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3 years ago

Need help with this, finding the length of the third side

Mathematics

1 answer:

Advocard [28]3 years ago

5 0

Answer:

its sqrt (50)

Step-by-step explanation:

The triangle has a right angle, therefore we can use the Pythagorean theorem to find the hypotenuse. According to the theorem the sum of the squares of both other sides is equal to the square of the hypotenuse.

We have 5^2+5^2=c^2

=> 25+25=c^2

=> 50=c^2

Therefore c or the missing side is equal to sqrt(50)

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CAN SOMEONE HELP ME PLS!! im confused

neonofarm [45]

Answer:

the angle in upper and the lower corners have to be the same, so we can write an equation

5p-62 = 6p-88

(this picture is indeed complicated to read)

rearrange the equation

-62 = 1p-88

26 = p

hope this helped you

4 0

3 years ago

Part A: Solve −np − 70 < 40 for n. Show your work. (4 points) Part B: Solve 4w − 7k = 28 for k. Show your work.

kondaur [170]

Hey!

For part A, let's write the problem,
$-np-70\ \textless \ 40$
Add $70$ to both sides...
$-np-70+70\ \textless \ 40+70$
$-np\ \textless \ 110$
We are going to reverse the inequality, to do this we are going to multiply both sides by $-1$ .
$\left(-np\right)\left(-1\right)\ \textgreater \ 110\left(-1\right)$
$pn\ \textgreater \ -110$
Divide both sides by $p$ .
$\frac{pn}{p}\ \textgreater \ \frac{-110}{p}$
Our answer would be,
$n\ \textgreater \ -\frac{110}{p}$

Let's go with Part B. Let's write the problem,
$4w-7k=28$
Subtract $4w$ from both sides,
$4w-7k-4w=28-4w$
$-7k=28-4w$
Divide both sides by $-7$ .
$\frac{-7k}{-7}=\frac{28}{-7}-\frac{4w}{-7}$
If we simplified it, we would be left with,
$k=-\frac{28-4w}{7}$

Thanks!
-TetraFish

4 0

3 years ago

Two numbers, +10 and -10, are 10 units from 0 on the number line. Enter the correct answer.

neonofarm [45]

Answer:

nzfgfgnfsfhreaheyrzyhrsyhar5haqrtyqre4ger

3 0

3 years ago

Find the orthocenter for the triangle described by each set of vertices.

givi [52]

Answer:

The orthocentre of the given vertices ( 2 , -3.5)

Step-by-step explanation:

Step(i):-

The orthocentre is the intersecting point for all the altitudes of the triangle.

The point where the altitudes of a triangle meet is known as the orthocentre.

Given Points are K (3.-3), L (2,1), M (4,-3)

The Altitudes are perpendicular line from one side of the triangle to the opposite vertex

The altitudes are MN , KO , LP

step(ii):-

Slope of the line

$KL = \frac{y_{2}-y_{1} }{x_{2}-x_{1} } = \frac{1-(-3)}{2-3} = -4$

The slope of MN =

The perpendicular slope of KL

= $\frac{-1}{m} = \frac{-1}{-4} = \frac{1}{4}$

The equation of the altitude

$y - y_{1} = m( x-x_{1} )$

$y - (-3) = \frac{1}{4} ( x-4 )$

4y +12 = x -4

x - 4 y -16 = 0 ...(i)

Step(iii):-

Slope of the line

$LM = \frac{y_{2}-y_{1} }{x_{2}-x_{1} } = \frac{-3-1}{4-2} = -2$

The slope of KO =

The perpendicular slope of LM

= $\frac{-1}{m} = \frac{-1}{-2} = \frac{1}{2}$

The equation of the altitude

$y - y_{1} = m( x-x_{1} )$

The equation of the line passing through the point K ( 3,-3) and slope

m = 1/2

$y - (-3) = \frac{1}{2} ( x-3 )$

2y +6 = x -3

x - 2y -9 =0 ....(ii)

Solving equation (i) and (ii) , we get

subtracting equation (i) and (ii) , we get

x - 4y -16 -( x-2y-9) =0

- 2y -7 =0

-2y = 7

y = - 3.5

Substitute y = -3.5 in equation x -4y-16=0

x - 4( -3.5) - 16 =0

x +14-16 =0

x -2 =0

x = 2

The orthocentre of the given vertices ( 2 , -3.5)

4 0

3 years ago

Determine the values of n and m so that the following system have infinite number of solutions

cupoosta [38]

Answer:

n = 3

m = -1/2

Step-by-step explanation:

Multiply the second equation by 4

4(n - 2)x + y = 4(n + m) Y is now correct.

4(n - 2) = 4 Remove the brackets

4n - 8 = 4 Add 8 to both sides

4n = 8 + 4 Combine

4n = 12 Divide by 4

n = 3 x is now correct

4(n - 2) = 4

4(3 -2) = 4

4 = 4

Now you have to worry about m

4(n + m) = 10

n = 3

4(3 + m) = 10 Divide by 4

m + 3 = 10/4

m + 3 = 2.5 Subtract 3 from both sides.

m = 2.5 - 3

m = - 1/2

6 0

3 years ago