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2 years ago

USLI5

Mathematics

1 answer:

Reptile [31]2 years ago

4 0

For #3.

Zeros are 3 and -1 (plot them on the x-axis as a point)

Axis of symmetry is halfway between those so it’s: x=1

Vertex is (1, -4)

Y-intercept is (0,-3)

Domain is always all real numbers
Range is y is greater than or equal to -3

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The radius of a circle is 1 foot. What is the circle's area?

nikdorinn [45]

Answer:

3.14ft²

Step-by-step explanation:

Area of a circle= πr²

= 3.14 x 1²

= 3.14 x 1

= 3.14 ft²

8 0

3 years ago

Please help me out please

natita [175]

Answer:

RC = 40

Step-by-step explanation:

Note that the circumcentre is equally distant from the triangle's 3 vertices.

That is : PC = RC = QC

Equate any pair and solve for x

Using RC = PC, then

5x - 15 = 3x + 7 ( subtract 3x from both sides )

2x - 15 = 7 ( add 15 to both sides )

2x = 22 ( divide both sides by 2 )

x = 11

Hence

RC = (5 × 11) - 15 = 55 - 15 = 40 units

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3 years ago

In triangle ABC, m is the value of x?

Nezavi [6.7K]

Answer:

lol if a = 90 then b = 75 so C is 105

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Step-by-step explanation:

6 0

3 years ago

Find the equation of the line that is parallel to x=5y-8 and pass through (6, -5) in slope intercept form

DochEvi [55]

Answer:

y=1/5x-31/5 (or y=0.2x-6.2 in decimals)

Step-by-step explanation:

x=5y-8 (original equation)

x+8=5y (Addition Property of Equality)

1/5x+8/5=y (Division Property of Equality, slope of original equation is 1/5)

y-y1=m(x-x1) (point-slope formula)

y-(-5))=1/5(x-(6)) (plug in the slope that was found earlier and the point given in the question)

y+5=1/5(x-6)

y+5=1/5x-6/5 (Distributive Property of Multiplication) Note: 5 = 25/5

y=1/5x-31/5 (Subtraction Property of Equality, and there's your answer)

y=0.2x-6.2 (This is the same answer, but written with decimals)

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3 years ago

Use triangle ABC drawn below & only the sides labeled. Find the side of length AB in terms of side a, side b & angle C o

Brrunno [24]

Answer:

$AB = \sqrt{a^2 + b^2-2abCos\ C}$

Step-by-step explanation:

Given:

The above triangle

Required

Solve for AB in terms of a, b and angle C

Considering right angled triangle BOC where O is the point between b-x and x

From BOC, we have that:

$Sin\ C = \frac{h}{a}$

Make h the subject:

$h = aSin\ C$

Also, in BOC (Using Pythagoras)

$a^2 = h^2 + x^2$

Make $x^2$ the subject

$x^2 = a^2 - h^2$

Substitute $aSin\ C$ for h

$x^2 = a^2 - h^2$ becomes

$x^2 = a^2 - (aSin\ C)^2$

$x^2 = a^2 - a^2Sin^2\ C$

Factorize

$x^2 = a^2 (1 - Sin^2\ C)$

In trigonometry:

$Cos^2C = 1-Sin^2C$

So, we have that:

$x^2 = a^2 Cos^2\ C$

Take square roots of both sides

$x= aCos\ C$

In triangle BOA, applying Pythagoras theorem, we have that:

$AB^2 = h^2 + (b-x)^2$

Open bracket

$AB^2 = h^2 + b^2-2bx+x^2$

Substitute $x= aCos\ C$ and $h = aSin\ C$ in $AB^2 = h^2 + b^2-2bx+x^2$

$AB^2 = h^2 + b^2-2bx+x^2$

$AB^2 = (aSin\ C)^2 + b^2-2b(aCos\ C)+(aCos\ C)^2$

Open Bracket

$AB^2 = a^2Sin^2\ C + b^2-2abCos\ C+a^2Cos^2\ C$

Reorder

$AB^2 = a^2Sin^2\ C +a^2Cos^2\ C + b^2-2abCos\ C$

Factorize:

$AB^2 = a^2(Sin^2\ C +Cos^2\ C) + b^2-2abCos\ C$

In trigonometry:

$Sin^2C + Cos^2 = 1$

So, we have that:

$AB^2 = a^2 * 1 + b^2-2abCos\ C$

$AB^2 = a^2 + b^2-2abCos\ C$

Take square roots of both sides

$AB = \sqrt{a^2 + b^2-2abCos\ C}$

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3 years ago