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1 year ago

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If a₁ = a + 2a +3 and d₂ = 2a + a + 1, what is the value of 2 (d1- d2)?

1 answer:

maria [59]1 year ago

6 0

With the help of equations, we know that the value of equation 2(d1- d2) is 4.

<h3>What is an equation?</h3>

A mathematical expression with an equals sign is referred to as an equation.
Algebra is widely used in equations.
When performing calculations but unsure of the precise amount, algebra is used.
The definition of an equation in algebra is a mathematical statement that demonstrates the equality of two mathematical expressions.
For instance, the equation 3x + 5 = 14 consists of the two equations 3x + 5 and 14, which are separated by the 'equal' sign.

The equations we have are as follows:

d1 = a+2a+3
d2 = 2a+a+1

After simplifying, d1= 3a+3:

d2 = 3a+1
2(d1-d2) = 2(3a+3-3a-1)
=2(2)
= 4

Therefore, with the help of equations, we know that the value of equation 2(d1- d2) is 4.

To know more about the equation, visit:

brainly.com/question/28937794

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What would you expect the nominal rate of interest to be if the real rate is 4.4 percent and the expected inflation rate is 7.5

ANTONII [103]

Answer:

Nominal Interest rate=11.9%

Step-by-step explanations:

The Fisher effect is a theory propounded by an economist named Irving Fisher.

Fisher's equation shows the relationship between real Interest rate, expected inflation rate and nominal Interest rate.

It can be calculated by subtracting the expected inflation rate from the nominal Interest rate to give the real Interest rate.

Real Interest rate= nominal Interest rate - expected inflation rate

Given,

Real Interest rate= 4.4%=0.044

Expected inflation rate=7.5%=0.075

Nominal Interest rate=?

Therefore,

Real Interest rate=nominal Interest rate - expected inflation rate

Nominal Interest rate=Real Interest rate+expected inflation rate

Nominal Interest rate=0.044+0.075

Nominal Interest rate=0.119

Nominal Interest rate=11.9%

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Find all values of the angle θ (in radians, with 0 ≤ θ < 2π) for which the matrix a = cos θ −sin θ sin θ cos θ has real eigen

harina [27]

The matrix

$A=\begin{bmatrix}\cos\theta&-\sin\theta\\\sin\theta&\cos\theta\end{bmatrix}$

has eigenvalues $\lambda$ such that

$\det(A-\lambda I)=\begin{vmatrix}\cos\theta-\lambda&-\sin\theta\\\sin\theta&\cos\theta-\lambda\end{vmatrix}=0$

$(\cos\theta-\lambda)^2+\sin^2\theta=0$

$(\cos\theta-\lambda)^2=-\sin^2\theta$

$\cos\theta-\lambda=\pm\sqrt{-\sin^2\theta}$

$\lambda=\cos\theta\pm\sqrt{-\sin^2\theta}$

$\sin^2\theta\ge0$ for all values of $\theta$ , so we need to have $\sin\theta=0$ in order for $\lambda$ to be real-valued. This happens for

$\sin\theta=0\implies\theta=n\pi$

where $n$ is any integer, and over the given interval we have $\theta=0$ and $\theta=\pi$ .

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2 years ago