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Doss [256]
3 years ago
7

HELP HELP HELP Read carefully before answering The cargo weight in Morten's truck cannot be greater then 3 tons. He has 5,000 po

unds of cargo in his truck. What is the greatest amount of cargo Morten can add without going over the weight limit. A. 2 ton B. 5.5 ton C. 1,000 lbs. D. 2,000 lbs.
Mathematics
1 answer:
Zigmanuir [339]3 years ago
4 0
The answer is C as D is too much as 3 tons equal around 6000 pounds

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Thu wants to play "guess my number".she states,"when I triple my number and add five,I get twenty-six. What's my number?" what i
Mama L [17]
Let's break this up:
let x = the number
"when I triple my number" is the same as
3x
"and add five" is the same as
+5
We combine it to get
3x+5
"I get twenty-six"
=26
combine to get
3x+5=26
Now we just solve:
subtract 5 from both sides to get
3x=21
divide by 3 to get
x=7
6 0
3 years ago
15 points. What is the value of k?
Lubov Fominskaja [6]

Answer:

k = 10

Step-by-step explanation:

The exterior angle of a triangle is equal to the sum of the 2 opposite interior angles.

∠ ZYM is an exterior angle of the triangle, thus

4k + 5 + 6k + 10 = 115, that is

10k + 15 = 115 ( subtract 15 from both sides )

10k = 100 ( divide both sides by 10 )

k = 10

7 0
3 years ago
Read 2 more answers
Mathew just got a job. He receives $100 bonus when he starts. The table shows the total number of hours (h) and the total amount
Tanzania [10]
T=10H+100

T being your total, and H is the hours worked.
5 0
3 years ago
2. When Joey dives off a diving board, the equation of his pathway can be modeled by h = -16- + 15 + 12.
Sergeu [11.5K]

Answer:

a) The maximum height is approx 15.5 unit.

b) The time it will take for Joey to reach the water is 1.45 hour.

Step-by-step explanation:

Given : When Joey dives off a diving board, the equation of his pathway can be modeled by h(t)= -16t^2+15t + 12

To find : a) Find Joey's maximum height.

b) Find the time it will take for Joey to reach the water.

Solution :

Modeled  h(t)= -16t^2+15t + 12 ....(1)

a) To find maximum height

Derivate (1) w.r.t. t,

h'= -32t+15

For critical point put h'=0,

-32t+15=0

t=\frac{15}{32}

t=0.46875

Derivate again w.r.t. t,

h''= -32

It is maximum at t=0.46875.

Substitute t in equation (1),

h(0.46875)= -16(0.46875)^2+15(0.46875)+12

h(0.46875)= -3.515625+7.03125+12

h(0.46875)= 15.515625

The maximum height is approx 15.5 unit.

b) To find the time it will take for Joey to reach the water.

Put h=0 in equation (1),

-16t^2+15t + 12=0

Apply quadratic formula, x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}

Here, a=-16 , b=15, c=12

t=\frac{-15\pm\sqrt{(15)^2-4(-16)(12)}}{2(-16)}

t=\frac{-15\pm\sqrt{225+768}}{-32}

t=\frac{-15\pm\sqrt{993}}{-32}

t=\frac{-15+\sqrt{993}}{-32},\frac{-15-\sqrt{993}}{-32}

t=−0.515,1.453

Reject negative value.

The time is t=1.45.

The time it will take for Joey to reach the water is 1.45 hour.

5 0
3 years ago
What is the answer to 20+x>70
choli [55]
20+x>70
-20. -20. on both sides to even it out

x>50 is the answer
7 0
3 years ago
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