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nlexa [21]
3 years ago
15

Can somebody help me with this ASAP

Mathematics
2 answers:
madam [21]3 years ago
6 0

Answer:

Slope is -2/5 and y-intercept is 4

\

Because of y=mx+b

b is always going to be the y-intercept and mx is going to be the slope.

Step-by-step explanation:

This is because  y=mx+b

Harman [31]3 years ago
5 0

Answer:

Slope is -2/5 and y-intercept is 4

Step-by-step explanation:

Because of y=mx+b

b is always going to be the y-intercept and mx is going to be the slope.

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Please answer this question now
soldier1979 [14.2K]

Answer:

6 km

Step-by-step explanation:

You use Pythagoras

{QN}^2 = (PN)^{2}+(PQ)^{2} \\PQ = \sqrt{(QN)^{2}-(PN)^{2}  }\\ PQ = 6 km

7 0
3 years ago
The freezing point of methanol is -97.6. What is the freezing point of methanol in Fahrenheit
motikmotik

Answer:

-143.68°F

Step-by-step explanation:

Celsius to Fahrenheit Conversion Formula: F = 1.8C + 32

Simply plug in <em>c</em> as -97.6:

F = 1.8(-97.6) + 32

F = -175.68 + 32

F = -143.68

3 0
3 years ago
Read 2 more answers
Does anyone know how to solve this problem? I am stuck. It is using solving systems of equations
bearhunter [10]
They both would need to get two toppings to cost the same amount which would be $8.60
8 0
3 years ago
Write an expression that is equivalent to (4x+11)+7x
Juliette [100K]

Answer:

Step-by-step explanation:

11x+11

or you could do

11(x+1)

4 0
3 years 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
3 years ago
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