Ask question

Ask question

All categories

3 years ago

11

Who wrote most definitive text on geometry

1 answer:

djverab [1.8K]3 years ago

6 0

Answer:

Euclid

Step-by-step explanation:

Euclid was a Greek geometer who wrote the Elements, the world's most definitive text on geometry. The book synthesized earlier knowledge about geometry and was used for centuries in Western Europe as a geometry textbook.

You might be interested in

To what power must you raise the expression 7^1/2 to get 7 as the result?

Anvisha [2.4K]

Answer:

2

Step-by-step explanation:

(7^1/2)^2=7^(1/2*2)=7

4 0

2 years ago

Are these right before I go on I have a 100 in math #wannakeepit

Serjik [45]

They are all correct

4 0

2 years ago

Find the solution in slope-intercept form y+7=-3(x-1) and 3x+y=-4

Svetach [21]

Answer:

The slope intercept form of both given equations is : y = - 3 x - 4.

Step-by-step explanation:

Here, the given equations are:

y +7 = -3 ( x - 1 )

and 3 x + y = - 4

Now,the SLOPE INTERCEPT FORM of any given equation is given as:

y = m x + C : here, C = Y - intercept, m = Slope

Consider equation (1):

y +7 = -3 ( x - 1 ) ⇒ y + 8 = - 3 x + 3

or, y = -3x + 3 - 7 = -3x - 4

⇒ y = -3x -4

Hence, the slope-intercept form of the given equation is y = -3x -4.

Consider equation (2):

3 x + y = - 4 ⇒ y = -4 - 3 x

⇒ y = -3 x - 4

Hence, the slope-intercept form of the given equation is y = -3x -4.

6 0

2 years ago

Given that El bisects ZCEA, which statements must be

Alexxx [7]

Question: Given that BE bisects ∠CEA, which statements must be true? Select THREE options.

(See attachment below for the figure)

m∠CEA = 90°

m∠CEF = m∠CEA + m∠BEF

m∠CEB = 2(m∠CEA)

∠CEF is a straight angle.

∠AEF is a right angle.

Answer:

m∠CEA = 90°

∠CEF is a straight angle.

∠AEF is a right angle

Step-by-step explanation:

Line AE is perpendicular to line CF, which is a straight line. This creates two right angles, <CEA and <AEF.

Angle on a straight line = 180°. Therefore, m<CEA + m<AEF = m<CEF. Each right angle measures 90°.

Thus, the three statements that must be TRUE are:

m∠CEA = 90°

∠CEF is a straight angle.

∠AEF is a right angle

3 0

2 years ago

Please help me for the love of God if i fail I have to repeat the class

Elena-2011 [213]

$\theta$ is in quadrant I, so $\cos\theta>0$ .

$x$ is in quadrant II, so $\sin x>0$ .

Recall that for any angle $\alpha$ ,

$\sin^2\alpha+\cos^2\alpha=1$

Then with the conditions determined above, we get

$\cos\theta=\sqrt{1-\left(\dfrac45\right)^2}=\dfrac35$

and

$\sin x=\sqrt{1-\left(-\dfrac5{13}\right)^2}=\dfrac{12}{13}$

Now recall the compound angle formulas:

$\sin(\alpha\pm\beta)=\sin\alpha\cos\beta\pm\cos\alpha\sin\beta$

$\cos(\alpha\pm\beta)=\cos\alpha\cos\beta\mp\sin\alpha\sin\beta$

$\sin2\alpha=2\sin\alpha\cos\alpha$

$\cos2\alpha=\cos^2\alpha-\sin^2\alpha$

as well as the definition of tangent:

$\tan\alpha=\dfrac{\sin\alpha}{\cos\alpha}$

Then

1. $\sin(\theta+x)=\sin\theta\cos x+\cos\theta\sin x=\dfrac{16}{65}$

2. $\cos(\theta-x)=\cos\theta\cos x+\sin\theta\sin x=\dfrac{33}{65}$

3. $\tan(\theta+x)=\dfrac{\sin(\theta+x)}{\cos(\theta+x)}=-\dfrac{16}{63}$

4. $\sin2\theta=2\sin\theta\cos\theta=\dfrac{24}{25}$

5. $\cos2x=\cos^2x-\sin^2x=-\dfrac{119}{169}$

6. $\tan2\theta=\dfrac{\sin2\theta}{\cos2\theta}=-\dfrac{24}7$

7. A bit more work required here. Recall the half-angle identities:

$\cos^2\dfrac\alpha2=\dfrac{1+\cos\alpha}2$

$\sin^2\dfrac\alpha2=\dfrac{1-\cos\alpha}2$

$\implies\tan^2\dfrac\alpha2=\dfrac{1-\cos\alpha}{1+\cos\alpha}$

Because $x$ is in quadrant II, we know that $\dfrac x2$ is in quadrant I. Specifically, we know $\dfrac\pi2$ , so $\dfrac\pi4$ . In this quadrant, we have $\tan\dfrac x2>0$ , so

$\tan\dfrac x2=\sqrt{\dfrac{1-\cos x}{1+\cos x}}=\dfrac32$

8. $\sin3\theta=\sin(\theta+2\theta)=\dfrac{44}{125}$

6 0

3 years ago