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artcher [175]
3 years ago
8

AB is a diameter of a circle, center O. C is a point on the circumference of the circle, such that

Mathematics
1 answer:
Solnce55 [7]3 years ago
7 0

Given:

AB is the diameter of a circle.

m∠CAB = 26°

To find:

The measure of m∠CBA.

Solution:

Angle formed in the diameter of a circle is always 90°.

⇒ m∠ACB = 90°

In triangle ACB,

Sum of the angles in the triangle = 180°

m∠CAB + m∠ACB + m∠CBA = 180°

26° + 90° + m∠CBA = 180°

116° + m∠CBA = 180°

Subtract 116° from both sides.

116° + m∠CBA - 116° = 180° - 116°

m∠CBA = 64°

The measure of m∠CBA is 64°.

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Evaluate f ( - 5 ) =
Firdavs [7]

Answer: −5

Step-by-step explanation:

simplify: (−5)

Re-order terms so constants are on the left:

(−5)f(−5)−5

6 0
1 year ago
Find a polynomial function with leading coefficient 1 that has the given zeros, multiplicities, and degree. zero: 1, multiplicit
vfiekz [6]

Polynomial expression for the given  conditions is formulated in the equation

y(x)=1*(x-2)^3(x-0)^2

The polynomial for the conditions with leading coefficient 1 that has the given zeros, multiplicities, and degree. Zero: 2, multiplicity: 3 Zero: 0, multiplicity: 2 Degree: 5 is given as

According to the given condition

The given conditions to write the polynomial equations are as follows

Zero 2  and Multiplicity 3

Zero 0 and Multiplicity 2  

Degree of polynomial expression  = 5

The leading coefficient of polynomial expression = 1

Let us consider the polynomial equation has only one variable and hence in the general form we can write the equation of polynomial as follows

From the general form of polynomial expression as shown in equation (1) we can write

Polynomial expression for the given  conditions is formulated in the equation

y(x)=1*(x-2)^3(x)^2-------(2)

So the polynomial for the conditions given in the question is expressed as

y(x)=1*(x-2)^3(x)^2

For more information on polynomials click on the link below:

brainly.com/question/15702527

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8 0
1 year ago
How do i solve 3x² - 8x + 2 = 0 using quadratic formula?
DENIUS [597]
3x^2 - 8x + 2 = 0\\ \\a=3 , \ \ b=-8 , \ \ c=2 \\ \\x_{1}=\frac{-b-\sqrt{b^2-4ac}}{2a} =\frac{8-\sqrt{ (-8)^2-4 \cdot 3\cdot 2}}{2 \cdot 3} =\frac{8-\sqrt{ 64-24 }}{6} =\\ \\ =\frac{8-\sqrt{40 }}{6} = \frac{ 8-\sqrt{4\cdot 10 } }{6} = \frac{ 8-2\sqrt{ 10 }}{6} = \frac{2 (4-\sqrt{ 10 })}{6} = \frac{ 4-\sqrt{ 10 } }{3}

x_{2}=\frac{-b+\sqrt{b^2-4ac}}{2a} =\frac{8+\sqrt{ (-8)^2-4 \cdot 3\cdot 2}}{2 \cdot 3} = \frac{ 4+\sqrt{ 10 } }{3}



7 0
2 years ago
Read 2 more answers
I need help with this
Pachacha [2.7K]

Answer:8a-28b+15c

Step-by-step explanation:

3 0
2 years ago
I'm having trouble with #2. I've got it down to the part where it would be the integral of 5cos^3(pheta)/sin(pheta). I'm not sur
Butoxors [25]
\displaystyle\int\frac{\sqrt{25-x^2}}x\,\mathrm dx

Setting x=5\sin\theta, you have \mathrm dx=5\cos\theta\,\mathrm d\theta. Then the integral becomes

\displaystyle\int\frac{\sqrt{25-(5\sin\theta)^2}}{5\sin\theta}5\cos\theta\,\mathrm d\theta
\displaystyle\int\sqrt{25-25\sin^2\theta}\dfrac{\cos\theta}{\sin\theta}\,\mathrm d\theta
\displaystyle5\int\sqrt{1-\sin^2\theta}\dfrac{\cos\theta}{\sin\theta}\,\mathrm d\theta
\displaystyle5\int\sqrt{\cos^2\theta}\dfrac{\cos\theta}{\sin\theta}\,\mathrm d\theta

Now, \sqrt{x^2}=|x| in general. But since we want our substitution x=5\sin\theta to be invertible, we are tacitly assuming that we're working over a restricted domain. In particular, this means \theta=\sin^{-1}\dfrac x5, which implies that \left|\dfrac x5\right|\le1, or equivalently that |\theta|\le\dfrac\pi2. Over this domain, \cos\theta\ge0, so \sqrt{\cos^2\theta}=|\cos\theta|=\cos\theta.

Long story short, this allows us to go from

\displaystyle5\int\sqrt{\cos^2\theta}\dfrac{\cos\theta}{\sin\theta}\,\mathrm d\theta

to

\displaystyle5\int\cos\theta\dfrac{\cos\theta}{\sin\theta}\,\mathrm d\theta
\displaystyle5\int\dfrac{\cos^2\theta}{\sin\theta}\,\mathrm d\theta

Computing the remaining integral isn't difficult. Expand the numerator with the Pythagorean identity to get

\dfrac{\cos^2\theta}{\sin\theta}=\dfrac{1-\sin^2\theta}{\sin\theta}=\csc\theta-\sin\theta

Then integrate term-by-term to get

\displaystyle5\left(\int\csc\theta\,\mathrm d\theta-\int\sin\theta\,\mathrm d\theta\right)
=-5\ln|\csc\theta+\cot\theta|+\cos\theta+C

Now undo the substitution to get the antiderivative back in terms of x.

=-5\ln\left|\csc\left(\sin^{-1}\dfrac x5\right)+\cot\left(\sin^{-1}\dfrac x5\right)\right|+\cos\left(\sin^{-1}\dfrac x5\right)+C

and using basic trigonometric properties (e.g. Pythagorean theorem) this reduces to

=-5\ln\left|\dfrac{5+\sqrt{25-x^2}}x\right|+\sqrt{25-x^2}+C
4 0
2 years ago
Read 2 more answers
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