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Alexeev081 [22]
3 years ago
11

a delivery truck is transporting boxes of two sizes: large and small. The combined weight of a large box and a small box is 90 p

ounds. The truck is transporting 70 large boxes and 65 small boxes. If the truck is carrying a total of 6150 pounds in boxes, how much does each type of box weigh?
Mathematics
2 answers:
frozen [14]3 years ago
6 0

Answer:

45

Step-by-step explanation:

70 + 65 = 135 boxed both large and small.

6150 / 135 = 45 pounds estimated.

Fofino [41]3 years ago
6 0

Answer:

<em>A small box weighs 30 lb; a large box weight 60 lb.</em>

Step-by-step explanation:

Let L = weight of large box

Let S = weight of small box

"The combined weight of a large box and a small box is 90 pounds."

L + S = 90

"The truck is transporting 70 large boxes and 65 small boxes. If the truck is carrying a total of 6150 pounds in boxes"

L = 90 - S    Equation 1

The truck carries 65 small boxes and 65 large boxes and another 5 large boxes.

65(90) + 5(90 - S) = 6150   Equation 2

We use these two equations (Eq. 1 and Eq. 2) as a system of equations.

L = 90 - S

5850 + 450 - 5S = 6150

6300 - 5S = 6150

-5S = -150

S = 30

L = 90 - S = 90 - 30 = 60

S = 30; L = 60

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egoroff_w [7]
B equals c hope this helps :)
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2 years ago
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Can somebody please help me with 1 and 2 and explain why that’s the answer thanks
Masteriza [31]

Answer:

1) A

2) C

Step-by-step explanation:

The range is all real y values, in this case it includes zero and continues going downward towards negative infinity.

The domain is all real x values. In this case it includes zero and continues increasing to positive infinity.

Hope this helps!

5 0
3 years ago
The midpoint of GH are M(4,3) and D(5,-6). The coordinates of the endpoint are?
Phantasy [73]

Answer:

The endpoint coordinates are (3,12)

Step-by-step explanation:\frac{5+x^{2} }{2} =4 \frac{-6+y^{2} }{2} = 3

You multiply 4 by 2 which gives you 8 and write the equation5+x^{2} =8 then you subtract 5 from 8 to get 3 for the x coordinate.

You multiply 3 by 2 which gives 6 and writhe the equation-6+y^{2}=6 then you substitute the -6 for +6 and add +6 and 6 to get 12 for the y coordinate.

To check use the midpoint formula with coordinates D and your endpoint coordinates\frac{5+3}{2}=\frac{8}{2} =4 \\  \frac{-6+12}{2} =\frac{6}{2} =3 and you get the answer for the midpoint which is (4,3)

6 0
2 years ago
How many terms of the arithmetic sequence {1,22,43,64,85,…} will give a sum of 2332? Show all steps including the formulas used
MA_775_DIABLO [31]

There's a slight problem with your question, but we'll get to that...

Consecutive terms of the sequence are separated by a fixed difference of 21 (22 = 1 + 21, 43 = 22 + 21, 64 = 43 + 21, and so on), so the <em>n</em>-th term of the sequence, <em>a</em> (<em>n</em>), is given recursively by

• <em>a</em> (1) = 1

• <em>a</em> (<em>n</em>) = <em>a</em> (<em>n</em> - 1) + 21 … … … for <em>n</em> > 1

We can find the explicit rule for the sequence by iterative substitution:

<em>a</em> (2) = <em>a</em> (1) + 21

<em>a</em> (3) = <em>a</em> (2) + 21 = (<em>a</em> (1) + 21) + 21 = <em>a</em> (1) + 2×21

<em>a</em> (4) = <em>a</em> (3) + 21 = (<em>a</em> (1) + 2×21) + 21 = <em>a</em> (1) + 3×21

and so on, with the general pattern

<em>a</em> (<em>n</em>) = <em>a</em> (1) + 21 (<em>n</em> - 1) = 21<em>n</em> - 20

Now, we're told that the sum of some number <em>N</em> of terms in this sequence is 2332. In other words, the <em>N</em>-th partial sum of the sequence is

<em>a</em> (1) + <em>a</em> (2) + <em>a</em> (3) + … + <em>a</em> (<em>N</em> - 1) + <em>a</em> (<em>N</em>) = 2332

or more compactly,

\displaystyle\sum_{n=1}^N a(n) = 2332

It's important to note that <em>N</em> must be some positive integer.

Replace <em>a</em> (<em>n</em>) by the explicit rule:

\displaystyle\sum_{n=1}^N (21n-20) = 2332

Expand the sum on the left as

\displaystyle 21 \sum_{n=1}^N n-20\sum_{n=1}^N1 = 2332

and recall the formulas,

\displaystyle\sum_{k=1}^n1=\underbrace{1+1+\cdots+1}_{n\text{ times}}=n

\displaystyle\sum_{k=1}^nk=1+2+3+\cdots+n=\frac{n(n+1)}2

So the sum of the first <em>N</em> terms of <em>a</em> (<em>n</em>) is such that

21 × <em>N</em> (<em>N</em> + 1)/2 - 20<em>N</em> = 2332

Solve for <em>N</em> :

21 (<em>N</em> ² + <em>N</em>) - 40<em>N</em> = 4664

21 <em>N</em> ² - 19 <em>N</em> - 4664 = 0

Now for the problem I mentioned at the start: this polynomial has no rational roots, and instead

<em>N</em> = (19 ± √392,137)/42 ≈ -14.45 or 15.36

so there is no positive integer <em>N</em> for which the first <em>N</em> terms of the sum add up to 2332.

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2 years ago
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Bond [772]

Answer:

Whichever one shows >24

Step-by-step explanation:

Which ever one that is greater than 6

7 0
3 years ago
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