El documento critica a quienes culpan al gobierno y a otros por los problemas en lugar de asumir la responsabilidad personal de hacer el cambio, como recoger la basura en la calle en lugar de esperar a que otros lo hagan. También señala que dejar que otros resuelvan los problemas resulta en un sistema estúpido donde nadie asume la responsabilidad.
The document lists key areas of cancer research including cell signaling, DNA repair, cancer stem cells, cancer genomics, cancer metabolomics, cell cycle, cancer therapy, translational cancer research, tumour immunology, cancer epigenomics, and cancer proteomics. It also mentions animal modeling and imaging as areas related to cancer research.
This document discusses developing human potential in nursing administration. It defines human potential as all of one's knowledge, talents, capacities and abilities. It discusses how education can help maximize human potential through lifelong learning. Nursing administrators can foster human potential by encouraging continued learning, matching staff to educational opportunities, and providing a motivating work environment that allows staff to meet intrinsic needs like achievement, responsibility and growth. Theories on motivation, behavior change, and leadership are presented to help administrators develop human potential in nurses.
The document discusses the Professional Practice of Nursing Administration by outlining the goals of the book which are to present nursing administration as part of professional nursing practice, provide an approach to nursing administration that integrates practice, education, and research, and provide a conceptual framework for nursing administration in various settings. The book is intended to provide knowledge about factors that influence nursing administration and forecast emerging trends in professional nursing practice.
El documento critica a quienes culpan al gobierno y a otros por los problemas en lugar de asumir la responsabilidad personal de hacer el cambio, como recoger la basura en la calle en lugar de esperar a que otros lo hagan. También señala que dejar que otros resuelvan los problemas resulta en un sistema estúpido donde nadie asume la responsabilidad.
The document lists key areas of cancer research including cell signaling, DNA repair, cancer stem cells, cancer genomics, cancer metabolomics, cell cycle, cancer therapy, translational cancer research, tumour immunology, cancer epigenomics, and cancer proteomics. It also mentions animal modeling and imaging as areas related to cancer research.
This document discusses developing human potential in nursing administration. It defines human potential as all of one's knowledge, talents, capacities and abilities. It discusses how education can help maximize human potential through lifelong learning. Nursing administrators can foster human potential by encouraging continued learning, matching staff to educational opportunities, and providing a motivating work environment that allows staff to meet intrinsic needs like achievement, responsibility and growth. Theories on motivation, behavior change, and leadership are presented to help administrators develop human potential in nurses.
The document discusses the Professional Practice of Nursing Administration by outlining the goals of the book which are to present nursing administration as part of professional nursing practice, provide an approach to nursing administration that integrates practice, education, and research, and provide a conceptual framework for nursing administration in various settings. The book is intended to provide knowledge about factors that influence nursing administration and forecast emerging trends in professional nursing practice.
This document summarizes details about a property called the Preserve at Rock Creek Inn Site located in Sapphire, North Carolina. The property consists of 10 acres that could be used for an inn surrounded by 65 acres of conservation land for a total of 75 acres. It is listed for $2,000,000 or $26,667 per acre. The property has some existing infrastructure and utilities from previously having 21 beds. It is located directly across the highway from the Burlingame Country Club golf course. The larger Preserve at Rock Creek residential development of 410 acres and 60 lots is located to the west, with 10 lots and 8 homes already sold. Indian Falls, a 70 acre parcel, is located to the east of the
This document discusses decision making as a fundamental component of nursing administration. It describes the decision process and different models for decision making. The prescriptive model involves optimizing decisions by identifying all alternatives and consequences. The descriptive model, called "satisficing", involves finding an acceptable rather than optimal solution due to limitations of time and information. Nurse administrators must balance clinical and organizational priorities in their decision making.
This chapter discusses mobilizing existing nursing resources by assigning nurses according to their level of expertise and experience. It identifies key components that make up nursing resources, explains factors related to recruitment, retention, and turnover, and examines reasons for the perceived nursing shortage. The chapter emphasizes using data to select nursing assignment patterns that meet patient and organizational needs in order to optimize nursing resources.
This document introduces Rosalina Berliani from Pemalang, Indonesia. It provides information about her hobbies, which include listening to music, watching films and anime, and playing with her cats. It also lists the schools she attended, from elementary school through high school in Jepara. Contact information is provided at the end.
