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3 years ago

Multiply the following ?

Mathematics

1 answer:

Harlamova29_29 [7]3 years ago

4 0

Answer:

$\frac{3x^2}{x^2-6x-7}$

Step-by-step explanation:

The first step is to find the multiplied out numerator and denominator, and then to simplify if possible. For the numerator, 3x*x=3x^2. For the denominator, (x+1)(x-7)=x^2-6x-7. This gives a final fraction of $\frac{3x^2}{x^2-6x-7}$ . Hope this helps!

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What is the value of x to the nearest tenth?

Colt1911 [192]

Answer:

C. 17.6

General Formulas and Concepts:

Math

Rounding

Pre-Algebra

Order of Operations: BPEMDAS

Brackets
Parenthesis
Exponents
Multiplication
Division
Addition
Subtraction

Left to Right

Equality Properties

Multiplication Property of Equality
Division Property of Equality
Addition Property of Equality
Subtraction Property of Equality

Trigonometry

[Right Triangles Only] SOHCAHTOA
[Right Triangles Only] tanθ = opposite over adjacent

Step-by-step explanation:

Step 1: Identify Variables

Angle θ = 40°

Opposite Leg = x

Adjacent Leg = 21

Step 2: Solve for x

Substitute in variables [tangent]: $\displaystyle tan(40^\circ) = \frac{x}{21}$
[Multiplication Property of Equality] Isolate x: $\displaystyle 21tan(40^\circ) = x$
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Evaluate: $\displaystyle x = 17.6211$
Round: $\displaystyle x \approx 17.6$

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2 years ago

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What is the simplified form of the expression? (b/7)^2

MakcuM [25]

(b/7)^2=(b/7)(b/7)= $\frac{b^2}{49}$

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3 years ago

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What eigen value for this matix (1 -2) (-2 0)

natali 33 [55]

You find the eigenvalues of a matrix A by following these steps:

Compute the matrix $A' = A-\lambda I$ , where I is the identity matrix (1s on the diagonal, 0s elsewhere)
Compute the determinant of A'
Set the determinant of A' equal to zero and solve for lambda.

So, in this case, we have

$A = \left[\begin{array}{cc}1&-2\\-2&0\end{array}\right] \implies A'=\left[\begin{array}{cc}1&-2\\-2&0\end{array}\right]-\left[\begin{array}{cc}\lambda&0\\0&\lambda\end{array}\right]=\left[\begin{array}{cc}1-\lambda&-2\\-2&-\lambda\end{array}\right]$

The determinant of this matrix is

$\left|\begin{array}{cc}1-\lambda&-2\\-2&-\lambda\end{array}\right| = -\lambda(1-\lambda)-(-2)(-2) = \lambda^2-\lambda-4$

Finally, we have

$\lambda^2-\lambda-4=0 \iff \lambda = \dfrac{1\pm\sqrt{17}}{2}$

So, the two eigenvalues are

$\lambda_1 = \dfrac{1+\sqrt{17}}{2},\quad \lambda_2 = \dfrac{1-\sqrt{17}}{2}$

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3 years ago

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(Geometry HELP) The perimeter of WXY is ___ cm?

jeka57 [31]

I hope this helps you

QR=WS=2.93

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QS=XR =2.28

RS=XQ=2.04

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XW=XQ+QW=2.04+2.04=4.08

WY=WS+SY=2.93+2.93=5.86

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4 years ago

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$(y-3)=(x-0)$

$(y-3)=(x)$

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4 years ago

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