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DaniilM [7]
2 years ago
8

How many different 5-digit whole

Mathematics
1 answer:
Lerok [7]2 years ago
4 0

Answer:

120

Step-by-step explanation:

so

<h2 /><h2><em><u> ?????</u></em></h2>

this 5 interrogative marks are our 5 digit number

for the first place you have 5 choices(2, 3, 4, 5, 6)

for the second:4

for the third:3

for the fourth:2

and for the fifth:1

you multiply this choices

5×4×3×2=120

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salantis [7]

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"The dog" doesn't belong in the sentence

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Svetach [21]

Answer:

16.4

Step-by-step

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Read 2 more answers
The width of a rectangle field is 3h yards and its length is 3 yards longer than the width. The field has a fence around its per
d1i1m1o1n [39]

Answer:

Part a) P=(12h+2)\ yd

Part b) The expression that represent the cost of fencing the field is (336h+56)

Part c) \$1,736

Step-by-step explanation:

<u>The complete question in the attached figure</u>

Part a) we know that

The perimeter of the rectangular field (excluding the width of the gate) is equal to

P=2(L+W)-4

we have

L=(3h+3)\ yd\\W=3h\ yd

substitute the given values

P=2(3h+3+3h)-4

P=2(6h+3)-4\\P=12h+6-4\\P=(12h+2)\ yd

Part b) we know that

To find out the cost of fencing the field (excluding the gate) , multiply the perimeter by the cost of $28 per yard

so

28(12h+2)=336h+56

Part c) we have

h=5

substitute the value of h in the expression of the cost

336(5)+56=\$1,736

8 0
3 years ago
Factor and solve 121r^2-64t^2
dexar [7]

Answer:

(11r + 8t)(11r - 8t)

Step-by-step explanation:

difference of two squares

6 0
2 years ago
Select the best answer for the question
Klio2033 [76]

Answer:

C. \left[\begin{array}{ccc}-24&0&-3\\8&-48&56\\25&-6&10\end{array}\right]

Step-by-step explanation:

The given matrices are;

F=\left[\begin{array}{cc}-2&0\\0&8\\2&1\end{array}\right]

and

C=\left[\begin{array}{ccc}12&0&\frac{3}{2}\\1&-6&7\end{array}\right]

FC=\left[\begin{array}{cc}-2&0\\0&8\\2&1\end{array}\right]\left[\begin{array}{ccc}12&0&\frac{3}{2}\\1&-6&7\end{array}\right]

FC=\left[\begin{array}{ccc}12(-2)+0(1)&0(-2)+-6(0)&\frac{3}{2}(-2)+7(0)\\12(0)+8(1)&0(0)+8(-6)&0(\frac{3}{2})+8(7)\\2(12)+1(1)&2(0)+1(-6)&2(\frac{3}{2})+1(7)\end{array}\right]

This simplifies to;

FC=\left[\begin{array}{ccc}-24&0&-3\\8&-48&56\\25&-6&10\end{array}\right]

The correct answer is C

5 0
2 years ago
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