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2 years ago

Please feel me the algebraic expression of this. Added a picture!

Mathematics

1 answer:

Lina20 [59]2 years ago

8 0

Answer:

I think the answer is = a^3+5c Step-by-step explanation:

first put "a" to the 3rd power then increse it by 5 times c.

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Work out m and c for the line:<br> 2x + 3y +1=0 <br><br>m=<br>c =

Sergio [31]

Answer:

m = - $\frac{2}{3}$ , c = - $\frac{1}{3}$

Step-by-step explanation:

The equation of a line in slope- intercept form is

y = mx + c ( m is the slope and c the y- intercept )

Given

2x + 3y + 1 = 0 ( subtract 2x + 1 from both sides )

3y = - 2x - 1 ( divide terms by 3 )

y = - $\frac{2}{3}$ x - $\frac{1}{3}$ ← in slope- intercept form

with m = - $\frac{2}{3}$ and c = - $\frac{1}{3}$

4 0

2 years ago

Solve the equation 6(z + 8.5) = 123

kakasveta [241]

I believe it’s z = 12

4 0

3 years ago

Read 2 more answers

You always need some time to get up after the alarm has rung. You get up from 10 to 20 minutes later, with any time in that inte

Mama L [17]

Answer:

a) P(x<5)=0.

b) E(X)=15.

c) P(8<x<13)=0.3.

d) P=0.216.

e) P=1.

Step-by-step explanation:

We have the function:

$f(x)=\left \{ {{\frac{1}{10},\, \, \, 10\leq x\leq 20 } \atop {0, \, \, \, \, \, \, otherwise }} \right.$

a) We calculate the probability that you need less than 5 minutes to get up:

$P(x$

Therefore, the probability is P(x<5)=0.

b) It takes us between 10 and 20 minutes to get up. The expected value is to get up in 15 minutes.

E(X)=15.

c) We calculate the probability that you will need between 8 and 13 minutes:

$P(8\leq x\leq 13)=P(10\leqx\leq 13)\\\\P(8\leq x\leq 13)=\int_{10}^{13} f(x)\, dx\\\\P(8\leq x\leq 13)=\int_{10}^{13} \frac{1}{10} \, dx\\\\P(8\leq x\leq 13)=\frac{1}{10} \cdot [x]_{10}^{13}\\\\P(8\leq x\leq 13)=\frac{1}{10} \cdot (13-10)\\\\P(8\leq x\leq 13)=\frac{3}{10}\\\\P(8\leq x\leq 13)=0.3$

Therefore, the probability is P(8<x<13)=0.3.

d) We calculate the probability that you will be late to each of the 9:30am classes next week:

$P(x>14)=\int_{14}^{20} f(x)\, dx\\\\P(x>14)=\int_{14}^{20} \frac{1}{10} \, dx\\\\P(x>14)=\frac{1}{10} [x]_{14}^{20}\\\\P(x>14)=\frac{6}{10}\\\\P(x>14)=0.6$