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3 years ago

Tell which number is prime: 39, 69, 89, 90.

Mathematics

2 answers:

zmey [24]3 years ago

6 0

The answer is 89 is a prime number because it has only two distinct divisors: 1 and itself

Sever21 [200]3 years ago

5 0

Answer:

Step-by-step explanation:

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A line has a slope of -13. There is a second line that passes through points (-7,-5) and (6,-4) . Would these lines be parallel

Elza [17]

<h3>Answer: The lines are perpendicular</h3>

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Work Shown:

Use the slope formula to find the slope of the line through the two given points.

m = (y2 - y1)/(x2 - x1)

m = (-4 - (-5))/(6 - (-7))

m = (-4 + 5)/(6 + 7)

m = 1/13

The slope of the line through the two given points is 1/13.

This is not equal to -13, which was the slope of the first line, so the two lines are not parallel.

However, the two lines are perpendicular because multiplying the two slope values leads to -1

(slope1)*(slope2) = (-13)*(1/13) = -1

note: It might help to think of -13 as -13/1 to help multiply the fractions.

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2 years ago

(a)(−(−b))(−c) HELP PLEASE

Liono4ka [1.6K]

Answer:

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Step-by-step explanation:

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3 years ago

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What is the value of x?

Nonamiya [84]

Answer:

9x-3x(2-4)=6x(3x+4)-6(3x-2)

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2 years ago

What is 450 increased by 18%

Blababa [14]

450*(1+0.18)=531 hence 450 increased by 18% is 531

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2 years ago

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PLEASE HELP GUYS i am struggling so much, two questions

ankoles [38]

Answer: the equation of the standard parabola

1) $(y-6)^2 = 4 (x-1)$

The equation of the standard parabola

2) (x+5)^2 = 16(y-2)

Step-by-step explanation:

Explanation

Parabola:-

The set of points in a plane whose distance from a fixed point and a constant ratio to their corresponding perpendicular distance from a fixed straight line is a conic.

Let S be a fixed point and l be a fixed straight line from any point P,the perpendicular PM is drawn to the line 'l'

The locus of P such that $\frac{SP}{PM} = constant$
The fixed point 'S' is called the Focus.
The fixed line'l 'is called the directrix of the conic
The constant ratio is known as the eccentricity, denoted by 'e'
If e=1 , the conic is called a parabola

1) Step 1 :-

Given the focus S = (2,6) and directrix is x=0

we know that $\frac{SP}{PM}=1$

now cross multiplication , we get

$SP = PM$

squaring on both sides,we get

$SP^{2} = PM^2$

step 2:-

now using distance formula is

$\sqrt(((x_{2}-x_{1})^2+(y_{2} -y_{1} )^2)$

Given S =(2,6) and P(x,y) be any point on parabola

$SP^2 = (x-2)^2+(y-6)^2$ ........(1)

Now using perpendicular distance formula

let P(x , y ) be any point on the parabola

$\frac{ax_{1}+by_{1}+c }{\sqrt{a^2+b^2} }$

Given the directrix is x =0 and P(x,y) be any point on parabola

$PM^2 = \frac{x^2}{\sqrt{1}^2 }$ ......(2)

equating equation(1) and equation (2), on simplification

we get $(x-2)^2+(y-6)^2 = x^2$ .....(3)

apply $(a-b)^2 = a^2+ b^2+2 ab$

now the equation (3) is

$(y-6)^2 = 4 x-4$

now the standard form of parabola is

$(y-k)^2 = 4 a(x-h)$

Final answer:-

$(y-6)^2 = 4 (x-1)$

2) Explanation:-

step 1:

Given vertex of a parabola is A(-5,2) and its focus is S(-5,6)

here the given points of 'x'co- ordinates are equal

Therefore the axis AS is parallel to y- axis

now the standard equation of parabola

$(x-h)^2 = 4 a (y-k)$

now you have to find' a' value

Given vertex of a parabola is A(-5,2) and its focus is S(-5,6)

The distance of $AS = \sqrt{(-5-(-5)^2+(2-6)^2}$

on simplification we get a =4

Final answer :-

the vertex (h,k) = (-5,2) and a=4

$(x-h)^2 = 4 a (y-k)$

The standard parabola is $(x+5)^2 = 16 (y-2)$

5 0

2 years ago