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Lady_Fox [76]
3 years ago
11

Find the value of p so the line that passes through (4, -2) and (p, -1) has a slope of negative 1/7

Mathematics
1 answer:
Colt1911 [192]3 years ago
6 0
Can u show what ur saying on a graph so i can solve it for u

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What is 287.9412 rounded to the nearest tenth?
Westkost [7]

Hello!


If we want to round 287.9412 to the nearest tenth, the answer would be 287.9


Since rounding to the nearest tenth would mean rounding the first decimal, that's why it gives us that answer.


Also:


287.9412

2 = hundreds

8 = tens

7 = ones

. = decimal

9 = tenths

4 = hundredths

1 = thousandths

2 = millionths


Hope this helped! c:

4 0
3 years ago
Read 2 more answers
Jaylyn rode her bike Ina a bike thon. she rode 33.48 miles in 2.7 hours. of she rode at the same speed, what was her speed in mi
andrey2020 [161]
She was going 14.4 mph. hope it helps!
5 0
3 years ago
For the figure below, give the following.
Zepler [3.9K]

Answer:

3

Step-by-step explanation:

Just c cuz i smart

3 0
3 years ago
=
ivann1987 [24]

9514 1404 393

Answer:

  • large: 55 lb
  • small: 30 lb

Step-by-step explanation:

Let x and y represent the weights of the large and small boxes, respectively. The problem statement gives rise to the system of equations ...

  x + y = 85 . . . . . combined weight of a large and small box

  70x +50y = 5350 . . . . combined weight of 70 large and 50 small boxes

We can subtract 50 times the first equation from the second to find the weight of a large box.

  (70x +50y) -50(x +y) = (5350) -50(85)

  20x = 1100 . . . . simplify

  x = 55 . . . . . . . divide by 20

Using this in the first equation, we can find the weight of a small box.

  55 +y = 85

  y = 30 . . . . . . . subtract 55

A large box weighs 55 pounds; a small box weighs 30 pounds.

4 0
2 years ago
HELP PRECALC DO NOT UNDERSTAND WILL GIVE BRAINLIEST
Olegator [25]

Answer:

  the lower right matrix is the third correct choice

Step-by-step explanation:

Your problem statement shows that you have correctly selected the matrices representing the initial problem setup (middle left) and the problem solution (middle right).

Of the remaining matrices, the upper left is an incorrect setup, and the lower left is an incorrect solution matrix.

__

We notice that in the remaining matrices on the right that the (2,3) term is 0, and the (3,2) and (3,3) terms are both 1.

The easiest way to get a 0 in the 3rd column of row 2 is to add the first row to the second. When you do that, you get ...

  \left[\begin{array}{ccc|c}1&1&1&29000\\1+2&1-3&1-1&1000(29+1)\\0&0.15&0.15&2100\end{array}\right] =\left[\begin{array}{ccc|c}1&1&1&29000\\3&-2&0&30000\\0&0.15&0.15&2100\end{array}\right]

Already, we see that the second row matches that in the lower right matrix.

The easiest way to get 1's in the last row is to divide that row by 0.15. When we do that, the (3,4) entry becomes 2100/0.15 = 14000, matching exactly the lower right matrix.

The correct choices here are the two you have selected, and <em>the lower right matrix</em>.

3 0
3 years ago
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