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3 years ago

Find the midpoint between 7, -2 and -4,2

Mathematics

2 answers:

Oduvanchick [21]3 years ago

7 0

Answer:

Step-by-step explanation:

(7 + (-4))/2 = 3/2

(-2 + 2)/2 = 0/2 = 0

(3/2, 0)

SpyIntel [72]3 years ago

3 0

Answer:

The midpoint is (3/2,0)

Step-by-step explanation:

To find the x coordinate of the midpoint, add the x coordinate of the endpoints and divide by 2

(7+-4)/2 = 3/2

To find the y coordinate of the midpoint, add the y coordinate of the endpoints and divide by 2

(-2+2)/2 = 0

The midpoint is (3/2,0)

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Find the value of tan theta if sin theta = 12/13 and theta is in quadrant 2

8090 [49]

Answer:

tanΘ = - $\frac{12}{5}$

Step-by-step explanation:

Using the trigonometric identities

• sin²x + cos²x = 1, hence

cosx = ± √(1 - sin²x )

• tanx = $\frac{sinx}{cosx}$

given sinΘ = $\frac{12}{13}$ , then

cosΘ = ± $\sqrt{1-(12/13)^2}$

Since Θ is in the second quadrant where cosΘ < 0, then

cosΘ = - $\sqrt{1-\frac{144}{169} }$

= - $\sqrt{\frac{25}{169} }$ = - $\frac{5}{13}$

tanΘ = $\frac{\frac{12}{13} }{\frac{-5}{13} }$

= $\frac{12}{13}$ × - $\frac{13}{5}$ = - $\frac{12}{5}$

5 0

3 years ago

The word geometry comes from the greek words geo, which means "earth," and metron, which means "a measuring of." given this info

Leokris [45]

The geometry was developed by a Greek mathematician called Euclid and developed to meet some practical need in surveying, construction, astronomy, and various crafts.

In this question,

Early geometry was a collection of empirically discovered principles concerning lengths, angles, areas, and volumes, which were developed to meet some practical need in surveying, construction, astronomy, and various crafts.

It was developed by a Greek mathematician called Euclid. This branch of geometry deals with terms like points, lines, surfaces, dimensions of the solids, etc.,

In several ancient cultures there developed a form of geometry suited to the relationships between lengths, areas, and volumes of physical objects. This geometry was codified in Euclid's Elements about 300 bce on the basis of 10 axioms, or postulates, from which several hundred theorems were proved by deductive logic.

Hence we can conclude that the geometry was developed by a Greek mathematician called Euclid and developed to meet some practical need in surveying, construction, astronomy, and various crafts.

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6 0

2 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

Write an equation of the line using the given information-2/3 and a y intercept =4

sashaice [31]

In y = mx + b form, the slope will be in the m position and the y int will be in the b position.

y = mx + b......slope(m) = -2/3....y int (b) = 4
y = -2/3x + 4

4 0

3 years ago

CAN AN EXPERT< ACE< MODERATOR OR GENIUS HELPP ME WITT THIS PLSSSSSSSSSSSSSS!!!!!!!!!!!!!!!!!!!!!!!!

Mamont248 [21]

Answer:

red box 5, 3, 0, -9, 4, 0, 2, -37

Step-by-step explanation:

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3 years ago