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This document discusses various types of addictions including smoking, drugs, alcohol, hookah/shisha, tobacco, and energy drinks. It provides definitions and health effects for each type. Smoking is defined as burning tobacco for inhalation and notes the active substances like nicotine that are absorbed. Drugs are substances that may have medical, intoxicating or performance effects. Alcohol is drinks containing ethanol. Hookah is an instrument for vaporizing flavored tobacco called shisha. Tobacco is processed from dried Nicotiana plants. Energy drinks contain stimulants like caffeine that are marketed as providing mental or physical stimulation. The document encourages saying no to all of these addictions to live a long and healthy life.
1. WWW.VNMATH.COM
SỞ GIÁO DỤC VÀ ĐÀO TẠO KIỂM TRA CHẤT LƯỢNG HỌC KỲ I
ĐỒNG THÁP
WWW.VNMATH.COM Năm học: 2012-2013
Môn thi: TOÁN- Lớp 10
Thời gian: 90 phút (không kể thời gian phát đề)
Ngày thi: 20/12/2012
ĐỀ ĐỀ XUẤT
(Đề gồm có 01 trang)
Đơn vị ra đề: THPT CAO LÃNH 1
Câu 1: (1đ)
Cho các tập hợp:
5
|
x
R
x
A và
7
3
|
x
R
x
B
Tìm B
A
B
A
;
Câu 2: (2,0 điểm)
1.Tìm giao điểm đường thẳng 2
3
:
)
(
x
y
d và parabol 1
4
2
:
)
( 2
x
x
y
P .
2. Xác định hàm số c
bx
ax
y
2
: , biết đồ thị của nó đi qua ba điểm
6
;
1
,
0
;
1
,
2
;
0
C
B
A .
Câu 3: (2đ)
Giải các phương trình
x
x
x
b
x
x
x
x
a
3
2
1
2
/
1
3
3
5
3
2
/
2
Câu 4: (2,0 điểm)
Trong mặt phẳng tọa độ Oxy, cho ba điểm
4
;
3
,
4
;
1
,
1
;
1
C
B
A .
1)Chứng minh rằng ba điểm A, B, C lập thành một tam giác.
2)Chứng minh tam giác ABC là tam giác vuông. Tính chu vi và diện tích của tam giác
ABC
II. PHẦN RIÊNG (3 điểm) (học sinh chọn một trong hai phần sau )
I) Theo chương trình chuẩn
Câu 5a (2,0 điểm)
1) Không dùng máy tính gỉai hệ phương trình.
2 3 4
3 5 5
x y
x y
2) Với mọi a, b, c > 0 Chứng minh:
1 1 1
2
a b c
bc ca ab a b c
Câu 6a (1,0 điểm)
Trong mặt phẳng tọa độ Oxy cho hai điểm A(3; 1), B(4, 2). Tìm tọa độ điểm M sao cho:
AM = 2 và 0
; 135
AB AM
II) Theo chương trình nâng cao
Câu 5b (2,0 điểm)
1) Xác định m để hệ
( 1) 2
( 1) 2
m x y m
mx m y
có nghiệm là (2; yo)
2) Tìm điều kiện của tham số m để pt :(m-1)x2 – 4x + 3 = 0 có 2 nghiệm phân biệt
Câu 6b (1,0 điểm)
Cho tam giác ABC có góc A nhọn ; D và E là 2 điểm nằm ngoài tam giác sao cho ABD và ACE
2. WWW.VNMATH.COM
vuông cân tại A .M là trung điểm BC .Chứng minh AM DE .
____________________ HẾT______________________
SỞ GIÁO DỤC VÀ ĐÀO TẠO KIỂM TRA CHẤT LƯỢNG HỌC KÌ I
ĐỒNG THÁP Năm học: 2012-2013
Môn thi: TOÁN – Lớp 10
HƯỚNG DẪN CHẤM ĐỀ ĐỀ XUẤT
(Hướng dẫn chấm gồm có 03 trang)
Đơn vị ra đề: THPT Cao Lãnh I.
