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3 years ago

What is the equation that is perpendicular and passes through 8,3

Mathematics

1 answer:

ss7ja [257]3 years ago

6 0

Y = 4x + 5
slope = 4
perpendicular, slope = - 1/4

passing thru (8,3)

equation
y - 3 = -1/4 (x - 8)
y - 3 = -1/4x + 2
y = -1/4x +5

answer is D. y = -1/4x +5

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Is 25/4 grater then 40/8

Tomtit [17]

Answer:

yes it is greater:)

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2 years ago

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HELPPPPPPPPPPPPPP PLZZZZZZZZZZZ clock arithmetic Will mark brainiest .

GarryVolchara [31]

Answer:

a) 29

b) 0

c) 7

d) 3/8

Step-by-step explanation:

Whenever you're facing a clock maths problem, the solution always have to be < to the number of hours in the given clock. If it's > the number of hours of the given clock, you subtract the number of hours until you get a result <= the number of clock hours.

If the result is negative, you add the clock hours.

a) 21 - 33 = -12 , so -12 + 41 = 29

b) 13 * 4 = 52, then do 52 - 52 = 0, since answer has to be < 52.

c) 11+19 = 30, 30 - 23 = 7

d) 3/8 = 3/8, since 3/8 <= 15, you're also fine.

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2 years ago

11 nickels to 11 dimes are what ratio?

castortr0y [4]

Answer: 11:11

Step-by-step explanation:

The ratio is 11 nickles to 11 dimes, Take away the names of the coins and there is your answer. (If i'm correct)

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3 years ago

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Can please someone help me? The question is in the picture. Thank you

maxonik [38]

Answer:

C. $9\pi$

Step-by-step explanation:

We proceed to solve for $x$ by the Pythagorean Theorem:

$(x+1)^{2} = (x-1)^{2} + (2\cdot x - 4)^{2}$ (1)

$x^{2} + 2\cdot x + 1 = (x^{2}-2\cdot x + 1) + (4\cdot x^{2}-16\cdot x + 16)$

$x^{2} + 2\cdot x + 1 = 5\cdot x^{2}-18\cdot x + 17$

$4\cdot x^{2} -20\cdot x + 16 = 0$

$(2\cdot x)^{2} -10\cdot (2\cdot x) + 16 = 0$

$(2\cdot x -8)\cdot (2\cdot x -2) = 0$

There are two different solutions:

$x_{1} = 4$ , $x_{2} = 1$

A solution of $x$ is considered realistic when the length of every side of the triangle is a not negative number.

$x_{1} = 4$

$UV = 2\cdot (4) - 4$

$UV = 4$

$WU = 4 - 1$

$WU = 3$

$WV = 4 + 1$

$WV = 5$

$x_{2} = 1$

$UV = 2\cdot (1) - 4$

$UV = -2$

$WU = 1 - 1$

$WU = 0$

$WV = 1 + 1$

$WV = 2$

The only realistic set of solutions is for $x = 4$ , and the radius of the circle is represented by the line segment $WU$ : $r = 3$

And the area of the circle ( $A$ ) is calculated from the following formula:

$A = \pi\cdot r^{2}$

$A = \pi\cdot (9)$

$A = 9\pi$

The correct answer is C.

6 0

2 years ago

Plz help me!!!!!!!!!

Over [174]

Answer: $\bold{\dfrac{4\pm \sqrt{6}}{2}}$

<u>Step-by-step explanation:</u>

$\dfrac{3}{y-2}-2=\dfrac{1}{y-1}\\\\\\\text{Multiply by the LCD (y-2)(y-1) to clear the denominator:}\\\\\dfrac{3}{y-2}(y-2)(y-1)-2(y-2)(y-1)=\dfrac{1}{y-1}(y-2)(y-1)\\\\\\3(y-1)-2(y-2)(y-1)=1(y-2)\\\\3y-3-2(y^2-3y+2)=y-2\\\\3y-3-2y^2+6y-4=y-2\\\\-2y^2+9y-7=y-2\\\\0=2y^2-8y+5\quad \rightarrow \quad a=2,\ b=-8,\ c=5\\\\\\\text{Quadratic formula is: }x=\dfrac{-b\pm \sqrt{b^2-4ac}}{2a}\\\\\\x=\dfrac{-(-8)\pm \sqrt{(-8)^2-4(2)(5)}}{2(2)}\\\\\\.\ =\dfrac{8\pm \sqrt{64-40}}{2(2)}$

$.\ =\dfrac{8\pm \sqrt{24}}{2(2)}\\\\\\.\ =\dfrac{8\pm 2\sqrt{6}}{2(2)}\\\\\\.\ =\dfrac{4\pm \sqrt{6}}{2}$

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2 years ago