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2 years ago

Task 7: In the triangle, AC = 10, CB = 7, and YZ = 4. Find the values:

Mathematics

1 answer:

Aleksandr-060686 [28]2 years ago

5 0

Answer:

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Step-by-step explanaI DONT KNOOOOOOOOOOOONIGAAAAAAAAAAAAAAxD

You might be interested in

Compare the decimals 16.30 and 16.3

pshichka [43]

16.30 and 16.3 equal. If you ever have a problem like that then you just add on a zero.

Example:
19.4500 = 19.45 Just add two zeros on the end
19.45 + two zeros = 1.4500

8 0

3 years ago

Triangle ABC has vertices at A(2,3),B(-4,-3) and C(2,-3) find the coordinates of each point of concurrency.

dem82 [27]

Answer:

Circumcenter =(-1,0)

Orthocenter =(2,-3)

Step-by-step explanation:

Given : Points A = (2,3), B = (-4,-3), C = (2,-3)

Formula used :

→Mid point of two points- $(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2})$

→Slope of two points - $\frac{y_2-y_1}{x_2-x_1})$

→Perpendicular of a line = $\frac{-1}{slope of line})$

Circumcenter- The point where the perpendicular bisectors of a triangle meets.

Orthocenter-The intersecting point for all the altitudes of the triangle.

To find out the circumcenter we have to solve any two bisector equations.

We solve for line AB and AC

So, mid point of AB = $(\frac{2-4}{2},\frac{3-3}{2})=(-1,0)$

Slope of AB = $\frac{-3-3}{-4-2}=1$

Slope of the bisector is the negative reciprocal of the given slope.

So, the slope of the perpendicular bisector = -1

Equation of AB with slope -1 and the coordinates (-1,0) is,

(y – 0) = -1(x – (-1))

y+x=-1………………(1)

Similarly, for AC

Mid point of AC = $(\frac{2+2}{2},\frac{3-3}{2})=(2,0)$

Slope of AC = $\frac{-3-3}{2-2}=\frac{-6}{0}$

Slope of the bisector is the negative reciprocal of the given slope.

So, the slope of the perpendicular bisector = 0

Equation of AC with slope 0 and the coordinates (2,0) is,

(y – 0) = 0(x – 2)

y=0 ………………(2)

By solving equation (1) and (2),

put y=0 in equation (1)

y+x=-1

0+x=-1

⇒x=-1

So the circumcenter(P)= (-1,0)

To find the orthocenter we solve the intersections of altitudes.

We solve for line AB and BC

So, mid point of AB = $(\frac{2-4}{2},\frac{3-3}{2})=(-1,0)$

Slope of AB = $\frac{-3-3}{-4-2}=1$

Slope of the bisector is the negative reciprocal of the given slope.

So, the slope of CF = -1

Equation of AB with slope -1 and the coordinates (-1,0) gives equation CF

(y – 0) = -1(x – (-1))

y+x=-1………………(3)

Similarly, mid point of BC = $(\frac{-4+2}{2},\frac{-3-3}{2})=(-1,-3)$

Slope of AB = $\frac{-3+3}{-4-2}=0$

Slope of the bisector is the negative reciprocal of the given slope.

So, the slope of AD = 0

Equation of AB with slope 0 and the coordinates (-1,-3) gives equation AD

(y-(-3)) = 0(x – (-1))

y+3=0

y=-3………………(4)

Solve equation (3) and (4),

Put y=-3 in equation (3)

y+x=-1

-3+x=-1

x=2

Therefore, orthocenter(O)= (2,-3)

7 0

3 years ago

When the function below is graphed, how many x-intercepts does it have?y = 4x2 - 12x + 9?

Sladkaya [172]

Only x-intercept because if you want to know about x-intercepts so y will be zero (y=0)
then 0 =4x^2 -12x + 9
0=(2x-3)(2x-3)
0=2x-3
so x=3/2 that is x-intercept

7 0

3 years ago

Read 2 more answers

A curve is given by y=(x-a)√(x-b) for x≥b, where a and b are constants, cuts the x axis at A where x=b+1. Show that the gradient

ankoles [38]

<u>Answer:</u>

A curve is given by y=(x-a)√(x-b) for x≥b. The gradient of the curve at A is 1.

<u>Solution:</u>

We need to show that the gradient of the curve at A is 1

Here given that ,

$y=(x-a) \sqrt{(x-b)}$ --- equation 1