пожалуйста помогите решить что-нибудь! алгебра 9 класс (фото). Created by Zhanna0405. algebra-ru

1.2.a)б)3.Что и требовалось доказать.4.α - угол 3-ей четверти.sinα и cosα имеют знак "-".не подходит5*.6*.7*.x - золотые медали в прошлом годуу - серебряные медали в прошлом годух+у=12020%=0,2х+0,2х=1,2х - золотые медали в этом годуу-0,2у=0,8у - серебряные медали в этом году1,2х+0,8у=120-101,2х+0,8у=110{x+y=120 | умножим на "-1,2"{1.2x+0.8y=110{-1.2x-1.2y=-144{1.2x+0.8y=110Складываем уравнения системы:-1.2y+0.8y=-144+110-0.4y=-34y=85x=120-85=351.2*35=42 - золотые медали в этом году0,8*85=68 - серебряные медали в этом годуОтвет: 42 золотые и 68 серебряных медалей.Новые вопросы в Алгебра52-0= (-61) - (- 12) = 66-76= 85 - (-16) = 95-45= 10-5= (-66) - (-38) = 55 - (-18) = (-31) - 52= (-24)-(-41) = (-84) + 75 =90+ (-77) = (-46) - 69 = (-25) - (-70) = (-6) - (-85) = 18+ (-68) = (-86) + 71= (-82) - (-58) = (-81) + 70 = (-16) + (-38) = 49- (-31) = 89-60= 85-26=…0 + (-3) = 8- (-98) = 14+ (-19) = 82 - 18 = 94 + 12 = 53 + 12 = (-81) + 99 = (-59) + 8 = 77- (-23) = 53 + (-73) =РЕШЕНИЕ СРОЧНО!!!!!!!!РЕШЕНИЕ ПОЖАЛУЙСТА!!!!!!!!Допоможіть будь ласка, це срочно. Даю 35 балів Доказать что : фото снизуУпростить выражениеа+в+а^2-в^2 розкласти на множникивычислить 4 корень 0,4^4 умножить5^8Помогите решить это вычеслениеПредыдущийСледующийВместе мы знаем большеЭтот сайт использует cookies Политика Cookies . 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пожалуйста помогите решить что-нибудь! алгебра 9 класс (фото)...

Zhanna0405 @Zhanna0405

August 2022 1 5 Report

пожалуйста помогите решить что-нибудь! алгебра 9 класс (фото)

Answers & Comments

m11m

Verified answer

1.
$2cos \frac{3 \pi }{4}+tg \frac{ \pi }{3}=2cos( \pi - \frac{ \pi }{4} )+ \sqrt{3}= \\ \\ 
=-2cos \frac{ \pi }{4}+ \sqrt{3}= -2* \frac{ \sqrt{2} }{2}+ \sqrt{3}= \sqrt{3}- \sqrt{2}$

2.
a)
$\frac{1}{tg^2 \alpha +1}+sin^2 \alpha =cos^2 \alpha +sin^2 \alpha =1$
б)
$sin( \frac{ \pi }{2}- \alpha )-cos( \frac{3 \pi }{2}- \alpha )+sin(- \alpha )-cos(- \alpha )= \\ 
=cos \alpha -(-sin \alpha )-sin \alpha -cos \alpha = \\ 
=cos \alpha +sin \alpha -sin \alpha -cos \alpha =0$

3.
$\frac{1}{1-sin \alpha }+ \frac{1}{1+sin \alpha }-2= \frac{1+sin \alpha +1-sin \alpha -2(1-sin \alpha )(1+sin \alpha )}{(1-sin \alpha )(1+sin \alpha )}= \\ \\ 
= \frac{2-2(1-sin^2 \alpha )}{1-sin^2 \alpha }= \frac{2-2cos^2 \alpha }{cos^2 \alpha }= \frac{2(1-cos^2 \alpha )}{cos^2 \alpha }= \frac{2sin^2 \alpha }{cos^2 \alpha }=2tg^2 \alpha \\ \\ 
2tg^2 \alpha =2tg^2 \alpha$
Что и требовалось доказать.

