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3 years ago

5

CAN SOMEONE PLEASE HELP ME WITH THIS ?

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?

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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

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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?

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3 years ago

Why is Point , Line , and Plane considered the 3 building blocks of geometry??? Please answer fast

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These 3 objects are used to make all of the other objects that we will use in geometry

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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

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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

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The fifth side is 9 inches

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