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Lisa [10]
2 years ago
13

If 4.2km is 42cm on a map, what was the scale of the map?

Mathematics
2 answers:
Inessa05 [86]2 years ago
8 0

Answer: I need da points

Step-by-step explanation: I’m just like that sry

serious [3.7K]2 years ago
6 0

Answer:

Step-by-step explanation:

njjvsiljsdkjgklsnmxl kcml;kfjms;lf

hope this helped

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Can u guys help me with #1
Furkat [3]

Answer: 4 1/2 cups


Step-by-step explanation: Half a cup per flower, one cup per 2 flowers. 2x4 = 8 flowers so thats 4 cups plus the half a cup

5 0
3 years ago
A) Find a polynomial, which, when added to the polynomial 5x2–3x–9, is equivalent to: 18
VMariaS [17]

Answer:

a) -5x^{2}+3x+27

b) -5x^{2}+3x+9

Step-by-step explanation:

a) Let the required polynomial be p(x).

We have the relation, 5x^{2}-3x-9 + p(x) = 18

i.e. p(x) = 18 -5x^{2}+3x+9

i.e. p(x) = -5x^{2}+3x+27


b) Let the required polynomial be q(x).

We have the relation, 5x^{2}-3x-9 + q(x) = 0

i.e. q(x) = 0 -5x^{2}+3x+9

i.e. q(x) = -5x^{2}+3x+9

3 0
3 years ago
Read 2 more answers
Which of the following questions describes the equation = a/4 -20?
Gekata [30.6K]
1) Negative five -5 x 4 = -20 2) Negative eighty -80 / -4 = 20 3) Negative 1/5 (-1/5) x (-20) = 4 4) Negative eighty (-80) / 4 = -20
6 0
2 years ago
Can you plz answer this for 10pt
Harrizon [31]
Round to the nearest thousands

8,276 = 8,000
2,451 = 2,000

8,000 + 2,000 = 10,000

answer
Maya traveled by plane about 10.000 miles
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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