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2 years ago

How do you find the equivalent ratios

Mathematics

1 answer:

larisa86 [58]2 years ago

8 0

Answer:

by multiplying or dividing

Step-by-step explanation:

Two ratios that have the same value are called equivalent ratios. To find an equivalent ratio, multiply or divide both quantities by the same number. It is the same process as finding equivalent fractions.

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What are the multiplication sentences for factors of 6?

mart [117]

Answer:

keep doubling the answer by 6

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2 years ago

Find the range of the inequality 2e-3< 3e-1

serg [7]

Answer:

$x = { - 1, 0,1 ,2 ...}$

Step-by-step explanation:

$2e - 3 < 3e - 1 = 2e - 3e < - 1 + 3 = - 1e < 2 = e > - 2$

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6 0

2 years ago

At the store I can buy a 2 pack of shirts for $22.78 or a 6 pack of shirts for $67.80. Which is the better deal?

mina [271]

Answer: A 6 pack of shirts for $67.80 is the better deal than the other one.

Step-by-step explanation: Is that one because you need to divide each of the prices to the number of shirts. First, lets start of with 22.78/2, 22.78 divided by the 2 pack of shirts is a quotient of $11.39.

Next lets do $67.80 / 6, So 67.80 divided by 6 equals $11.30. Meanwhile, 11.30 < 11.39 because of the sale. So that's why the 6 pack of shirts for $67.80 is the better deal.

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8 0

1 year ago

Help me out its division

Ber [7]

4/2=2
4/3=1.33
4/4=1
4/5=0.8
4/0= undefined

5 0

1 year ago

Suppose that the trace of a 2×2 matrix a is tr(a)=15 and the determinant is det(a)=50. find the eigenvalues of

IrinaK [193]

Recall that the characteristic polynomial of a 2x2 matrix $\mathbf A=\begin{bmatrix}a&b\\c&d\end{bmatrix}$ is

$\det(\mathbf A-\lambda\mathbf I)=\begin{vmatrix}a-\lambda&b\\c&d-\lambda\end{vmatrix}=(a-\lambda)(d-\lambda)-bc=\lambda^2-(a+d)\lambda+(ad-bc)$

but $\det(\mathbf A)=ad-bc$ and $\mathrm{tr}(\mathbf A)=a+d$ , so the characteristic polynomial for $\mathbf A$ is

$\lambda^2-\mathrm{tr}(\mathbf A)\lambda+\det(\mathbf A)$

We're given that the trace is 15 and determinant is 50, so the characteristic polynomial for the matrix in question is

$\lambda^2-15\lambda+50$

and the eigenvalues are those $\lambda$ for which the characteristic polynomial evaluates to 0.

$\lambda^2-15\lambda+50=(\lambda-5)(\lambda-10)=0\implies\lambda=5,\lambda=10$

5 0

2 years ago