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OlgaM077 [116]
3 years ago
12

Find the difference. (-9ab-a+9) - (-ab+1)

Mathematics
2 answers:
scoray [572]3 years ago
6 0

Answer:

-8ab + a + 8

Step-by-step explanation:

i simplified

is tht ok

dybincka [34]3 years ago
3 0
-8ab -a + 8
Hope that helps. If you need me to explain, I will.
You might be interested in
El entrenamiento que hace marco 10 dias antes de su carrera es constante. Diariamente corre 45 minutos a una velocidad de 30 km-
alexira [117]

Responder:

225 kilometros

Explicación paso a paso:

Dado que :

Velocidad de funcionamiento = 30 km / h

Duración = 45 minutos

Número de días = 10 días

Distancia total recorrida por día:

Duración en horas:

45 minutos / 60 = 0,75 hora (s)

Así, la distancia recorrida por día:

Tiempo de velocidad

30 km / h * 0,75 h = 22,5 km

Por lo tanto, la distancia total recorrida durante 10 días:

Distancia recorrida diariamente * 10

22,5 kilometros * 10 = 225 kilometros

6 0
3 years ago
Determine if each graph represent a direct proportional or not
Troyanec [42]

Answer:

i hope everyone has a good day keep your head up at all times and dont give up

Step-by-step explanation:

5 0
3 years ago
What is the difference 1/9 - 5/6 a 13/18 b 1/18 c - 1/18 d -13/18
Murrr4er [49]
1/9 - 5/6

Convert both of them into denominators of 18.

2/18 - 15/18

Subtract the numerators and keep the denominator:

-13/18
3 0
3 years ago
Given: r is inversely proportional to s. If r = 16 when s = 3, then write the formula for the relation between r and s.
riadik2000 [5.3K]

Answer:

B

Step-by-step explanation:

Given that r is inversely proportional to s then the equation relating them is

r = \frac{k}{s} ← k is the constant of proportion

To find k use the condition r = 16 when s = 3

k = rs = 16 × 3 = 48

r = \frac{48}{s} ← equation of variation → B

3 0
3 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
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