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Zina [86]
3 years ago
9

Which class of trailer hitch is best suited for a boat and its equipment weighing less than 2,000 pounds?

Mathematics
1 answer:
alexira [117]3 years ago
4 0
Class I

Trailer hitches are divided into different classes on the basis of the weight that they will be supporting. The classes are Class I - V, after which the classes are Xtra Duty and finally the Commercial Duty class, which has the ability to support 20,000 lbs.
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jenny has 68 1/2 pages of her left to read. How many pages will she have left after she reads 26 1/6 pages?
jekas [21]

Answer:

42 1/3

Step-by-step explanation:

We subtract 26 1/6 from 68 1/2. We do this by finding the LCM of 6 and 2. That is 6. We multiply 1/2 by 3/3 to get 3/6. Now, we subtract. 68 - 26 is 42, and 3/6 - 1/6 is 2/6. Our answer is 42 2/6. That can be simplified to 42 1/3.

6 0
3 years ago
Which numbers are "rational" and which numbers are "irrational"?
taurus [48]

the 3rd one. hope this helps

7 0
2 years ago
Read 2 more answers
What are the four points of intersection between 4x^2 + y^2 - 4y - 32 = 0 and x^2 - y - 7 = 0 ? Solve algebraically
tensa zangetsu [6.8K]
Remember (a²-b²)=(a-b)(a+b)

solve for  a single variable
solve for y in 2nd

add y to both sides
x²-7=y
sub (x²-7) for y in other equaiton

4x²+(x²-7)²-4(x²-7)-32=0
expand
4x²+x⁴-14x²+49-4x²+28-32=0
x⁴-14x²+45=0
factor
(x²-9)(x²-5)=0
(x-3)(x+3)(x-√5)(x+√5)=0
set each to zero

x-3=0
x=3

x+3=0
x=-3

x-√5=0
x=√5

x+√5=0
x=-√5


sub back to find y

(x²-7)=y

for x=3
9-7=2
(3,2)

for x=-3
9-7=2
(-3,2)

for √5
5-7=-2
(√5,-2)

for -√5
5-7=-2
(-√5,-2)


the intersection points are

(3,2)
(-3,2)
(√5,-2)
(-√5,-2)
8 0
3 years ago
What is the solution of log3x − 5 16 = 2?
Stolb23 [73]

Answer:

log (16)=2 in base 3x-5

==>16=2^(3x-5)

==>2^4=2^(3x-5)

==>4=3x-5

==>3x=9

==>x=3

Step-by-step explanation:

4 0
2 years ago
Kindly answer this one. Thank you.​
steposvetlana [31]

Answer:

See below.

Step-by-step explanation:

I'm assuming these questions are about the Midline Theorem (segment AL joins the midpoints of the non-parallel sides.

♦  The midline's length is the average of the lengths of the top and bottom parallel sides.

AL=\frac{OR+CE}{2}

Use this equation and substitute values given in each problem, then solve for the missing information.

1. AL = x, CE = 9, OR = 5

x=\frac{9+5}{2}=7

2. AL = <em>m</em> - 4, CE = 15, OR = 17

m-4=\frac{15+17}{2}=16\\\\m-4=16\\\\\\m=20\\\\AL=20-4=16

3. OR = y + 5, AL = 15, CE = 18

15=\frac{(y+5)+18}{2}\\\\15=\frac{y+23}{2}\\\\30=y+23\\\\7=y\\\\OR = 7+5=12

4 0
2 years ago
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