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2 years ago

Is 8/10 mile longer than 1/2

2 answers:

6 0

Yes because if you find common denominators (in this case, "10") you would have to compare 8/10 to 5/10. 8>5 so 8/10 is longer than 1/2.

Blizzard [7]2 years ago

3 0

Yes, 8/10 mile is longer than 1/2 because you can convert 1/2 into a half of 10, 5/10, and compare the fractions, since 8 is greater than 5, 8/10 mile is longer.

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How many real solutions are there for f(x) = (x + 3)^2 – 8?

Molodets [167]

There are 2 solutions. the highest degree of a polynomial is 2

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2 years ago

Bob does 10 miles on his bike every day. If he continues to ride his bike at this pace, how many miles will he go in h days?

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3 years ago

PQRS is a parallelogram. Find the m

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2 years ago

Read 2 more answers

MARK AS BRAINLIEST

Arisa [49]

Answer:

The answer to your question is 25%

Step-by-step explanation:

Data

Initial price = $2.00

Final price = $2.50

First we subtract the final price to the inicial price

increment = 2.50 - 2.00

= $0.50 It's the amount that the school increases to the community members

Now we use a rule of three to find the percent

$2.00 ---------------------- 100%

$0.50 --------------------- x

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4 0

3 years ago

The radius of a spherical balloon is measured as 20 inches, with a possible error of 0.03 inch. Use differentials to approximate

soldi70 [24.7K]

Answer:

a) $V = 33510.322\,in^{3}$ , b) $A_{s} = 5026.548\,in^{2}$ , c) $\% V = 0.450\,\%$ , $\%A_{s} = 0.300\,\%$ .

Step-by-step explanation:

The volume and the surface area of the sphere are, respectively:

$V = \frac{4}{3}\pi \cdot r^{3}$

$A_{s} = 4\pi \cdot r^{2}$

a) The volume of the sphere is:

$V = \frac{4}{3}\pi \cdot (20\,in)^{3}$

$V = 33510.322\,in^{3}$

b) The surface area of the sphere is:

$A_{s} = 4\pi \cdot (20\,in)^{2}$

$A_{s} = 5026.548\,in^{2}$

c) The total differentials for volume and surface area of the sphere are, respectively:

$\Delta V = 4\pi\cdot r^{2}\,\Delta r$

$\Delta V = 4\pi \cdot (20\,in)^{2}\cdot (0.03\,in)$

$\Delta V = 150.796\,in^{3}$

$\Delta A_{s} = 8\pi\cdot r \,\Delta r$

$\Delta A_{s} = 8\pi \cdot (20\,in)\cdot (0.03\,in)$

$\Delta A_{s} = 15.080\,in^{2}$

Relative errors are presented hereafter:

$\%V = \frac{\Delta V}{V}\times 100\%$

$\%V = \frac{150.796 \,in^{3}}{33510.322\,in^{3}}\times 100\,\%$

$\% V = 0.450\,\%$

$\% A_{s} = \frac{\Delta A_{s}}{A_{s}}\times 100\,\%$

$\% A_{s} = \frac{15.080\,in^{2}}{5026.548\,in^{2}}\times 100\,\%$

$\%A_{s} = 0.300\,\%$

4 0

2 years ago