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tekilochka [14]
3 years ago
7

Solve for x Solver for x

Mathematics
1 answer:
Vlada [557]3 years ago
5 0

Answer:

x = 4

Step-by-step explanation:

since it is an equilateral triangle, each angle is 60°. To find x, set up the equation 17x-8=60

17x = 68

x=4

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Mrs. Clements has a table with a top that is shaped like the trapezoid shown. Which expression can Mrs. Clements use to find the
Lunna [17]

Answer:

60

Step-by-step explanation:

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Compare √7 and 1.9? <br><br> &gt;<br> &lt;<br> =<br> Help pls <br><br> 
Paraphin [41]

Answer:

that answer is 7 is bigger then 1.9

Step-by-step explanation:

so 7 is a whole and 1.9 is a decimal

8 0
3 years ago
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dalvyx [7]

Answer:

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Step-by-step explanation:

4 0
3 years ago
Gravel is being dumped from a conveyor belt at a rate of 35 ft3/min, and its coarseness is such that it forms a pile in the shap
Mazyrski [523]

Answer:

  dh/dt ≈ 0.55 ft/min

Step-by-step explanation:

The volume is given by the formula ...

  V = (1/3)πr²h

We have r = h/2, so the volume as a function of height is ...

  V = (1/3)π(h/2)²h = (π/12)h³

Then the rates of change are related by ...

  dV/dt = (π/4)h²·dh/dt

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6 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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