Câu 1
(1,0 đ)
5
|
x
R
x
A
5
;
A 0.25
7
3
|
x
R
x
B
7
;
3
B 0.25
5
;
3
B
A 0.25
7
;
B
A 0.25
Câu 2
1
1.0đ
Hoành độ giao điểm của (P) và (d) là nghiệm của pt:
2
3
1
4
2 2
x
x
x 0.25
0
3
7
2 2
x
x
2
1
3
x
x 0.25
2
1
7
y
y 0.25
Vậy giao điểm cần tìm:
2
1
;
2
1
,
7
;
3
0.25
2
1.0đ
Hàm số qua ba điểm A, B, C nên ta có:
6
0
2
c
b
a
c
b
a
c
4
2
2
b
a
b
a
c
0.25x2
2
3
1
c
b
a
. vậy: 2
3
: 2
x
x
y
0.25x2
CÂU 3 1(đ)
a/ 1
3
3
5
3
2
x
x
x
x
(*)
3
:
x
ĐK 0.25
(*) 9
)
3
).(
3
5
(
)
3
(
2 2
x
x
x
x
x 0.25
0
6
6 2
x
x 0.25
3. WWW.VNMATH.COM
)
(
0
)
(
1
n
x
n
x
0.25
b/ x
x
x 3
2
1
2 2
2
2
3
2
1
4
0
3
2
x
x
x
x
0.25
2
2
9
12
4
1
4
3
2
x
x
x
x
x 0.25
0
16
5
3
2
2
x
x
x 0.25
)
(
5
16
)
(
0
3
2
l
x
n
x
x
0.25
Câu 4 1
1.0đ
)
3
;
4
(
),
3
;
0
(
AC
AB 0.25
3
3
4
0
0.25
AC
AB,
không cùng phương C
B
A ,
,
không thẳng hàng 0.25
Vậy ba điểm A,B,C lập thành một tam giác. 0.25
2
1.0đ
4
,
5
,
3
BC
AC
AB 0.25
Ta có: ABC
AC
BC
AB
2
2
2
25 vuông tại B. 0.25
Chu vi tam giác: 3+5+4=12 0.25
6
.
2
1
BC
AB
SABC
0.25
Câu 5a
1)(1,0đ)
6 9 12
6 10 10
x y
HPT
x y
0.25đ
2
2 3 4
y
x y
0.25đ
2
2 6 4
y
x
0.25đ
5
2
x
y
0.25đ
2)(1,0 đ) 1 1 1
2
a b c
bc ca ab a b c
(1)
2 2 2 1 1 1
2
a b c abc
a b c
0.25đ
4. WWW.VNMATH.COM
2 2 2
2 2 2 0
a b c bc ac ab
0.25đ
2
( ) 0
a b c
: đúng nên (1) đúng 0.25đ
Đẳng thức xãy ra a b c
0.25đ
Câu 6a:
(1,0 đ)
Gọi M( x; y )
(1;1)
( 3; 1)
AB
AM x y
0.25đ
2 2
2 ( 3) ( 1) 4
AM x y
(1) 0.25đ
0 3 1 2
( ; ) 135 2
2
1 1. 4
x y
AB AM x y
Thế vào (1)
0.25đ
2 2
(2 3) ( 1) 4
1 1
1 3
y y
y x
y x
Vậy có hai điểm M1(1; 1) và M2(-1; 3)
0.25đ
Câu 5b(2đ) 1) Hệ có nghiệm là (2; yo )
2 2 2
2 ( 1) 2
o
o
m y m
m m y
0.25
(1)
2 ( 1) 2 (2)
o
o
y m
m m y
0.25
Thế yo = m vào (2) ta được : m2- m – 2 = 0 0.25
Vậy m = - 1 ; m = 2 0.25
2) (m-1)x2 – 4x + 3 = 0 có 2 nghiệm phân biệt
4 3( 1) 0
1 0
m
m
0.5
7
3
1
m
m
0.5
5. WWW.VNMATH.COM
Câu 6b(1đ)
2 ( )( )
AM DE AB AC AE AD
0.25
= AB AE AB AD AC AE AC AD
0.25
= AB AE AC AD
(vì AB AD và AC AE ) 0.25
= AB.AE.cos(90o +A) – AC.AD.cos(90o +A)
= 0 (vì AB.AE = AC.AD) 0.25
Vậy : AM DE