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Rus_ich [418]
3 years ago
12

What is -15.41 round to the nearest hundredth

Mathematics
1 answer:
Hoochie [10]3 years ago
5 0

It is already rounded to the nearest hundredth.

Rounded to the nearest tenth: -15.4

Rounded to the nearest hundredth: -15.41

Hope this helps you!

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Solve for x using technology 2 log 3(x-1)=log(4 x 2-25)
bagirrra123 [75]
<span>2 log 3(x-1)=log(4 x 2-25)
1.  Determine the domain.  Since the input to the log function cannot be zero or negative, 4x^2-25 must be </span>≥ 0.  Thus, x^2 must be >0, or x>0.  Same domain applies to log (3(x-1); x must be > 0.
2.  Rewrite <span>2 log 3(x-1) as log 3(x-1)^2.
3.  Then we have </span>log 3(x-1)^2 = log(4 x 2-25).  We can discard the operator "log" from both sides:   3(x-1)^2 = 4 x 2-25.  There are various ways in which to solve this.  Since you're supposed to "use technology," 
graph y = 3(x-1)^2 and y = 4x^2 - 25 on the same set of axes.  Determine, using visual estimation or your calculator's tools, the value or values of x that satisfy this equation.  My result was x=3, y =11.


3 0
3 years ago
PLEASE PLEASE PLEASE HELP grade 9 work!!!!
11Alexandr11 [23.1K]

Answer:

P = 2(l+w)\\Length = 5x\\Width = 3x+7\\\\P = 2(5x +(3x+7))

Step-by-step explanation:

Further Solving :

P = 2(5x +(3x+7))\\P = 2(5x+3x+7)\\P = 2(8x+7)\\\\Perimeter = 16x + 14

5 0
3 years ago
Write the quadratic equation whose roots are 3 and -1, and whose leading coefficient is 5.
Assoli18 [71]

Answer:

5x² - 10x - 15 = 0

Step-by-step explanation:

Given that the roots are x = 3 and x = - 1, then the factors are

(x - 3) and (x + 1) and the quadratic is the product of the factors, that is

f(x) = a(x - 3)(x + 1) ← a is a multiplier

Here a = 5, thus

f(x) = 5(x - 3)(x + 1) ← expand factors using FOIL

     = 5(x² - 2x - 3) ← distribute parenthesis by 5

     = 5x² - 10x - 15

Thus equation is

5x² - 10x - 15 = 0

6 0
3 years ago
A meal coats $16.64. What's the total cost with a 15% tip?​
kotegsom [21]

Answer:

$19.136

Step-by-step explanation:

if 100% ----> $16.64

then 115% ----> $19.136

hope this helps :)

5 0
2 years ago
Read 2 more answers
Linear Algebra question! Please help!
kozerog [31]

Answers:

  1. false
  2. false
  3. true
  4. false
  5. True

==================================================

Explanation:

Problem 1

This is false because the A and B should swap places. It should be (AB)^{-1} = B^{-1}A^{-1}.

The short proof is to multiply AB with its inverse (AB)^{-1}  and we get: (AB)*(AB)^{-1} = (AB)*(B^{-1}A^{-1}) = A(B*B^{-1})*A^{-1} = A*A^{-1} = I

The fact we get the identity matrix proves that we have the proper order at this point. The swap happens so that B matches up its corresponding inverse B^{-1} and the two cancel each other out.

Keep in mind matrix multiplication is <u>not</u> commutative. So AB is not the same as BA.

-------------------------

Problem 2

This statement is true if and only if AB = BA

(A+B)^2 = (A+B)(A+B)

(A+B)^2 = A(A+B) + B(A+B)

(A+B)^2 = A^2 + AB + BA + B^2

(A+B)^2 = A^2 + 2AB + B^2 ... only works if AB = BA

However, in most general settings, matrix multiplication is <u>not</u> commutative. The order is important when multiplying most two matrices. Only for special circumstances is when AB = BA going to happen. In general,  AB = BA is false which is why statement two breaks down and is false in general.

-------------------------

Problem 3

This statement is true.

If A and B are invertible, then so is AB.

This is because both A^{-1} and B^{-1} are known to exist (otherwise A and B wouldn't be invertible) and we can use the rule mentioned in problem 1. Make sure to swap the terms of course.

Or you can use a determinant argument to prove the claim

det(A*B) = det(A)*det(B)

Since A and B are invertible, their determinants det(A) and det(B) are nonzero which makes the right hand side nonzero. Therefore det(A*B) is nonzero and AB has an inverse.

So if we have two invertible matrices, then their product is also invertible. This idea can be scaled up to include things like A^4*B^3 being also invertible.

If you wanted, you can carefully go through it like this:

  1. If A and B are invertible, then so is AB
  2. If A and AB are invertible, then so is A*AB = A^2B
  3. If A and A^2B are invertible, then so is A*A^2B = A^3B

and so on until you build up to A^4*B^3. Therefore, we can conclude that A^m*B^n is also invertible. Be careful about the order of multiplying the matrices. Something like A*AB is different from AB*A, the first of which is useful while the second is not.

So this is why statement 3 is true.

-------------------------

Problem 4

This is false. Possibly a quick counter-example is to consider these two matrices

A = \begin{bmatrix}1 & 0\\0 & 1\end{bmatrix} \text{ and } B = \begin{bmatrix}-1 & 0\\0 & -1\end{bmatrix}

both of which are invertible since their determinant is nonzero (recall the determinant of a diagonal matrix is simply the product along the diagonal entries). So it's not too hard to show that the determinant of each is 1, and each matrix shown is invertible.

However, adding those two mentioned matrices gets us the 2x2 zero matrix, which is a matrix of nothing but zeros. Clearly the zero matrix has determinant zero and is therefore not invertible.

There are some cases when A+B may be invertible, but it's not true in general.

-------------------------

Problem 5

This is true because each A pairs up with an A^{-1} to cancel out (similar what happened with problem 1). For more info, check out the concept of diagonalization.

5 0
2 years ago
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