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Mathematics in the first grade of junior high school A and B climb a mountain. A climbs 10 meters per minute, and starts 30 minutes first. B climbs 15 meters per minute, and they climb the top of the mountain at the same time. How much time does a spend climbing? How high is the mountain?
Let this mountain be x meters high. X / 10-30 = x / 15, x = 900, 900 / 10 = 90 minutes
Question 5 on page 102 of the first volume of junior high school mathematics
A and B climb a mountain. A climbs 10 meters per minute and starts for 30 minutes. B climbs 15 meters per minute. They climb the mountain at the same time. How long does it take a to climb? How high is the mountain?
Set the time to X minutes
10x+30*10=15x
10x+300=15x
5x=300
x=60
Mountain height: 60 * 15 = 900
Seven exercises of solving equations in Mathematics in Volume 1 of junior high school
X is the unknown X
Formula: if you don't understand, look here:
(1) 7x-5 / 4 = 3 / 8 7x of 4 minus 5 = 3 of 8
(2) 2x-1 / 6 = 5x + 1 / 8 2x of 6 minus 1 = 5x of 8 plus 1
(3) 1 / 2x-7 = 9x-2 / 6 1 / 2 times x minus 7 = 9x minus 2 / 6
(4) 1 / 5x-1 / 2 (3-2x) = 1 / 5 times x minus 1 / 2 times bracket 3-2x, that is, (3 minus 2x) = 1
(5) 2X + 1 / 3-5x-1 / 6 = 1 / 3 of 2x plus 1 minus 6 of 5x minus 1 = 1
(6) 1 / 7 (2x + 14) = 4-2x 1 / 7 times parentheses 2x + 14, that is (2x + 14) = 4 minus 2x
(7) 3 / 10 (200 + x) - 2 / 10 (300-x) = 300x9 / 25 3 / 10 times bracket 200 plus x, that is (200 plus x) minus 2 / 10 times bracket 300 minus x, that is (300 minus x) = 300 times 9 / 25
(1)、2(7x-5)=314x=13X=13/14(2)、4(2x-1)=3(5x+1)-7x=7X=-1(3)、3x-42=9x-2-6x=40X=-20/3(4)、2x-5(3-2x)=102x-15+10x=1012x=25X=25/12(5)、2(2x+1)-(5x-1)=64x+2-5x+1=6-x=3X=-3(6)、2x+14=28-14x16x=14X=8/7(7)、...
As shown in the figure, the circumference of the known quadrilateral ABCD is 30cm, AE ⊥ BC and point E, AF ⊥ CD are at point F, and AE: AF = 2:3, ∠ C = 120 °,
Can I find the area of parallelogram ABCD without trigonometric function?
No, I'm afraid not
Area s = AE * BC = CD * AF ∵ AE ∶ AF = 2 ∶ 3 ∵ BC: CD = 3:2
And ∵ the circumference of ABCD is 30cm ∵ 2 (BC + CD) = 30 ∵ ad = BC = 9cm AB = CD = 6cm
And ∵ - C = 120 ° so ∵ - B = 60 °, AE = AB * sin60 ° = 3 √ 3
Area s = AE * BC = 27 √ 3cm & # 178;
How to calculate 3 / 17 × 5 / 37 + 5 / 17 × 34 / 37? The formula is clear·····
3/17×5/37+5/17×34/37
=5/17×3/37+5/17×34/37
=5/17×(3/37+34/37)
=5/17×1
=5/17
How to write? And Qin Guan's huanxisha?
Bu Suan Zi Yong Mei
Wind and rain send spring home
Spring is coming with flying snow
It's a cliff full of ice
There are still beautiful flowers
Beauty does not fight for spring
Just spring
When the flowers are blooming
She was laughing in the bush
Silk-Washing Stream
Yan Shu
A new song, a cup of wine,
Last year, the weather was bad for the old Pavilion
When will the sun set?
I can't help it,
A familiar bird returns
The fragrant path of the small garden wanders alone
As shown in the figure, the known point F is the midpoint of the side BC of the square ABCD, CG bisects ∠ DCE, GF ⊥ AF
It is proved that: take the midpoint m of AB, connect FM. ∵ point F is the midpoint of the side BC of the square ABCD, ∵ BF = BM, ∵ BMF = 45 °, ∵ AMF = 135 °. ∵ CG bisection ∵ DCE, ∵ GCE = 45 °, ∵ FCG = 135 °, ∵ AMF = ∵ FCG. ∵ B = 90 °, ∵ fam = 90 ° - ∵ AFB, ∵ GF ⊥ AF, ∵
Simple calculation of (12 + 4 / 5) × (6 + 2 / 5) + (8 + 3 / 25) × 64
（12+0.8）×（6+0.4）+（8+0.12）×64
=12.8×6.4+8.12×64
=12.8×6.4+81.2×6.4
=6.4×（12.8+81.2）
=6.4×94
=601.6
Why is the open way of quadratic radical nonnegative
In the field of real number, the number of even radical's root must be nonnegative, otherwise it is meaningless
When extended to the complex number field (including real and imaginary numbers), because the square of the imaginary number unit I = - 1, it is still meaningful to open even root. For example, the root sign - 1 = positive and negative I
We can't work out the formula here, so we have to replace it with words in some places
As shown in the figure, let m and n be the midpoint of AD and CB at the waist of rectangular trapezoid ABCD respectively, AB on De is at point E, and △ ade is folded along De, and m and N coincide exactly, then AE: be is equal to ()
A. 2：1B. 1：2C. 3：2D. 2：3
Let de and Mn intersect at point F, ∵ m and n be the midpoint of AD and CB respectively, ∵ Mn ∥ AB, and ∵ m be the midpoint of AD, ∵ MF = 12ae, and ∵ m and N coincide, ∵ NF = be, MF = NF, ∵ AE: be = 2mf: NF = 2:1, so a
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