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ad-work [718]
2 years ago
8

The table shows the cost of a taxicab ride for several driving distances.

Mathematics
1 answer:
s2008m [1.1K]2 years ago
7 0

Answer:

oops didnt mean to click answer again

Step-by-step explanation:

so it asked me go through the verification and yk how it be hella irriuitating kuz all im tryna do is

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Suppose that a is a square matrix with characteristic polynomial (λ − 3)2(λ − 6)3(λ + 1). (a) what are the dimensions of a? (giv
Gnesinka [82]

The problem statement gives the correct answers for parts (a) and (b). The total number of roots of the characteristic polynomial is the dimension of the matrix: 6. The eigenvalues are the zeros of the characteristic polynomial, 3 (multiplicity 2), 6 (multiplicity 3), and -1.

(c) The matrix is not invertible when one or more eigenvalues is zero. None of yours are zero, so the matrix is invertible.

3 0
2 years ago
(b) 3m - 2n=19<br> 5m + 7n=11​
Korvikt [17]

3m-2n=19

5m+7n=11

21m-14n=133

10m+14n=22

Now add equations

31m=155

m=5

Now let's find n

3×5-2n=19

-2n=4

n=-2

Answer: (5; -2)

6 0
2 years ago
How can you write the expression 7(b – 2) in words?
brilliants [131]
The answer is the product of 7 and the difference of b and - 2
8 0
2 years ago
Read 2 more answers
X1/3y1/6 rewrite expression in radical form
arsen [322]

Answer:

Step-by-step explanation:

Please, use the symbol " ^ " to denote exponentiation:

x^(1/3) * y^(1/6)

In radical form, this would be:

∛x*(6th root of y)    (the index of the second root is 6).

Alternatively, you could write:

∛x * √((∛y)).

8 0
2 years ago
Read 2 more answers
Prove that (sec 12A-1)/(sec 6A-1)=tan 12A/tan 3A
adelina 88 [10]

Let x=3A. Recall the following identities,

\cos^2\theta=\dfrac{1+\cos2\theta}2

\sin^2\theta=\dfrac{1-\cos2\theta}2

\sin2\theta=2\sin\theta\cos\theta

Now,

\dfrac{\sec12A-1}{\sec6A-1}=\dfrac{\sec4x-1}{\sec2x-1}

=\dfrac{\cos2x(1-\cos4x)}{\cos4x(1-\cos2x)}

=\dfrac{2\cos2x\sin^22x}{\cos4x(1-\cos2x)}

=\dfrac{2\cos2x\sin^22x(1+\cos2x)}{\cos4x(1-\cos^22x)}=\dfrac{2\cos2x\sin^22x(1+\cos2x)}{\cos4x\sin^22x}=\dfrac{2\cos2x(1+\cos2x)}{\cos4x}

=\dfrac{4\cos2x\cos^2x}{\cos4x}

=\dfrac{4\cos2x\cos^2x\sin4x}{\cos4x\sin4x}=\dfrac{4\cos2x\cos^2x\tan4x}{\sin4x}

=\dfrac{4\cos2x\cos^2x\tan4x}{2\sin2x\cos2x}=\dfrac{2\cos^2x\tan4x}{\sin2x}

=\dfrac{2\cos^2x\tan4x}{2\sin x\cos x}=\dfrac{\cos x\tan4x}{\sin x}

=\dfrac{\tan4x}{\frac{\sin x}{\cos x}}=\dfrac{\tan4x}{\tan x}=\dfrac{\tan12A}{\tan3A}

QED

5 0
2 years ago
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