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abruzzese [7]
3 years ago
14

If 9 is highlighted to b the place value wat place were do the 9 sit at 50,907,652

Mathematics
1 answer:
Zinaida [17]3 years ago
3 0
The nine would be set in the hundred-thousand place.
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Why do you add a zero when multiplying two digits by two digits
malfutka [58]
When you add a zero to the problem, it shows you are now multiplying in the double-digits. Let's say you are multiplying 12x12. First you cover up the one then multiply them. Then cover up the two and you have 10. When you multiply them, it ends in a zero. Don't think of it as adding a zero but adding to products.
6 0
3 years ago
Please help I need the answer ASAP!
stepladder [879]

Answer:

the second answer choice x greater than or equal to -6

Step-by-step explanation:

because u have to find the where the equation is defined. the range is the set  of values that correspond to the domain.

6 0
3 years ago
I need help with all the ones that don't have answers.
zlopas [31]
<em>- What two consecutive odd integers have a sum of 48</em>?

23 + 25 = 48

- <em>Two negative consecutive integers have a sum of -45. What are the integers?
</em>
<em />-22 + -23

- <em>The sum of two consecutive integers 75. What are the two integers?</em>

37 + 38 = 75

- <em>What <u>three</u> consecutive odd  integers have a sum of 81?
</em>
<em />25 + 27 + 29 = 81

Hope I could help! Have a good one. I believe that is all of the unanswered questions. If I missed one, let me know!<em />
7 0
3 years ago
Someone please help me answer this!!
dybincka [34]

Answer:

17.9 units is the answer

Step-by-step explanation:

points are (-7,-2) and (9,6)

Pythagorean theorem can be written as,

d=\sqrt{(x_2-x_1)^2 + (y_2-y_1)^2} \\\\d=\sqrt{9-(-7))^2 + (6-(-2))^2}\\\\d=\sqrt{(16)^2+(8)^2} \\\\d=\sqrt{256+64} \\\\d=\sqrt{320} \\\\d=17.88

Rounding off,

d = 17.9 units

7 0
3 years ago
Determine (without solving the problem) an interval in which the solution of the given initial value problem is certain to exist
yuradex [85]

Answer:

0 < t < 5 is the required interval for the differential equation (t - 5)y' + (ln t)y = 6t to have a solution.

Step-by-step explanation:

Given the differential equation

(t - 5)y' + (ln t)y = 6t

and the condition y(1) = 6

We can rewrite the differential equation by dividing it by (t - 5) as

y' + [(ln t)/(t - 5)]y = 6t/(t - 5)

(ln t)/(t - 5) is continuous on the interval (0, 5) and (5, +infinity).

6t/(t - 5) is continuous on (-infinity, 5) and (5, +infinity)

We see that for these expressions, we have continuity at the intervals (0, 5) and (5, +infinity).

But the initial condition is y = 6, when t = 1.

The solution to differential equation is certain to exist at (0, 5)

Which implies that

0 < t < 5

is the required interval.

3 0
3 years ago
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