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3 years ago

Match the name of the shape to its word description:

Mathematics

1 answer:

HACTEHA [7]3 years ago

4 0

Https://youtu.be/dQw4w9WgXcQ that helps too

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The distance of the point (5, -1) from the y axis is *

pashok25 [27]

Answer:

5 units

Step-by-step explanation:

draw a horizontal line from the point (5, -1) to the y-axis: y=-1, it hits the y-axis at (0, -1). the distance between the two is 5 units

4 0

3 years ago

Convert 14.25% to a ratio

Varvara68 [4.7K]

1425/100=(57•25)/(4•25)
=57/4
=14 1/4

7 0

3 years ago

A lab is trying to determine if a new medication is effective at reducing acne breakouts. The results are displayed in the Venn

In-s [12.5K]

Answer:

its %60 i just took the test and got it right

Step-by-step explanation:

7 0

3 years ago

Carter draws one side of equilateral △PQR on the coordinate plane at points P(-3,2) and Q(5,2). Which ordered pair is a possible

Law Incorporation [45]

Step-by-step explanation:

Hey, there!!!

Let me simply explain you about it.

We generally use the distance formula to get the points.

let the point R be (x,y)

As it an equilateral triangle it must have equal distance.

now,

let's find the distance of PQ,

we have, distance formulae is;

$pq = \sqrt{( {x2 - x1)}^{2} + ( {y2 - y1)}^{2} }$

$or \: \sqrt{( {5 + 3)}^{2} + ( {2 - 2)}^{2} }$

By simplifying it we get,

$8$

Now,

again finding the distance between PR,

$pr = \sqrt{( {x2 - x1}^{2} + ( {y2 - y1)}^{2} }$

or,

$\sqrt{( {x + 3)}^{2} + ( {y - 2)}^{2} }$

By simplifying it we get,

$= \sqrt{ {x}^{2} + {y}^{2} + 6x - 4y + 13 }$

now, finding the distance of QR,

$qr = \sqrt{( {x - 5)}^{2} + ( {y - 2)}^{2} }$

or, by simplification we get,

$\sqrt{ {x}^{2} + {y}^{2} - 10x - 4y + 29 }$

now, equating PR and QR,

$\sqrt{ {x}^{2} + {y}^{2} + 6x - 4y + 13} = \sqrt{ {x}^{2} + {y}^{2} - 10x - 4y + 29 }$

we cancelled the root ,

${x}^{2} + {y}^{2} + 6x - 4y + 13 = {x}^{2} + {y}^{2} -10x - 4y + 29$

or, cancelling all like terms, we get,

6x+13= -10x+29

16x=16

x=16/16

Therefore, x= 1.

now,

equating, PR and PQ,

$\sqrt{ {x}^{2} + {y}^{2} + 6x - 4y + 13 } = 8}$

cancel the roots,

${x}^{2} + {y}^{2} + 6x - 4y + 13 = 8$

now,

(1)^2+ y^2+6×1-4y+13=8

or, 1+y^2+6-4y+13=8

y^2-4y+13+6+1=8

or, y(y-4)+20=8

or, y(y-4)= -12

either, or,

y= -12 y=8

Therefore, y= (8,-12)

by rounding off both values, we get,

x= 1

y=(8,-12)

So, i think it's (1,8) is your answer..

Hope it helps...

3 0

3 years ago

Simplify the expression. 12x^-6 y^10 times 3x^7 y

Vlada [557]

We need to simplify the expression:

$12x^{-6}y^{10} \times (3x^{7}y)$

Now, we know that we can resolve the exponents of the variables with the like terms only and we can multiply the coefficients independently:

Now,

$12x^{-6}y^{10} \times 3x^{7}y=(12 \times 3)\times (x^{-6}\times x^{7})\times (y^{10}\times y)$

On simplifying the above expression we get:

$36\times x^{(-6+7)}\times y^{(10+1)}$

$=36 \times x^{1}\times y^{11}$

$=36 \times x \times y^{11}$

$=36xy^{11}$

So the simplified form of the expression $12x^{-6}y^{10} \times 3x^{7}y=36xy^{11}$ .

7 0

3 years ago