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irina1246 [14]
3 years ago
14

A 24 hour digital watch shows 19:29:00 on its face. The first two digitals adicate

Mathematics
1 answer:
liraira [26]3 years ago
8 0

Answer:

271

Step-by-step explanation:

4 hours and 31 minutes = 240 + 31 = 271 minutes

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*15 points help easy question*
DIA [1.3K]

Answer:

x = 12

Step-by-step explanation:

The segment from the vertex to the base is a perpendicular bisector.

Using Pythagoras' identity on the right triangle on the right.

(\frac{1}{2} x )² + 3² = (\sqrt{45} )² , that is

\frac{1}{4} x² + 9 = 45 ( subtract 9 from both sides )

\frac{1}{4} x² = 36 ( multiply both sides by 4 )

x² = 144 ( take the square root of both sides )

x = \sqrt{144} = 12

8 0
3 years ago
Simplify these expressions: 12x - 3x​
kkurt [141]

Answer:

12x-3x

= 9x

Step-by-step explanation:

Hope it is helpful....

4 0
3 years ago
The GPA you earn in a particular semester is your​ ________ GPA, and your cumulative GPA for all completed semesters is your​ __
Mariulka [41]

Answer:

The GPA you earn in a particular semester is your Marginal GPA, and your cumalative GPA for all completed semesters is your average GPA.

Step-by-step explanation:

8 0
3 years ago
A baby elephant that is 4 ft tall casts a
Kazeer [188]

Answer:

15 ft

Step-by-step explanation:

10/4

=2.5

6*2.5

=15 ft

PLS GIVE BRAINLIEST

7 0
3 years ago
Read 2 more answers
Solve dis attachment and show all work ( I got it all wrong and I want to know how to solve it )
DedPeter [7]
(a) First find the intersections of y=e^{2x-x^2} and y=2:

2=e^{2x-x^2}\implies \ln2=2x-x^2\implies x=1\pm\sqrt{1-\ln2}

So the area of R is given by

\displaystyle\int_{1-\sqrt{1-\ln2}}^{1+\sqrt{1-\ln2}}\left(e^{2x-x^2}-2\right)\,\mathrm dx

If you're not familiar with the error function \mathrm{erf}(x), then you will not be able to find an exact answer. Fortunately, I see this is a question on a calculator based exam, so you can use whatever built-in function you have on your calculator to evaluate the integral. You should get something around 0.5141.

(b) Find the intersections of the line y=1 with y=e^{2x-x^2}.

1=e^{2x-x^2}\implies 0=2x-x^2\implies x=0,x=2

So the area of S is given by

\displaystyle\int_0^{1-\sqrt{1-\ln2}}\left(e^{2x-x^2}-1\right)\,\mathrm dx+\int_{1-\sqrt{1-\ln2}}^{1+\sqrt{1-\ln2}}(2-1)\,\mathrm dx+\int_{1+\sqrt{1-\ln2}}^2\left(e^{2x-x^2}-1\right)\,\mathrm dx
\displaystyle=2\int_0^{1-\sqrt{1-\ln2}}\left(e^{2x-x^2}-1\right)\,\mathrm dx+\int_{1-\sqrt{1-\ln2}}^{1+\sqrt{1-\ln2}}\mathrm dx

which is approximately 1.546.

(c) The easiest method for finding the volume of the solid of revolution is via the disk method. Each cross-section of the solid is a circle with radius perpendicular to the x-axis, determined by the vertical distance from the curve y=e^{2x-x^2} and the line y=1, or e^{2x-x^2}-1. The area of any such circle is \pi times the square of its radius. Since the curve intersects the axis of revolution at x=0 and x=2, the volume would be given by

\displaystyle\pi\int_0^2\left(e^{2x-x^2}-1\right)^2\,\mathrm dx
5 0
3 years ago
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