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4 years ago

Calcule senx, sabendo que cosx= -√3/3 e π

1 answer:

8 0

Sendo sen x= -3/5 e π < x < 3π/2, calcule:? ... Pedem-se várias funções trigonométricas,sabendo-se que senx = -3/5 e que "x' está no seguinte intervalo: PI < x < 3PI/2 ---Veja que esse ... cosx = +-4/5 -----mas, como no 3º quadrante o cosseno é negativo, então temos que: .... Se cos x = 3/5 e 3π/2 < x < 2π , calcular sen(3x)?.

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Which of the following always passes through a vertex of a triangle . choose all that apply.

Helga [31]

Just finished the test and the write answers are :

1) A,D,E

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3)B,E

4) D

5) A

6) C

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9) part A is C

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7 0

3 years ago

the scale of a model railroad set is 3.5 millimeters to one foot if the length of a model car in the set is 150 millimeters what

Montano1993 [528]

$\bf \begin{array}{ccll} \stackrel{mm}{model}&\stackrel{ft}{actual}\\ \cline{1-2} 3.5&1\\ 150&x \end{array}\implies \cfrac{3.5}{150}=\cfrac{1}{x}\implies x=\cfrac{150\cdot 1}{3.5}\implies x\approx 42.86$

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4 years ago

Read 2 more answers

Mandy can buy 4 containers of yogurt and 3 boxes of crackers for $9.55. She can buy 2 containers of yogurt and 2 boxes of cracke

Anna11 [10]

Answer:

The cost of a box of crackers is $2.25.

Step-by-step explanation:

Let the cost of a containers of yogurt is $x and cost of a box of crackers is $y.

It is given that Mandy can buy 4 containers of yogurt and 3 boxes of crackers for $9.55.

$4x+3y=9.55$ .... (1)

She can buy 2 containers of yogurt and 2 boxes of crackers for $5.90.

$2x+2y=5.90$ .... (2)

On solving (1) and (2), using graphing calculator, we get

$x=0.7$

$y=2.25$

Therefore the cost of a box of crackers is $2.25.

6 0

3 years ago

Read 2 more answers

What is the result of substituting for y in the bottom equation y=x+3 y=x^2+2x-4

8_murik_8 [283]

The solutions are (2.1925, 5.1925) and (-3.1925, -0.1925)

Solution:

Given that,

$y = x + 3 ------- eqn 1\\\\y = x^2 + 2x - 4 ----- eqn 2$

We have to substitute eqn 1 in eqn 2

$x + 3 = x^2 + 2x - 4$

$\mathrm{Switch\:sides}\\\\x^2+2x-4=x+3\\\\\mathrm{Subtract\:}3\mathrm{\:from\:both\:sides}\\\\x^2+2x-4-3=x+3-3\\\\\mathrm{Simplify}\\\\x^2+2x-7=x\\\\\mathrm{Subtract\:}x\mathrm{\:from\:both\:sides}\\\\x^2+2x-7-x=x-x\\\\\mathrm{Simplify}\\\\x^2+x-7=0$

$\mathrm{Solve\:with\:the\:quadratic\:formula}\\\\\mathrm{For\:a\:quadratic\:equation\:of\:the\:form\:}ax^2+bx+c=0\mathrm{\:the\:solutions\:are\:}\\\\x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}\\\\\mathrm{For\:}\quad a=1,\:b=1,\:c=-7\\\\x =\frac{-1\pm \sqrt{1^2-4\cdot \:1\left(-7\right)}}{2\cdot \:1}$