(1) What's the distance on the map with the actual distance of 450m and the scale of 1:1000000? (2) The actual distance is 400m and the scale is 1:1000000. What is the distance on the map? (3) What is the distance on the map with the actual distance of 600m and scale of 1:1000000?

0.045CM 0.04CM 0.06CM

The perimeter of the rectangle is 242 cm. If the width increases by 25%, the length decreases by one seventh and the perimeter remains unchanged, what is the area of the original rectangle?
Please hurry up

Let the length of the original rectangle be x, then the width is (121-x)
∴25%x=(121-x)/7
7x=484-4x
x=44
121-44=77
The area is 44 × 77 = 3388

The approximate value of 0.00326 is taken by the rounding method, and two significant numbers are retained, and expressed as 0.00326 by scientific notation

The approximate value of 0.00326 is taken according to the rounding method, and the two significant numbers are retained, which are expressed as 3.3 × 10 ^ - 3 by scientific notation

It takes 80 seconds for a train to pass a 600 meter long bridge, and 77 seconds to pass a 543 meter long tunnel at the same speed

Solution
From v = s / T
So s = vt
Then S1 = VT1
S2=Vt2
S1 = s Bridge + L = 600m + L T1 = 80s
S2 = s channel + L = 543m + L T2 = 77s
I.e. 600m + L = 80s V ①
543m+L =77s V …… ②
Solution 1, 2
We get v = 19 m / s
L=920 m
For reference only!

The distance between the corresponding sides of the large and small rectangles is 5cm, and the area between the two rectangles is 1000 square centimeters. What is the perimeter of the large rectangle?
That is to say, the big rectangle sets the small rectangle in it, showing the shape of a circle. The topic requires no equation, no matter whether it is used or not, it will be added. It is better not to use the equation

The area between the two rectangles is 1000 square centimeters, which is actually the side length of the large rectangle minus 5 cm and then multiplied by 5 cm to form four rectangles
Perimeter of large rectangle = 1000 / 5 + 4 * 5 = 220

The P power of 2 plus the P power of 3 is equal to the n power of A. P is a prime number and a is a positive integer

2 ^ P + 3 ^ P = a ^ n (n > 1) the left side of the equation is odd, so a is odd, so that a = 2K + 1. Obviously, when p = 2, n can't be larger than 1. Obviously, a > = 5. K > 1, because when 2 ^ P + 3 ^ PNP > 2: 2 ^ P + 3 ^ P = 2 ^ P + (2 + 1) ^ P, expand (1) 2 ^ P + (2 + 1) ^ P = 2 ^ P + 2 ^ P + p * 2 ^ P (p-1) + p * (p-1) /

There are two mobile phone charging schemes for a telecom company: scheme 1: monthly rent 50 yuan, local call fee 0.40 yuan / min; scheme 2: no monthly rent, local call fee 0.60 yuan / min

Your question reminds me of my exam in middle school
When the call time is 250 minutes
50+0.4×250（＝）0.6×250
When the call time is 300 minutes
50+0.4×300（＜）0.6×300
The first method is more cost-effective
I can't help but miss my campus time again

In rectangular paper ABCD, ab = 3, ad = 5. As shown in the figure, fold the paper so that point a falls at a 'on the edge of BC, and the crease is PQ. When point a' moves on the edge of BC, the end point P.Q of the crease also moves. If the limiting points P and Q move on the edge of AB and ad respectively, the maximum distance that point a 'can move on the edge of BC is ()
A. 1B. 2C. 3D. 4

As shown in Figure 1, when point d coincides with point Q, according to the folding symmetry, we can get a ′ d = ad = 5. In RT △ a ′ CD, a ′ D2 = a ′ C2 + CD2, that is 52 = (5-a ′ b) 2 + 32, we can get a ′ B = 1. As shown in Figure 2, when point P coincides with point B, according to the folding symmetry, we can get a ′ B = AB = 3, ∵ 3-1 = 2, and the maximum distance that a ′ can move on the edge of BC is 2

A pen is 5 yuan, a pencil is 1.5 yuan. Xiao Hong bought 6 pens and pencils with 16 yuan. Guess how many pens she bought each?

If you buy x pencils, there are (6-x) pens
1.5x+5（6-x）=16
14=3.5x
x=4
6-x=2
So I bought four pencils and two pens

Xiaoming walks 78 meters per minute. Xiaoming starts from home at 14 o'clock and walks to school in 15 minutes. The distance from the Grand Theater to school is three times that from Xiaoming's home to school. How far is the distance from Xiaoming's home to the Grand Theater?

78 * 15 is 1170 meters from Xiaoming's home to the school, then 1170 * 3 is 3510, which is the distance to the home. 1170 + 3510 is 4680, which is Xiaoming's home plus the distance from the school to the Grand Theater, which is 4680
1170*3=3510M
3510+1170=4680M

(1) What's the distance on the map with the actual distance of 450m and the scale of 1:1000000? (2) The actual distance is 400m and the scale is 1:1000000. What is the distance on the map? (3) What is the distance on the map with the actual distance of 600m and scale of 1:1000000?