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sergij07 [2.7K]

3 years ago

15

Chord RS intersects chord Tu at point C. The length of TC is 3 times the length of CU. Find the length of TC. If necessary,

2 answers:

Snowcat [4.5K]3 years ago

7 0

Answer:

D

Step-by-step explanation:

D

Amanda [17]3 years ago

7 0

Answer:

C) 24

Step-by-step explanation:

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Solve the quadratic equation by factorisation:<br>x2-3x - 10 = 0

frez [133]

Answer:

x=−2

Step-by-step explanation:

4 0

3 years ago

Identify all of the problems that would result in a product of 2 5 ?

mars1129 [50]

1 multiplied by 25 and 5 multiplied by 5

8 0

3 years ago

PLEASE HELP WITH #3 i’ll give brainliest

tekilochka [14]

Step-by-step explanation:

Lets begin with this 4x-25=x+14, 3x=39, so x=13, than we plug it in to side NP because that is the most simple one to plug it into to and that side is 27 since the triangle is equilateral all sides are the same so it is 27, 27, 27, for the sides and x=13

4 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

A roadside vegetable stand sells pumpkins for $5 each and tomatoes for $3 each. The total sales on a certain day was $98. If the

dexar [7]

Answer:

16 tomatoes

Step-by-step explanation:

50+3t=98

3t=48

t=16

8 0

3 years ago