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Greeley [361]
3 years ago
7

(6 x 10^2) divide (3 x 10^-5)

Mathematics
2 answers:
Hunter-Best [27]3 years ago
7 0

Answer:

=20000000 (let me know if it's incorrect)

Step-by-step explanation:

LenKa [72]3 years ago
6 0

Answer:

20,000,000

Step-by-step explanation:

(6 x 10^2) = 600

(3 x 10^(-5)) = 0.00003

600 divided by 0.00003 =

20,000,000

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Would these be linear combinations or quadratic equations
Komok [63]

Answer:

actually it is a simultaneous equation from linear combinations

5 0
3 years ago
To which interval would we restrict f(x) = cos (x-π/4) so that f(x) is invertible?
nadezda [96]

Answer:

The range of x values for which y is unique is 2·π

Step-by-step explanation:

For a function j: X → Y to be invertible, we have that for every y in Y, there is associated only one x which is an element of x

Hence, f(x) = cos(x - π/4) gives

the x intercept at two penultimate points of the graph of cos(x - π/4) are;

x = 2.36, and x = 8.64

x = 3/4·π, and x = 2.75·π = 2\tfrac{3}{4}\cdot \pi

Hence the range of x values for which y is unique is presented as follows

2\tfrac{3}{4}\cdot \pi - \frac{3}{4}\cdot \pi = 2}\cdot \pi

The range of x values for which y is unique = 2·π.

4 0
3 years ago
A. For 3 days in July the temperatures
ASHA 777 [7]

la temperatura promedio fue de 103°

4 0
3 years ago
An equation ___ has one solution <br><br> A. Always <br> B. Sometimes <br> C. Never
Doss [256]

im am pretty sure it is b.sometimes because an equation can have more then one answer

4 0
3 years ago
The perimeter of the trapezoid-shaped window frame is 23.59 feet. Write and solve an equation to find the unknown side length x
krok68 [10]

Answer:

5.62 + 3.65 + 5.62 + x = 23.59

x = 8.7 ft

Step-by-step explanation:

Perimeter of the trapezoid-shaped window = sum of all the sides of the window frame

Thus, the equation to find the unknown side length, x, would be:

✔️5.62 + 3.65 + 5.62 + x = 23.59

Solve for x

14.89 + x = 23.59

Subtract 14.89 from each side

14.89 + x - 14.89 = 23.59 - 14.89

✔️x = 8.7 ft

4 0
3 years ago
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