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3 years ago

6

EASY PLS HELP ASAP!! ANY QUESTION PLSSS

1 answer:

Lena [83]3 years ago

5 0

Rudy’s age today is 96.

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Hey, help me with this pls i only have 10 minutes. 30 points each :)

vlada-n [284]

Answer:

y=-2x

Step-by-step explanation:

First you want to remember rise over run. Your rise is -2 and your run is 1. So -2/1 is your answer. And -2/1 is equal to -2.

7 0

3 years ago

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PLEASE HELP ME OUT !!! WILL APPRECIATE IT SO MUCH !!

mash [69]

Start with the general formula. The key word is difference. So the general formula is

D(x) = W(x) - R(x) Now substitute in the values for W(x) and R(x)
D(x) = 0.002x^3 - 0.01x^2 - (x^2 - 4x + 13) Be very careful about the sign in front of the brackets.

D(x) = 0.002x^3 - 0.01x^2 - x^2 + 4x - 13 Do you see what that minus sign did? It affected all 3 terms.

D(x) = 0.002x^3 - 1.01x^2 + 4x - 13 <em>Note: -0.01 - 1x^2 = - 1.01 x^2</em>
That gives you a clear cut answer

C<<<<< answer.

That sign is the worst part of the question. Make sure you understand what it did.

4 0

3 years ago

Hey <br>what is 6/5 ÷ 10/7 in the simplest form possible

d1i1m1o1n [39]

Exact Form: 21/25

Decimal Form: 0.84

Step-by-step explanation: Reduce the expression, if possible, by cancelling the common factors.

5 0

3 years ago

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The hypotenuse of a right triangle has endpoints A(4, 1) and B(–1, –2). On a coordinate plane, line A B has points (4, 1) and (n

GarryVolchara [31]

Answer:

$(-1,1),(4,-2)$

Step-by-step explanation:

Given: The hypotenuse of a right triangle has endpoints A(4, 1) and B(–1, –2).

To find: coordinates of vertex of the right angle

Solution:

Let C be point $(x,y)$

Distance between points $(x_1,y_1),(x_2,y_2)$ is given by $\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

$AC=\sqrt{(x-4)^2+(y-1)^2}\\BC=\sqrt{(x+1)^2+(y+2)^2}\\AB=\sqrt{(4+1)^2+(1+2)^2}=\sqrt{25+9}=\sqrt{34}$

ΔABC is a right angled triangle, suing Pythagoras theorem (square of hypotenuse is equal to sum of squares of base and perpendicular)

$34=\left [ (x-4)^2+(y-1)^2 \right ]+\left [ (x+1)^2+(y+2)^2 \right ]$

Put $(x,y)=(-1,1)$

$34=\left [ (-1-4)^2+(1-1)^2 \right ]+\left [ (-1+1)^2+(1+2)^2 \right ]\\34=25+9\\34=34$

which is true. So, $(-1,1)$ can be a vertex

Put $(x,y)=(4,-2)$

$34=\left [ (4-4)^2+(-2-1)^2 \right ]+\left [ (4+1)^2+(-2+2)^2 \right ]\\34=9+25\\34=34$

which is true. So, $(4,-2)$ can be a vertex

Put $(x,y)=(1,1)$

$34=\left [ (1-4)^2+(1-1)^2 \right ]+\left [ (1+1)^2+(1+2)^2 \right ]\\34=9+4+9\\34=22$

which is not true. So, $(1,1)$ cannot be a vertex

Put $(x,y)=(2,-2)$

$34=\left [ (2-4)^2+(-2-1)^2 \right ]+\left [ (2+1)^2+(-2+2)^2 \right ]\\34=4+9+9\\34=22$

which is not true. So, $(2,-2)$ cannot be a vertex

Put $(x,y)=(4,-1)$

$34=\left [ (4-4)^2+(-1-1)^2 \right ]+\left [ (4+1)^2+(-1+2)^2 \right ]\\34=4+25+1\\34=30$

which is not true. So, $(4,-1)$ cannot be a vertex

Put $(x,y)=(-1,4)$

$34=\left [ (-1-4)^2+(4-1)^2 \right ]+\left [ (-1+1)^2+(4+2)^2 \right ]\\34=25+9+36\\34=70$

which is not true. So, $(-1,4)$ cannot be a vertex

So, possible points for the vertex are $(-1,1),(4,-2)$

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3 years ago

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I need hep plz quick

Harman [31]

.15 repeating which is the 2nd one

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3 years ago

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