4.
α - угол 3-ей четверти.
sinα и cosα имеют знак "-".

$tg^2 \alpha +1= \frac{1}{cos^2 \alpha } \\ \\ 
( \sqrt{3} )^2+1= \frac{1}{cos^2 \alpha } \\ \\ 
cos^2 \alpha = \frac{1}{4} \\ \\ \\[tex]sin^2 \alpha =1-cos^2 \alpha =1- \frac{1}{4}= \frac{3}{4}$

$sin \alpha = \frac{ \sqrt{3} }{2}$
не подходит

$sin \alpha =- \frac{ \sqrt{3} }{2}$

5*.
$-1 \leq \frac{ \sqrt{2} }{2}sin \alpha - \frac{ \sqrt{2} }{2}cos \alpha \leq 1 \\ \\ 
-1 \leq cos \frac{ \pi }{4}sin \alpha -sin \frac{ \pi }{4}cos \alpha \leq 1 \\ \\ 
-1 \leq sin( \alpha - \frac{ \pi }{4} ) \leq 1$

6*.
$tg \alpha =2 \\ 
tg(2* \frac{ \alpha }{2} )=2 \\ \\ 
 \frac{2tg \frac{ \alpha }{2} }{1-tg^2 \frac{ \alpha }{2} }=2 \\ \\ 
2tg \frac{ \alpha }{2}=2(1-tg^2 \frac{ \alpha }{2} ) \\ \\ 
tg \frac{ \alpha }{2}=1-tg^2 \frac{ \alpha }{2}$

$\frac{3sin \alpha +4cos \alpha }{5sin \alpha -3cos \alpha }= \frac{ \frac{3*2tg \frac{ \alpha }{2} }{1+tg^2 \frac{ \alpha }{2} }+ \frac{4(1-tg^2 \frac{ \alpha }{2} )}{1+tg^2 \frac{ \alpha }{2} } }{ \frac{5*2tg \frac{ \alpha }{2} }{1+tg^2 \frac{ \alpha }{2} } - \frac{3(1-tg^2 \frac{ \alpha }{2} )}{1+tg^2 \frac{ \alpha }{2} } }= \frac{6(1-tg^2 \frac{ \alpha }{2} )+4(1-tg^2 \frac{ \alpha }{2} )}{10(1-tg^2 \frac{ \alpha }{2} )-3(1-tg^2 \frac{ \alpha }{2}) }= \\ \\ \\ 
$
$= \frac{10(1-tg^2 \frac{ \alpha }{2} )}{7(1-tg^2 \frac{ \alpha }{2} )}= \frac{10}{7}$

7*.
x - золотые медали в прошлом году
у - серебряные медали в прошлом году
х+у=120

20%=0,2
х+0,2х=1,2х - золотые медали в этом году
у-0,2у=0,8у - серебряные медали в этом году
1,2х+0,8у=120-10
1,2х+0,8у=110

{x+y=120 | умножим на "-1,2"
{1.2x+0.8y=110

{-1.2x-1.2y=-144
{1.2x+0.8y=110

Складываем уравнения системы:
-1.2y+0.8y=-144+110
-0.4y=-34
y=85
x=120-85=35

1.2*35=42 - золотые медали в этом году
0,8*85=68 - серебряные медали в этом году

Ответ: 42 золотые и 68 серебряных медалей.

Новые вопросы в Алгебра

52-0= (-61) - (- 12) = 66-76= 85 - (-16) = 95-45= 10-5= (-66) - (-38) = 55 - (-18) = (-31) - 52= (-24)-(-41) = (-84) + 75 =

90+ (-77) = (-46) - 69 = (-25) - (-70) = (-6) - (-85) = 18+ (-68) = (-86) + 71= (-82) - (-58) = (-81) + 70 = (-16) + (-38) = 49- (-31) = 89-60= 85-26=…0 + (-3) = 8- (-98) = 14+ (-19) = 82 - 18 = 94 + 12 = 53 + 12 = (-81) + 99 = (-59) + 8 = 77- (-23) = 53 + (-73) =

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