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klasskru [66]
3 years ago
5

CAN SOMEONE PLEASE HELP ME WITH THIS ?

Mathematics
1 answer:
Rus_ich [418]3 years ago
8 0

Answer:b

Step-by-step explanation:

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What is the slope of the line?
bonufazy [111]

Answer:

4/5

Step-by-step explanation:

You look to where the line would meet a point on the graph ( you need 2) then you find rise over run or the y/x difference

4 0
3 years ago
Coefficient of b in expansion of (3+b)^4
Margarita [4]
Are you sure you want ONLY the coefficient of b?  If you expand this, you will have b in 3 of 4 terms.

According to Pascal's Triangle, the coefficients of (a+b)^4 are as follows:

                             1
                        1    2    1
                   1      3    3     1
              1       4     6     4     1

So (a+b)^4 would be 1a^4 + 4a^3b + 6a^2b^2 + 4ab^3 + b^4

Here, you want (3 + b)^4.  Here's what that looks like:

3^4 + 4[3^3*b] + 6[3^2*b^2] + 4[3*b^3] + 1[b^4]

Which coeff did you want?
4 0
3 years ago
Why is Point , Line , and Plane considered the 3 building blocks of geometry??? Please answer fast
tangare [24]
These 3 objects are used to make all of the other objects that we will use in geometry
3 0
3 years ago
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)

7 0
3 years ago
Read 2 more answers
Davia draws a shape with 5 sides . two sides are each 5 inches long. two other sides are each 4 inches long. the perimeter of th
soldier1979 [14.2K]
The fifth side is 9 inches
5 0
3 years ago
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