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3 years ago

Solve -2(t+2)+5t=6t+11

Mathematics

2 answers:

Snezhnost [94]3 years ago

6 0

Answer:

t=-5

Step-by-step explanation:

umka21 [38]3 years ago

4 0

Answer:

t = -5

Step-by-step explanation:

Distribute by multiplication -2(t + 2), obtaining

-2t - 4 + 5t = 6t + 11.

Combine the t terms; this yields -4 = 3t + 11. Solving for 3t, we get

3t = -4 -11, or 3t = -15. Thus t must be -5.

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Mr. Howe ate 1/3 of a pizza and then Mr. Kurt ate 1/8 of the same pizza. How

Nikolay [14]

12/24

Step One: We need to convert 1/3 and 1/8 so both have the same denominator, so we need to find the a number that is able to be multiply by 3 and 8 for the process.

Step Two: 1/3 x 8= 8/24 and 1/8x3= 3/24

Step Three: Add our new fractions: 3/24+8/24= 12/24

Step Four: Subtract 12 by 24: 24-12= 12; our answer is 12/24 or half the pizza was eaten

I hope I've help!

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3 years ago

What is the value of -2|6x -y| when x-3 and y =4

lisabon 2012 [21]

So, we have been given the following

$-2(6x -y)$

But, we have also been given heads up that $x=-3$ , and $y=4$ .

So, that would mean the new equation would be the following.

$-2(6-3 -4)$

After we solve it our answer should be $2$ !

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3 years ago

Read 2 more answers

Draw two rectangles that each have a perimeter of 8 units and different areas

Tanzania [10]

Answer:

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3 years ago

The price of milk has increased in year from rs.28 per liter to rs.32 per liter . find the percentage increase in the price....

Mice21 [21]

Answer:

14.28%

Step-by-step explanation:

here's your solution

=> price of milk before = ₹28

=> price of milk after increasing = ₹32

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=> now , increased percentage= 4/28*100

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4 0

3 years ago

A math class consists of 25 students, 14 female and 11 male. Three students are selected at random to participate in a probabili

vovangra [49]

Answer:

(a) The probability that a male is selected, then two females is 0.4352.

(b) The probability that a female is selected, then two males is 0.3348.

(c) The probability that two females are selected, then one male is 0.4352.

(d) The probability that three males are selected is 0.0717.

(e) The probability that three females are selected is 0.1583.

Step-by-step explanation:

We are given that a math class consists of 25 students, 14 female and 11 male. Three students are selected at random to participate in a probability experiment.

(a) The probability that a male is selected, then two females is given by;

Number of ways of selecting a male from a total of 11 male = $^{11}C_1$

Number of ways of selecting two female from a total of 14 female = $^{14}C_2$

Total number of ways of selecting 3 students from a total of 25 = $^{25}C_3$

So, the required probability = $\frac{^{11}C_1 \times ^{14}C_2}{^{25}C_3}$

= $\frac{\frac{11!}{1! \times 10!} \times \frac{14!}{2! \times 12!} }{\frac{25!}{3! \times 22!} }$ { $\because ^{n}C_r = \frac{n!}{r! \times (n-r)!}$ }

= $\frac{1001}{2300}$ = 0.4352

(b) The probability that a female is selected, then two males is given by;

Number of ways of selecting a female from a total of 14 female = $^{14}C_1$

Number of ways of selecting two males from a total of 11 male = $^{11}C_2$

Total number of ways of selecting 3 students from a total of 25 = $^{25}C_3$

So, the required probability = $\frac{^{14}C_1 \times ^{11}C_2}{^{25}C_3}$

= $\frac{\frac{14!}{1! \times 13!} \times \frac{11!}{2! \times 9!} }{\frac{25!}{3! \times 22!} }$ { $\because ^{n}C_r = \frac{n!}{r! \times (n-r)!}$ }

= $\frac{770}{2300}$ = 0.3348

(c) The probability that two females is selected, then one male is given by;

Number of ways of selecting two females from a total of 14 female = $^{14}C_2$

Number of ways of selecting one male from a total of 11 male = $^{11}C_1$

Total number of ways of selecting 3 students from a total of 25 = $^{25}C_3$

So, the required probability = $\frac{^{14}C_2 \times ^{11}C_1}{^{25}C_3}$

= $\frac{\frac{14!}{2! \times 12!} \times \frac{11!}{1! \times 10!} }{\frac{25!}{3! \times 22!} }$ { $\because ^{n}C_r = \frac{n!}{r! \times (n-r)!}$ }