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OleMash [197]
3 years ago
8

What is 1/100th of 882

Mathematics
2 answers:
masha68 [24]3 years ago
6 0
882 divided by 100 will give you 8.82
Lostsunrise [7]3 years ago
5 0
The answer is 8.82 because if you divide 882 by 100 it gives you 8.82
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Please help me on this question <br> 92x3<br> A. 79<br> B.42<br> C.276<br> D.267
victus00 [196]
The correct answer is
C.276
8 0
4 years ago
Can someone help..? thanks ​
Serjik [45]

Answer:

2.2*10^5

Step-by-step explanation:

to write a # in scientific notation change the # to decimal form.

so 2.2

then move the decimal point to the right until you get the original #

4 0
3 years ago
346 divided by 2 what is the answer
vladimir1956 [14]

Answer:

692


Step-by-step explanation:


3 0
3 years ago
Read 2 more answers
a jumping spider jumps from a log on the ground below. its height , h, in cm as a function of time ,t, in seconds since it jumpe
Anni [7]
Inorder to find the height, you miust plug in 0.05s for all times (t)
h(f) = -490t^2 + 75t + 12
h(f) = -490 • 0.05^2 + 75 • 0.05 + 12

Next follow PEMDAS from left to right.
(MD reversible, AS reversible)
(parentheses, exponents, multiply, divide, add, subtract)
parentheses~ nothing to simplify with parentheses
exponents~ h(f) = -490 • 0.0025 + 75 • 0.05 +12
multiply~  h(f) = -1.225 + 3.75 +12
divide~ nothing to simplify with division
add~ h(f) = 14.525
subtract~ nothing to simplify with subtraction

DON'T FORGET UNITS!

Answer: h(f) = 14.525 cm



5 0
3 years ago
Consider the following sets of matrices: M2(R) is the set of all 2 x 2 real matrices; GL2(R) is the subset of M2(R) with non-zer
defon

Answer:

Step-by-step explanation:

REcall the following definition of induced operation.

Let * be a binary operation over a set S and H a subset of S. If for every a,b elements in H it happens that a*b is also in H, then the binary operation that is obtained by restricting * to H is called the induced operation.

So, according to this definition, we must show that given two matrices of the specific subset, the product is also in the subset.

For this problem, recall this property of the determinant. Given A,B matrices in Mn(R) then det(AB) = det(A)*det(B).

Case SL2(R):

Let A,B matrices in SL2(R). Then, det(A) and det(B) is different from zero. So

\text{det(AB)} = \text{det}(A)\text{det}(B)\neq 0.

So AB is also in SL2(R).

Case GL2(R):

Let A,B matrices in GL2(R). Then, det(A)= det(B)=1 is different from zero. So

\text{det(AB)} = \text{det}(A)\text{det}(B)=1\cdot 1 = 1.

So AB is also in GL2(R).

With these, we have proved that the matrix multiplication over SL2(R) and GL2(R) is an induced operation from the matrix multiplication over M2(R).

7 0
3 years ago
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