Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
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Gọi \(x\) là số học sinh giỏi lớp 5A. Theo đề bài ta có:
\(x\) x \(\dfrac{3}{5}\) = 6 ⇒ \(x\) = 6 x \(\dfrac{5}{3}\) = 10 (học sinh)
Vậy lớp 5A có 10 học sinh giỏi
Số học sinh lớp 5A là:
6 : \(\dfrac{3}{5}\)
= 10 ( em)
Đ/S: 10 em
trọng lượng hòn gạch : P=10m=10.2=20N
trọng lượng hòn đá : P=10m=10.10=100N .
vậy hòn đá có trọng lượng lớn hơn .
Vì 2 kg < 10 kg => Hoàn gạch có khối lượng 2 kg nhẹ hơn hòn đá có khối lượng 10 kg
24 - 16(x - 1/2) = 23
=> 16(x - 1/2) = 24 - 23
=> 16(x - 1/2) = 1
=> x - 1/2 = 1/16
=> x = 1/16 + 1/2
=> x = 9/16
\(24-16(x-\frac{1}{2})=23\)
\(16(x-\frac{1}{2})=24-23\)
\(16(x-\frac{1}{2})=1\)
\(x-\frac{1}{2}=\frac{1}{16}\)
\(x=\frac{1}{16}+\frac{1}{2}\)
\(x=\frac{9}{16}\)
Vậy số thực x cần tìm là \(\frac{9}{16}\)
Chúc bạn hok tốt ~
a: Xét tứ giác ABDC có
M là trung điểm của BC
M là trung điểm của AD
Do đó:ABDC là hình bình hành
mà \(\widehat{BAC}=90^0\)
nên ABDC là hình chữ nhật
b: Xét tứ giác AMCE có
AM//CE
AE//CM
Do đó:AMCE là hình bình hành
mà MA=MC
nên AMCE là hình thoi
cảm ơn em trai nha, hồi đó mà có người trả lời thì tốt :v