional $75 in tips on Friday. She earned more than $150 total. Which inequa

krek1111 [17]

Answer:

the answer is c :)

Step-by-step explanation:

3 0

3 years ago

Read 2 more answers

6. Find LCM (10,14,63).

Katen [24]

Answer:

630 is the LCM

Step-by-step explanation:

You have to list all the multiples of each number till you find the lowest same value one which in this case it's 630.

Multiples of 10:

10, 20, 30, 40, 50, 60, 70, 80, 90, 100, 110, 120, 130, 140, 150, 160, 170, 180, 190, 200, 210, 220, 230, 240, 250, 260, 270, 280, 290, 300, 310, 320, 330, 340, 350, 360, 370, 380, 390, 400, 410, 420, 430, 440, 450, 460, 470, 480, 490, 500, 510, 520, 530, 540, 550, 560, 570, 580, 590, 600, 610, 620, 630, 640, 650

Multiples of 14:

14, 28, 42, 56, 70, 84, 98, 112, 126, 140, 154, 168, 182, 196, 210, 224, 238, 252, 266, 280, 294, 308, 322, 336, 350, 364, 378, 392, 406, 420, 434, 448, 462, 476, 490, 504, 518, 532, 546, 560, 574, 588, 602, 616, 630, 644, 658

Multiples of 63:

63, 126, 189, 252, 315, 378, 441, 504, 567, 630, 693, 756

3 0

3 years ago

Is the following an algebraic expression: 18r + (r ÷ 8) - 13

avanturin [10]

Answer:

nah

Step-by-step explanation:

4 0

2 years ago

Initially, there are 40 grams of A and 50 grams of B, and for each gram of B, 2 grams of A is used. It is observed that 15 grams

hram777 [196]

Answer:

$X(16)=25.71grams$

Step-by-step explanation:

let $X(t)$ denote grams of $C$ formed in $t$ mins.

For $X grams$ of $C$ we have:

$\frac{2}{3}Xg$ of A and $\frac{1}{3}Xg$ of B

Amounts of A,B remaining at any given time is expressed as:

$40-\frac{2}{3}Xg$ of A and $50-\frac{1}{3}Xg$ of B

Rate at which C is formed satisfies:

$\frac{dX}{dt} \infty(40-\frac{2}{3}X)(50-\frac{1}{3}X)->\frac{dX}{dt}=k(90-X)\\\therefore \frac{dX}{(90-X)^2}=kdt->\int{\frac{dX}{(90-X)^2}} \, =\int {k} \, dt \\\therefore \frac{1}{90-X}=kt+c->90-X=\frac{1}{kt+c}\\\\X(t)=90-\frac{1}{kt+c}$

Apply the initial condition, $X(0)=0$ ,to the expression above

$0=90-\frac{1}{c} \ \ ->c=\frac{1}{90}\\\therefore\\X(t)=90-\frac{1}{kt+\frac{1}{90}} \ \ ->X(t)=90-\frac{90}{90kt+c}$

Now at $X(8)=15$ :

$15=90-\frac{90}{90\times 8k+1} \ ->75=\frac{90}{720k+1}\\k=0.0002778$

Substitute in X(t) to get

$X(t)=90-\frac{90}{0.0002778t\times 90+1}\\X(t)=90-\frac{90}{0.25t+1}\\But \ t=16\\\therefore X(t)=90-\frac{90}{0.025\times16+1}\\X(t)=25.71$

5 0

2 years ago