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3 years ago

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

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

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

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

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

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

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

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

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 162mm^2

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

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 108mm^2.

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

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, 378mm^2

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

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

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

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

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