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AveGali [126]
3 years ago
10

Evaluate the function g(t) =2/3t when t = 30. 25 10 20 15

Mathematics
1 answer:
galben [10]3 years ago
6 0
G(30) = 2/3(30)
g(30) = 20

The answer is 20
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Please help it's urgent.
arlik [135]

Answer:

It is the second triangle.

Step-by-step explanation:

This is because the area of the square is given as 25^2 as the perpendicular, 144^2 as the base, and the hypotenuse is 169^2.

Length of the perpendicular = 5^2 (Because 25^2 / 5^2)

Length of the base = 12^2

Length of the hypotenuse = 13^2

So, the Pythagoras theorem is,

=> H^{2} = P^{2} + B^{2}

=> 13^{2} =12^{2} +5^{2}

This is true, because, 13*13 = 12*12 + 5*5

=> 169 = 144 + 25

=> 169 = 169

If my answer helped, kindly mark me as the brainliest!!

Thanks!!

3 0
3 years ago
What is the correct letter? (Giving brainliest)
vampirchik [111]

Answer:

d is the answer

Step-by-step explanation:

perimeter of rect = 2(7+3) = 20

perimeter of square = 4 a =20

a = 5

3 0
3 years ago
Write the point-slope form of the line that passes through (-8, 2) and is perpendicular to a line with a slope of -8
Schach [20]

Answer:

y - 2 = (1/8)(x + 8)

Step-by-step explanation:

The slope of the given line is -8.  That means the slope of any line perpendicular to the given line will be the negative reciprocal of -8, which would be (1/8).

Then the desired equation is

y - 2 = (1/8)(x + 8)

4 0
4 years ago
PLZ HELP, AND PLZ EXPLAIN
shutvik [7]

Answer:

C

Step-by-step explanation:

To make it easy let's start by organizing our information :

  • AC=12 AND BD=8
  • ABCD is a rhombus
  • K and L are the midpoints of sides AD and CD
  • we notice that the rhombus ABCD is divided into four right triangles

What do you think of when you hear a right triangle ?

  • The pythagorian theorem !

AC and BD  are khown so let's focus on them .

If we concentrated we can notice that AB and BD are cossing each other in the midpoints . why ?

Simply because they are the diagonals of a rhombus .

ow let's apply the pythagorian theorem :

  • (AC/2)² + (BD/2)² = BC²
  • 6²+4²=52
  • BC²= 52⇒\sqrt{52}=BC

Now we khow that : AB=BC=CD=AD=\sqrt{52}

This isn't enough . Let's try to figure out a way to calculate the length of KL  wich is the base of the triangle

  • KL is parallel to AC
  • k is the midpoint of AD and L of DC

I smell something . yes! Thales theorem

  • KL/AC=DL/DC=DK/AD WE4LL TAKE OLY ONE
  • KL/12=\sqrt{52}/2*\sqrt{52}  
  • KL/12=1/2⇒ KL=6

Now we have the length of the base kl

Now the big boss the height :

  • notice that you khow the length of KL
  • BD crosses kl from its midpoint and DL = \sqrt{52} /2

What I want to do is to apply the pythgorian thaorem to khow the lenght of that small part that is not a part of the height of the triangle . I will call it D

  • DL²=(KL/2)²+D²
  • 52/4= 9+ D²
  • D² = 52/4-9 +4 SO D=2

now the height of the trigle is H= BD-D= 8-2=6

NOw the area of the triangle is :

  • A=(KL*H)/2 ⇒ A= (6*6)/2=18

THE ANSWER IS 18 SQ.UN

5 0
3 years ago
BD is the angle bisector of
AlexFokin [52]

Note: Let us consider, we need to find the m\angle ABC and m\angle DBC.

Given:

In the given figure, BD is the angle bisector of ABC.

To find:

The m\angle ABC and m\angle DBC.

Solution:

BD is the angle bisector of ABC. So,

m\angle ABD=m\angle DBC

3x=x+20

3x-x=20

2x=20

Divide both sides by 2.

x=\dfrac{20}{2}

x=10

Now,

m\angle DBC=(x+20)^\circ

m\angle DBC=(10+20)^\circ

m\angle DBC=30^\circ

And,

m\angle ABC=(3x)^\circ+(x+20)^\circ

m\angle ABC=(4x+20)^\circ

m\angle ABC=(4(10)+20)^\circ

m\angle ABC=(40+20)^\circ

m\angle ABC=60^\circ

Therefore, m\angle DBC=30^\circ,m\angle ABD=30^\circ and m\angle ABC=60^\circ.

8 0
3 years ago
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