A problem of filling in the blanks in the first year of junior high school The lengths of the two sides are 5 and 8 respectively Then the value range of the third side is () I hope I can do it right, because the answer was wrong when I asked last time ~! Please tell me the process. I remember a theorem that the third side is greater than or less than Please say it! And fill it in!

A problem of filling in the blanks in the first year of junior high school The lengths of the two sides are 5 and 8 respectively Then the value range of the third side is () I hope I can do it right, because the answer was wrong when I asked last time ~! Please tell me the process. I remember a theorem that the third side is greater than or less than Please say it! And fill it in!

The sum of the two sides of the triangle is greater than the third side, and the difference between the two sides is less than the third side, so 8-5

Junior one mathematics, a fill in the blank problem!
If the solution set of (a + B + C) (a + B + C) ^ 2 is______ .

x

Ask four questions to fill in the blanks in the first year of junior high school~
1. Write a polynomial with degree 2, number of terms 3 and constant - 1
2. Given ∠ 1 = 71 ° 30 ', then the complementary angle of ∠ 1 is equal to () °
3. Simplification | B-A | - a = ()
4. The radius of the sun is about 69660km, which is about () m by scientific counting

1.3x²-1
2.108°30′
3. When B is greater than a, b-2a, when B is less than a, - B, when B is equal to a, - A
The 7th power of 4.6.966 * 10

How many microamperes is one ampere

1A (a) = 1000mA (MA) = 1000000 μ a (MA)

On Newton's second law
The pulling force F acts on the object with the weight of G to make it move at a constant speed along the horizontal ground. As shown in the figure, if the dynamic friction coefficient between the object and the ground is μ, what is the angle θ between the force and the ground when the pulling force is minimum
When θ = arctan μ, f is the smallest

Using orthogonal decomposition method: because the object is moving at a uniform speed, the resultant force is zero
Horizontal direction fcos θ = μ FN
Vertical direction fsin θ + FN = g
We obtain that f = μ g / (COS θ + μ sin θ) when the denominator is the largest, f is the smallest. The denominator = (1 + μ＾ 2) square root * sin (θ + a)
Among them, angle a satisfies Sina = 1 / (1 + μ＾ 2) square root, cosa = μ / (1 + μ＾ 2) square root
Obviously, when sin (θ + a) = 1, f is the smallest, then Tana = 1 / μ, θ = 90 degrees - arctan1 / μ = arctan μ

If the circumference of an isosceles triangle is 20cm and one side is 5cm, then the length of the other two sides is 5cm______ ．

∵ when the waist is 5cm, the length of the bottom edge is 20-5-5 = 10cm, and ∵ 5 + 5 = 10, so it can not form a triangle; when the bottom is 5cm, the waist of the triangle = (20-5) △ 2 = 7.5cm, and the length of the other two sides is 7.5cm, 7.5cm

A hollow copper ball with a volume of 1dm & # 179; is hung on a spring scale. When it is immersed in water and at rest, the indication of the spring scale is 5 / 6 of that when it is weighed in air,
(1) What is the buoyancy of the copper ball?
(2) What is the gravity of a copper ball?
(3) What is the volume of the hollow part of the copper ball?
We need a specific process
The hollow copper ball with the volume of 1dm & # 179; is hung on the spring scale. When it is immersed in water and still, the indication of the spring scale is 5 / 6 of that when it is weighed in air. The known density of copper is 9.0x10 & # 179; kg / M & # 179; [g is 10N / kg]
Ask for:
(1) What is the buoyancy of the copper ball?
(2) What is the gravity of a copper ball?
(3) What is the volume of the hollow part of the copper ball?
We need a specific process

(1) F floating = ρ water GV row = 10 & # 179; kg / M & # 179; × 10N / kg × 0.001M & # 179; = 10N (2) let the indication of the spring scale placed in the air be x n. then 5 / 6 × x = x-10n-1 / 6 × x = - 10nx = 60N ∵ two force balance ∵ g copper = f pull = 60N (3) V copper = m copper / ρ copper = (g copper / g) △ ρ copper = (60N / 10N

There are three simple operations in grade four of primary school, 100000 urgent!
（1）0.1 + 0.4 + 0.7 + 1 …… + 3.7 + 4
PS. simple calculation, earth equation
（2）1.05 + 0.97 + 0.91 + 1.14 + 0.98 + 0.95
PS. simple calculation, earth equation
（3）（ 36.51 - 15.93 ）- （ 4.07 - 3.49 ）
PS. simple calculation, earth equation

First question:
=（0.1+4）×7
=28.7
Second question:
=2+1+1+1+1
=6
Third question:
=（36.51+3.49）—（15.93+4.07）
=40—20
=20

A large disk with radius r rotates at angular velocity W. someone stands on the edge of the disk and rotates with the disk. He wants to shoot the target o in the center of the disk. If the velocity of the bullet is V, then ()
A. A bullet aimed at o must not hit the target
B. The bullet can hit the target only when its firing direction deviates a proper angle to the left of Po
C. The bullet can hit the target only when it is fired at a proper angle to the right of Po
D. The bullet is aimed at O and may hit the target
Excuse me, why can't we turn right? Isn't there a centripetal force? Isn't it possible to hit o?

It has nothing to do with the centripetal force. It's the combination of velocity, a velocity that deviates to the left, and the velocity in the tangent direction of the disk, which is the combination of velocity to o point

The first n terms and Sn, A2 + A7 = - 23, S10 = - 145 of the arithmetic sequence {an} are known
(1) The general term formula of the sequence {an}
(2) Let {BN / an} be an equal ratio sequence with the first term of 1 and the common ratio of C. find the first n terms and TN of {BN}

1
Because of the arithmetic sequence
a2+a7 +2d = a2 +a9 = a1+ a10
s10= 5*(a1 +a10 ) = 5(a2+a7 +2d) = -23*5 +5d
-23*5 +10d = -145
10d = -145 +115
d=-3
a2+a7 = a1+d + a1+6d = 2a1 +7d
2a1 + 7 * (-3) = -23
a1 = -1
The general term formula of sequence {an} an = - 1 + (n-1) (- 3) = - 3N + 2
two
bn/an = 1 *C ^(n-1)
bn = C^(n-1) * (-3n +2)
b1 = C^0 * (-3*1 +2) = -1
Finding the first n terms and TN of sequence {BN}
CTn - Tn = C(b1+b2+...+bn) -(b1+b2+...+bn)
= -b1 + (-d) (C + C^2 + ...+ C^(n-1)) + C bn
= -(-1) + 3*(C-C^n)/(1-C) +C^n* (-3n +2)
= (-3n +2)*C^n + 3*(C-C^n)/(1-C) +1