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UkoKoshka [18]
3 years ago
6

Which answer is closest to the true value of the expression:

Mathematics
1 answer:
d1i1m1o1n [39]3 years ago
6 0
(9.1\times10^4)(1.1\times10^{-5})(log10^{-13})(1000)\\\\=(9.1\times1.1)(10^4\times10^{-5}\times10^3)(-13log10)\\\\=10.01\times10^{4-5+3}(-13)\\\\=-130.13\times10^2\\\\=-13013\approx-13000\\\\Answer:(c).

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Which is the most accurate way to estimate 25% of 53?
Alex Ar [27]
You can calculate them out and check which one is the most accurate

so if one of the choices was 1/3 of 50, then you would do:
1/3 in the calculator (should be 0.333...) times 50

And you can compare it with the EXACT number, which is:
0.25*53=13.25

This way you can test all the options
8 0
4 years ago
The curves y = √x and y=(2-x) and the Cartesian axes form two distinct regions in the first quadrant. Find the volumes of rotati
makkiz [27]

Answer:

Step-by-step explanation:

If you graph there would be two different regions. The first one would be

y = \sqrt{x} \,\,\,\,, 0\leq x \leq 1 \\

And the second one would be

y = 2-x \,\,\,\,\,,  1 \leq x \leq 2.

If you rotate the first region around the "y" axis you get that

{\displaystyle A_1 = 2\pi \int\limits_{0}^{1} x\sqrt{x} dx = \frac{4\pi}{5} = 2.51 }

And if you rotate the second region around the "y" axis you get that

{\displaystyle A_2 = 2\pi \int\limits_{1}^{2} x(2-x) dx = \frac{4\pi}{3} = 4.188 }

And the sum would be  2.51+4.188 = 6.698

If you revolve just the outer curve you get

If you rotate the first  region around the x axis you get that

{\displaystyle A_1 =\pi \int\limits_{0}^{1} ( \sqrt{x})^2 dx = \frac{\pi}{2} = 1.5708 }

And if you rotate the second region around the x axis you get that

{\displaystyle A_2 = \pi \int\limits_{1}^{2} (2-x)^2 dx = \frac{\pi}{3} = 1.0472 }

And the sum would be 1.5708+1.0472 = 2.618

7 0
3 years ago
I’ll give you brainliest for the first person with an answer
raketka [301]

Answer:

1296mm^2

Step-by-step explanation:

Surface Area of the Rectangular Prism to the left on top of the box:

Area of square = lw, 9mm * 9mm = 81mm, then multiply by 2 because 2 of the squares are a part of the surface, giving <u>162mm^2</u>

Area of a rectangle =lw, 9mm * 9mm = 81mm, then multiply by 2 again, because 2 rectangles are a part of the surface, giving <u>162mm^2</u>

So, the total surface area of the rectangular prism to the left on top, is 324mm^2

Surface Area of the Triangular Prism:

Area of a triangle: 1/2bh, and our base length would be 12, because we have to subtract 9 from 21 since the base length of the triangle isn't stated. Anyways, A = 1/2bh, so A = 1/2(12)(9) = 54mm^2, but multiply by two, so we get <u>108mm^2</u>.

Area of the rectangle: lw, so 15mm * 9mm = <u>135mm^2</u>

So, the total surface area of the triangular prism is 243mm^2

Surface Area of the Rectangular Prism at the bottom:

Area of the long rectangles in front = lw, 21mm * 9mm = 189mm^2, multiply by 2, <u>378mm^2</u>

Area of the rectangles to the side = lw, 9mm * 9mm = 81mm^2, multiply by 2, <u>162mm^2 </u>

Area of the rectangle at the very bottom = lw, 21mm * 9mm = <u>189mm^2</u>

So, the total surface area of the rectangular prism at the bottom is 729mm^2

Add all the total surface areas of each shape to get the total surface area of the figure:

324mm^2 + 243mm^2 + 729mm^2 = 1296mm^2

The surface area of the figure above is 1296mm^2

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4 0
2 years ago
Which is the completely factored form of 12x^3-60x^2+4x-20​
Shkiper50 [21]

Answer:

(12x²+4)(x-5)

Step-by-step explanation:

To factor the expression factor by grouping.

(12x³ - 60x²) + (4x - 20)

12x²(x - 5) 4(x-5)

(12x²+4)(x-5)

6 0
3 years ago
What is the best result for h?
Kryger [21]

Answer: First option.

Step-by-step explanation:

To solve for "h" from the given the equation S=2\pi rh+2\pi r^2, you need to:

Apply the Subtraction property of equality and subtract 2\pi r^2 to both sides of the equation.

Then you need to apply the Division property of equality and divide both sides of the equation by 2\pi r.

Then:

S-2\pi r^2=2\pi rh+2\pi r^2-2\pi r^2\\\\S-2\pi r^2=2\pi rh\\\\\frac{S}{2\pi r}-\frac{2\pi r^2}{2\pi r}=\frac{2\pi rh}{2\pi r}\\\\\frac{S}{2\pi r}-r=h

6 0
3 years ago
Read 2 more answers
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