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3 years ago

1) Shannon and Kristoph are dividing numbers written in scientific notation.

Mathematics

1 answer:

mihalych1998 [28]3 years ago

5 0

Answer:

The answers will have the same value, beacuse both scientific notations have the same coefficient, 4.2 and 1.4. Also, if you mentally subtract the exponents of those powers, you would have the same exponent. So, equal coefficient and equal exponent will give the same result.

The value of each expression can be found by dividing coefficients and powers.

$\frac{4.2 \times 10^{9} }{1.4 \times 10^{4} }=3 \times 10^{9-4} = 3 \times 10^{5}$

The value of each expression in standard form is

$3 \times 10^{5}=300,000$

Remember that standard form refers to the number without the ten-power.

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Evaluate triple integral

kaheart [24]

Answer:

$\\ \frac{1}{8} e^{4a}-\frac{3}{4}e^{2a}+e^{a} -\frac{3}{8} \\\\or\\\\ \frac{e^{4a}-6e^{2a}+8e^{a}-3}{8}$

Step-by-step explanation:

$\\ \int\limits^{a}_{0} \int\limits^{x}_{0} \int\limits^{x+y}_{0} {e^{x+y+z}} \, dzdydx \\\\=\int\limits^{a}_{0} \int\limits^{x}_{0} [\int\limits^{x+y}_{0} {e^{x+y}e^z} \, dz]dydx \\\\\\=\int\limits^{a}_{0} \int\limits^{x}_{0} [e^{x+y}\int\limits^{x+y}_{0} {e^z} \, dz]dydx\\\\=\int\limits^{a}_{0} \int\limits^{x}_{0} [e^{x+y}e^z\Big|_0^{x+y}]dydx \\\\\\=\int\limits^{a}_{0} \int\limits^{x}_{0} [e^{x+y}e^{x+y}-e^{x+y}]dydx \\\\\\=\int\limits^{a}_{0} \int\limits^{x}_{0} e^{2x+2y}-e^{x+y}dydx \\\\\\$

$\\=\int\limits^{a}_{0} [\int\limits^{x}_{0} e^{2x}e^{2y}-e^{x+y}dy]dx \\\\\\=\int\limits^{a}_{0} [\int\limits^{x}_{0} e^{2x}e^{2y}dy- \int\limits^{x}_{0}e^{x}e^{y}dy]dx \\\\\\u=2y\\du=2dy\\dy=\frac{1}{2}du\\\\\\=\int\limits^{a}_{0} [\frac{e^{2x}}{2}\int e^{u}du- e^x\int\limits^{x}_{0}e^{y}dy]dx \\\\\\=\int\limits^{a}_{0} [\frac{e^{2x}}{2}\cdot e^{2y}\Big|_0^x- e^xe^{y}\Big|_0^x]dx \\\\\\=\int\limits^{a}_{0} [\frac{e^{2x+2y}}{2} - e^{x+y}\Big|_0^x]dx \\\\$

$\\=\int\limits^{a}_{0} [\frac{e^{4x}}{2} - e^{2x}-\frac{e^{2x}}{2} + e^{x}]dx \\\\\\=\int\limits^{a}_{0} \frac{e^{4x}}{2} -\frac{3e^{2x}}{2} + e^{x}dx \\\\\\=\int\limits^{a}_{0} \frac{e^{4x}}{2}dx -\int\limits^{a}_{0}\frac{3e^{2x}}{2}dx + \int\limits^{a}_{0}e^{x}dx \\\\\\u_1=4x\\du_1=4dx\\dx=\frac{1}{4}du_1\\\\\u_2=2x\\du_2=2dx\\dx=\frac{1}{2}du_2\\\\\\=\frac{1}{8}\int e^{u_1}du_1 -\frac{3}{4}\int e^{u_2}du_2 + \int\limits^{a}_{0}e^{x}dx \\\\\\$

$\\=\frac{1}{8}e^{u_1}\Big| -\frac{3}{4}e^{u_2}\Big| + e^{x}\Big|_0^a \\\\\\=\frac{1}{8}e^{4x}\Big|_{0}^a -\frac{3}{4}e^{2x}\Big|_{0}^a + e^{x}\Big|_0^a \\\\\\=\frac{1}{8}e^{4x} -\frac{3}{4}e^{2x} + e^{x}\Big|_0^a \\\\\\=\frac{1}{8}e^{4a} -\frac{3}{4}e^{2a} + e^{a}-\frac{1}{8} +\frac{3}{4} -1\\\\\\=\frac{1}{8}e^{4a} -\frac{3}{4}e^{2a} + e^{a}-\frac{3}{8}\\\\\\$

Sorry if that took a while to finish. I am in AP Calculus BC and that was my first time evaluating a triple integral. You will see some integrals and evaluation signs with blank upper and lower boundaries. I just had my equation in terms of u and didn't want to get any variables confused. Hope this helps you. If you have any questions let me know. Have a nice night.

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2 years ago

Which inequality is equivalent to |x-4|<9

Lorico [155]

Answer with explanation:

The Given Inequality is

|x-4|<9

Modulus of a number yields always a positive value.

→|x-4|= x-4, when, x-4≥0

That is , x≥4.

= -(x-4), when, x-4<0

means, x<4.

it can be Written as

→ -9<x-4<9

→ -9+4 < x -4+4 < 9+4

→ -5<x<13

→x∈(-5,13)

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3 years ago

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kvv77 [185]

Answer:

V= π r2h

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2 years ago

What is the answer to 8x - 6x = -18

SashulF [63]

8X - 6X = -18

2X = -18

X = -9.

Hope this helps!

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3 years ago

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Joanne buys a car for $1500 after 12 months Joanne is forced to sell the car for 10% less than what she brought it for, how much

MAVERICK [17]

Answer:

$1350

Step-by-step explanation:

Given she sold the car 10% less than what she bought it, amount she sold the car will be 10% of the amount she bought the car subtracted from the amount she bought he car .

She bought the car for $1500

Therefore,

10% of $1500

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0.1 x $1500

$150

Now , amount she sold the car = $1500 - $150

= $1350

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