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Daniel [21]
4 years ago
13

Assuming a truck container holds 24 cases, how many cases would 6 truck containers hold?

Mathematics
1 answer:
Montano1993 [528]4 years ago
6 0
If you multiply 6 by 24 for you get 144 containers.
6 truck will hold 144 containers in total
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Help answering this simple Series Question? I don't know what I did wrong!
Tatiana [17]
Looks like you just evaluated the summand for the given value of n, whereas the question is asking you to find the value of the sum for the first n terms.

Let S_k=\displaystyle\sum_{n=1}^k\frac3{(-2)^n}. Then S_k is the kth partial sum.

S_1 happens to be the first term in the series, which is why that box is marked correct:

S_1=\displaystyle\sum_{n=1}^1\frac3{(-2)^n}=\frac3{(-2)^1}=-1.5

But the next partial sum is not correct:

S_2=\displaystyle\sum_{n=1}^2\frac3{(-2)^n}=\frac3{(-2)^1}+\frac3{(-2)^2}=-0.75

and this is not the same notion as the second term (which indeed is 0.75) in the series.
5 0
3 years ago
Can anyone help<br> Me with this problem plz
seropon [69]
First you need common denominator
x/2 - 6/2 = 34/2
now , eliminate denominator 
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x=40
3 0
4 years ago
How many cups are in 1 kilogram
Lemur [1.5K]

Answer:

1 kilogram (kg) = 4.226752838 cup (cup).

Step-by-step explanation:

6 0
4 years ago
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Are there more or less than 5 pt of blood in a human adult. Please help ASAP
Mazyrski [523]
There are 8pt on blood on average in a human adult, which is more than 5.

Hope this helps!
7 0
3 years ago
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Find the lengths and slopes of the diagonals to name the parallelogram. Choose the most specific name. E (-2, -4), F(0, -1), G(-
Kay [80]

Answer:

1) d) Square

2) Proofs that PWRS is a rhombus are

Length of QS ≠ PR and

Slope of segment QR and PS is -1/2 and Slope of segment RS and QP is -2.

Step-by-step explanation:

The given points (x, y) of the parallelogram are;

E(-2, -4), F(0, -1), G(-3, 1), H(-5, -2)

The slope, m, of the segments are found as follows;

Slope, \, m =\dfrac{y_{2}-y_{1}}{x_{2}-x_{1}}

By computation, the slope of segment EF = 1.5

The slope of segment FG = -0.67

The slope of segment GH = 1.5

The slope of segment HE = -0.67

Therefore, EF is parallel to GH and FG is parallel to HE

The length of the sides are;

\sqrt{\left (y_{2}-y_{1}  \right )^{2}+\left (x_{2}-x_{1}  \right )^{2}}

By computation, the length of segment EF = 3.61

The length of segment FG = 3.61

The length of segment GH = 3.61

The length of segment HE = 3.61

The diagonals are;

EG and FH

The length of segment EG = 5.099

The length of segment FH = 5.099

Therefore, the diagonals are equal and the parallelogram is a square

2) The given dimensions are;

P(-1, 3), Q(-2, 5), R(0, 4), S(1, 2)

A rhombus has all sides equal

The length of segment PQ = 2.24

The length of segment QR = 2.24

The length of segment RS = 2.24

The length of segment PS = 2.24

The diagonals are;

QS and PR

The length of segment QS = 4.24

The length of segment PR = 1.41

The slope of segment QR = -0.5

The slope of segment PS = -0.5

The slope of segment RS = -2

The slope of segment QP = -2

Therefore, QS≠QR the parallelogram is a rhombus

The correct option ;

Length of QS ≠ PR and

Slope of segment QR and PS is -1/2 and Slope of segment RS and QP is -2.

Where there are acute angles in parallelogram PQRS, then the correct option is d) Length of QR and PS is 2.2 and Length of RS and QP is 2.2

8 0
3 years ago
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