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3 years ago

A Class 4 truck weighs between 14,000 and 16,000 pounds.

Mathematics

1 answer:

Nady [450]3 years ago

6 0

Answer:

Class 4: This class of truck has a GVWR of 14,001–16,000 pounds or 6,351–7,257 kilograms. Class 5: This class of truck has a GVWR of 16,001–19,500 pounds or 7,258–8,845 kilograms.

Step-by-step explanation:

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What are the 2 inequalities for 5x - 13 >2 2 -1 5 4

nordsb [41]

5 and 4 are two in inequalities because they make 5 bigger then 13

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Sharon paid $78 sales tax on a new camera. If the sales tax is 6.5%, what was the cost of the camera?

Irina18 [472]

Answer:

$1200

Step-by-step explanation:

$1200

Step-by-step explanation:

78 = 6.5%

This means that if you divide 78 by 6.5, you get the equivalent of 1% of the price:

78 ÷ 6.5 = 12

So 1% = 12

Now simply multiply this by 100 to get the full answer:

12 x 100 = 1200

So the camera cost $1200!

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At 7:00 p.m., there are 1,000 home-team fans and 700 away-team fans at a football game. After 7:00 p.m, 15 home team fans arrive

OleMash [197]

Answer:

Therefore the ratio of home-team fan to away team fan at 8:00 p.m is

=109:67

Step-by-step explanation:

Initially , total home team fans =1000

and away team fans = 700.

7:00 pm to 8:00 pm = 1 hour = 60 minutes

Given that

Every 10 minutes the number of arrival home team fans is 15.

Then in 1 minutes the number of arrival home team fans is $\frac{15}{10}$

and in 60 minutes the number of arrival home team fans is $\frac{15 \times 60}{10}$

=90

Therefore total home team fans = (90+1000)=1090

Given that

Every 10 minutes 5 fans leave of away team leave.

then in 1 minutes $\frac{5}{10}$ fans leave of away team leave.

and In 60 minutes $\frac{5\times 60}{10}$ =30 fans leave of away team leave.

Since 30 away team fans leave.

Total away team fans = (700-30)=670

Therefore the ratio of home-team fan to away team fan at 8:00 p.m is

=1090:670

=109:67

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3 years ago

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A, B, B,C for 8,9,10,11

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Find the intersection points using substitution or elimination for each system of equations:

natka813 [3]

Answer:

Step-by-step explanation:

Okay, so I think I know what the equations are, but I might have misinterpreted them because of the syntax- I think when you ask a question you can use the symbols tool to input it in a more clear way, otherwise you can use parentheses and such.

Problem 1:

(x²)/4 +y²= 1

y= x+1

*substitute for y*

Now we have a one-variable equation we can solve-

x²/4 + (x+1)² = 1

x²/4 + (x+1)(x+1)= 1

x²/4 + x²+2x+1= 1

*subtract 1 from both sides to set equal to 0*

x²/4 +x^2+2x=0

x²/4 can also be 1/4 * x²

1/4 * x² +1*x² +2x = 0

*combine like terms*

5/4 * x^2+2x+ 0 =0

now, you can use the quadratic equation to solve for x

a= 5/4

b= 2

c=0

the syntax on this will be rough, but I'll do my best...

x= (-b ± √(b²-4ac))/(2a)

x= (-2 ±√(2²-4*(5/4)*(0))/(2*(5/4))

x= (-2 ±√(4-0))/(2.5)

x= (-2±2)/2.5

x will have 2 answers because of ±

x= 0 or x= 1.6

now plug that back into one of the equations and solve.

y= 0+1 = 1

y= 1.6+1= 2.6

Hopefully this explanation was enough to help you solve problem 2.

Problem 2:

x² + y² -16y +39= 0

y²- x² -9= 0

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3 years ago