All categories

3 years ago

3t + 35=6 Enter your answer in the box.

Mathematics

2 answers:

IrinaVladis [17]3 years ago

8 0

3t+35−35=6−35

3t=−29

3t/3=-29/3

<u>Answer is t=-29/3</u>

Hope this helps!

GenaCL600 [577]3 years ago

7 0

Here is your answer:

$3t + 35 = 6 -35 -35 -35 + 6 = -29 or 
6 - 35 = -29 3t = -29 \div 3 \div 3 -29 \div 3 = -9.6 t = -9.6$

Therefore the answer is "-9.6 unless your rounding to the nearest whole number then it would be 10."

Hope this helps!

You might be interested in

An object is launched from ground level with an initial velocity of 120 meters per second. For how long is the object at or abov

babunello [35]

Answer:

D) 14 seconds

Step-by-step explanation:

First we will plug 500 in for y:

500 = -4.9t² + 120t

We want to set this equal to 0 in order to solve it; to do this, subtract 500 from each side:

500-500 = -4.9t² + 120t - 500

0 = -4.9t²+120t-500

Our values for a, b and c are:

a = -4.9; b = 120; c = -500

We will use the quadratic formula to solve this. This will give us the two times that the object is at exactly 500 meters. The difference between these two times will tell us when the object is at or above 500 meters.

The quadratic formula is:

$x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

Using our values for a, b and c,

$x=\frac{-120\pm \sqrt{120^2-4(-4.9)(-500)}}{2(-4.9)}\\\\=\frac{-120\pm \sqrt{14400-9800}}{-9.8}\\\\=\frac{-120\pm \sqrt{4600}}{-9.8}\\\\=\frac{-120\pm 67.8233}{-9.8}\\\\=\frac{-120+67.8233}{-9.8}\text{ or }\frac{-120-67.8233}{-9.8}\\\\=\frac{-57.1767}{-9.8}\text{ or }\frac{-187.8233}{-9.8}\\\\=5.3242\text{ or }19.1656$

The two times the object is at exactly 500 meters above the ground are at 5 seconds and 19 seconds. This means the amount of time it is at or above 500 meters is

19-5 = 14 seconds.

6 0

3 years ago

I really need help with this question please

love history [14]

Answer:

The answer is 6/24

Step-by-step explanation:

1/3 was multiplied by 3 so should the other one too.

3 0

3 years ago

Read 2 more answers

I need help Please help.

asambeis [7]

Answer:

Step-by-step explanation:

Number of students 10

Problem 1. $625 for the bus hire per friday, So 625*4=$2500

Problem 2. 2500/25=$100 each for the whole 4 weeks

Problem 3.

10 students tickets 220= 2200 for all tickets. The bus, 625/10 = $62.5*4= $250 dollars for the whole 4 weeks for the bus so in all each student pays $470 each

20 students, tickets 220=4400 for all tickets. The bus, 625/20=$31.25*4=$125 for the whole 4 weeks for the bus, so in all each student must pay $345 each

30 students, tickets 220 = 6600 for all tickets. The bus, 625/30 =$20.83*4=$83.32 for the whole 4 weeks for the bus, so in all each student must pay $303.32 each

41 students, tickets $160=$6560 for all tickets. The bus, because you need 2 buses at 625 each so $1250 for both buses 1250/41= 30.49*4=$121.96 for the whole 4 weeks for the bus. So in al each student must pay $281.96 each

Hope this is correct