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3 years ago

Complete the solution of the equation. Find the value of y when x equals 4. -3x+9y=-57

2 answers:

4 0

-3x + 9y = -57 x = 4

-3*4 + 9y = -57

-12 + 9y = -57

9y = -57 + 12

9y = 12 - 57

9y =-45 Divide through by 9.

y = -45/9

<span>y = -5

</span>Hope this helped.

rusak2 [61]3 years ago

4 0

-3x+9y=-57
-3(4)+9y=-57
-12+9y=-57
+12 +12
9y=-45
Y=-5

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Step2247 [10]

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Step-by-step explanation:

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3 years ago

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Zoe is solving the equation 3x – 4 = –10 for x. She used the addition property of equality to isolate the variable term as shown

kherson [118]

3x-4 = -10

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3 years ago

An n × n matrix B has characteristic polynomial p(λ) = −λ(λ − 3) 3 (λ − 2) 2 (λ + 1). Which of the following statements is false

asambeis [7]

Answer:

Only d) is false.

Step-by-step explanation:

Let $p=p(\lambda)=\lambda(\lambda-3)^3 (\lambda-2)^2 (\lambda+1)$ be the characteristic polynomial of B.

a) We use the rank-nullity theorem. First, note that 0 is an eigenvalue of algebraic multiplicity 1. The null space of B is equal to the eigenspace generated by 0. The dimension of this space is the geometric multiplicity of 0, which can't exceed the algebraic multiplicity. Then Nul(B)≤1. It can't happen that Nul(B)=0, because eigenspaces have positive dimension, therfore Nul(B)=1 and by the rank-nullity theorem, rank(B)=7-nul(B)=6 (B has size 7, see part e)

b) Remember that $p(\lambda)=\det(B-\lambda I)$ . 0 is a root of p, so we have that $p(0)=\det(B-0 I)=\det B=0$ .

c) The matrix T must be a nxn matrix so that the product BTB is well defined. Therefore det(T) is defined and by part c) we have that det(BTB)=det(B)det(T)det(B)=0.

d) det(B)=0 by part c) so B is not invertible.

e) The degree of the characteristic polynomial p is equal to the size of the matrix B. Summing the multiplicities of each root, p has degree 7, therefore the size of B is n=7.

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3 years ago

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3 years ago

How to solve this :') please help

blsea [12.9K]

Answer:

3/2

Step-by-step explanation:

sin(3π/4 - β) = sin(3π/4)cosβ - cos(3π/4)sinπ =

$\sin(\frac{3\pi}{4}-\beta)=\sin(\frac{3\pi}{4})\cos\beta-\cos\frac{3\pi}{4}\sin\beta\\=\frac{1}{\sqrt{2}}\cos\beta-(-\frac{1}{\sqrt{2}})\sin\beta\\\\=\frac{1}{\sqrt{2}}(\cos\beta +\sin\beta)\\\\\sin^2(\frac{3\pi}{4}-\beta)=\frac{1}{2}\cos\beta +\sin\beta)^2=\frac{1}{2}(\cos^2\beta +\sin^2\beta+2\sin\beta\cos\beta\\=\frac{1}{2}(1+\sin2\beta)=\frac{1}{2}(1-\frac{1}{5}) = \frac{2}{5}\\$

Use $\cot^2x=\csc^2x-1=\frac{1}{\sin^2x}-1$

$\cot^2(\frac{3\pi}{4}-\beta)=\frac{1}{\sin^2(\frac{3\pi}{4}-\beta)}-1 = \frac{1}{\frac{2}{5}}-1=\frac{5}{2}-1=\frac{3}{2}$

7 0

2 years ago