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3 years ago

Who wrote the Declaration of Independence

Mathematics

2 answers:

Alinara [238K]3 years ago

8 0

Answer:

Thomas Jefferson is cited with writing it.

Step-by-step explanation:

DochEvi [55]3 years ago

8 0

Answer:

TomasJeffrey

Step-by-step explanation:

You might be interested in

57 117find x triangle

alexandr402 [8]

Answer:

Step-by-step explanation:

x = 180 - [ 57 + ( 180 - 117 ) ]

= 180 - [ 57 + 63 ]

= 180 - 120

x = 60

6 0

3 years ago

Given f(x)=2x^2+3x-5 for what values of x is f(x) positive

kow [346]

Answer:

The function f(x) is positive in the interval (-≠,-2.5) ∪ (1,∞)

Step-by-step explanation:

we have

$f(x)=2x^{2}+3x-5$

This is a vertical parabola open upward (the leading coefficient is positive)

The vertex is a minimum

The coordinates of the vertex is the point (h,k)

step 1

Find the vertex of the quadratic function

Factor the leading coefficient 2

$f(x)=2(x^{2}+\frac{3}{2}x)-5$

Complete the square

$f(x)=2(x^{2}+\frac{3}{2}x+\frac{9}{16})-5-\frac{9}{8}$

$f(x)=2(x^{2}+\frac{3}{2}x+\frac{9}{16})-\frac{49}{8}$

Rewrite as perfect squares

$f(x)=2(x+\frac{3}{4})^{2}-\frac{49}{8}$

The vertex is the point (-\frac{3}{4},-\frac{49}{8})

step 2

Find the x-intercepts (values of x when the value of f(x) is equal to zero)

For f(x)=0

$2(x+\frac{3}{4})^{2}-\frac{49}{8}=0$

$2(x+\frac{3}{4})^{2}=\frac{49}{8}$

$(x+\frac{3}{4})^{2}=\frac{49}{16}$

take the square root both sides

$x+\frac{3}{4}=\pm\frac{7}{4}$

$x=-\frac{3}{4}\pm\frac{7}{4}$

$x_1=-\frac{3}{4}+\frac{7}{4}=1$

$x_2=-\frac{3}{4}-\frac{7}{4}=-2.5$

therefore

The function f(x) is negative in the interval (-2.5,1)

The function f(x) is positive in the interval (-≠,-2.5) ∪ (1,∞)

see the attached figure to better understand the problem

6 0

4 years ago

Helppp please it’s a test helppp

Sidana [21]

For this problem:
a= -6

b= -3

7 0

3 years ago

Heyo random person scrolling.Please help me?

Ierofanga [76]

Let's do

$\\ \sf\longmapsto 2sin^260cos60tan^245$

$\\ \sf\longmapsto 2\left(\dfrac{\sqrt{3}}{2}\right)^2\times \dfrac{1}{2}\times (1)^2$

$\\ \sf\longmapsto 2\times \dfrac{(\sqrt{3})^2}{2^2}\times \dfrac{1}{2}\times 1$

$\\ \sf\longmapsto \dfrac{3}{2}\times \dfrac{1}{4}$

$\\ \sf\longmapsto \dfrac{3}{4}$

Trigonometric values

$\Large{ \begin{tabular}{|c|c|c|c|c|c|} \cline{1-6} \theta & \sf 0^{\circ} & \sf 30^{\circ} & \sf 45^{\circ} & \sf 60^{\circ} & \sf 90^{\circ} \\ \cline{1-6} $ \sin $ & 0 & $\dfrac{1}{2 }$ & $\dfrac{1}{ \sqrt{2} }$ & $\dfrac{ \sqrt{3}}{2}$ & 1 \\ \cline{1-6} $ \cos $ & 1 & $ \dfrac{ \sqrt{ 3 }}{2} } $ & $ \dfrac{1}{ \sqrt{2} } $ & $ \dfrac{ 1 }{ 2 } $ & 0 \\ \cline{1-6} $ \tan $ & 0 & $ \dfrac{1}{ \sqrt{3} } $ & 1 & $ \sqrt{3} $ & $ \infty $ \\ \cline{1-6} \cot & $ \infty $ &$ \sqrt{3} $ & 1 & $ \dfrac{1}{ \sqrt{3} } $ &0 \\ \cline{1 - 6} \sec & 1 & $ \dfrac{2}{ \sqrt{3}} $ & $ \sqrt{2} $ & 2 & $ \infty $ \\ \cline{1-6} \csc & $ \infty $ & 2 & $ \sqrt{2 } $ & $ \dfrac{ 2 }{ \sqrt{ 3 } } $ & 1 \\ \cline{1 - 6}\end{tabular}}$

6 0

3 years ago

My weight is 120lbs

pav-90 [236]

Answer:

cool im fatter

Step-by-step explanation:

6 0

3 years ago