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QveST [7]
3 years ago
14

Please help the PHOTO IS HERE

Mathematics
2 answers:
Savatey [412]3 years ago
7 0

Answer:

  • t = 13 cm

Step-by-step explanation:

<u>ABC is isosceles, therefore:</u>

  • AC = BC

<u>Substitute values:</u>

  • 2t + 5 = 3t - 8
  • 3t - 2t = 5 + 8
  • t = 13
mario62 [17]3 years ago
6 0
The answer is t= 13cm I believe! :D
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quester [9]
I think 12 or 18 it’s one of them
6 0
3 years ago
What is sine in geometry for a right triangle
aksik [14]

Answer:

The ratio of the length opposite to the side of the triangle whose angle is opposite to the length of the hyptoneuse.

7 0
2 years ago
Please help me! I'm stuck and really confused, thank you so much!
Kryger [21]

Answer:

(a) 5 * (2x - 1) = 10x - 5

(b) 6 * x = x + 2x + 3x

(c) \frac{1}{2}(x - 6) = \frac{1}{2}x - 3

(d) y(3x + 4z) = 3xy + 4yz

(e) z(2xy - 3y + 4x) = 2xyz - 3yz + 4xz

Step-by-step explanation:

Solving (a):

Given

10x - 5

Required

Express as a product

Express 10x as 5 * 2x

10x - 5 = 5 * 2x - 5

Apply distributive property

10x - 5 = 5(2x - 1)

10x - 5 = 5 * (2x - 1)

So:

5 * (2x - 1) = 10x - 5

Solving (b):

Given

x + 2x + 3x

Required

Express as a product

Express 2x and 3x as 2 * x and 3 * x, respectively

x + 2x + 3x = x + 2*x + 3*x

Apply distributive property

x + 2x + 3x = x(1 + 2 + 3)

x + 2x + 3x = x(6)

x + 2x + 3x = 6*x

So:

6 * x = x + 2x + 3x

Solving (c):

Given

\frac{1}{2}(x - 6)

Required

Express as a sum/difference

Apply distributive property

\frac{1}{2}(x - 6) = \frac{1}{2}x - \frac{1}{2}*6

\frac{1}{2}(x - 6) = \frac{1}{2}x - \frac{6}{2}

\frac{1}{2}(x - 6) = \frac{1}{2}x - 3

Solving (d):

Given

y(3x + 4z)

Required

Express as a sum/difference

Apply distributive property

y(3x + 4z) = 3x * y + 4z * y

y(3x + 4z) = 3xy + 4yz

Solving (e):

Given

2xyz - 3yz + 4xz

Required

Express as a product

Factorize

2xyz - 3yz + 4xz = 2xy * z - 3y * z + 4x * z

Apply distributive property

2xyz - 3yz + 4xz = z(2xy - 3y + 4x)

So:

z(2xy - 3y + 4x) = 2xyz - 3yz + 4xz

8 0
3 years ago
What Is the.value of 5^3i^9
gregori [183]
5^3=5\cdot5\cdot5=125\\\\i=\sqrt{-1}\to i^2=-1\\\\i^9=i^8+1=i^8\cdot i^1=-1\cdot i=-i\\\\5^3i^9=125(-i)=\boxed{-125i}
4 0
4 years ago
Approximate the real zeros f(x)=3x^4+x^2-1 to the nearest tenth
Svet_ta [14]

Answer:

Option a is correct

Step-by-step explanation:

Given: 3x^4+x^2-1

To find: roots of the equation

Solution:

A number x is a root of an equation if it satisfies the equation. It is a real root if it is also a real number.

3x^4+x^2-1

Take y=x^2

3x^4+x^2-1=3y^2+y-1

For an equation of the form ay^2+by+c=0, roots are given by y=\frac{-1\pm \sqrt{1+12}}{6}

So,

x^2=\frac{-1\pm \sqrt{1+12}}{6}=\frac{-1\pm \sqrt{13}}{6}\\x=\pm \sqrt{\left ( \frac{-1\pm \sqrt{13}}{6} \right )}

Real zeroes:

x=\pm \sqrt{\left ( \frac{-1+\sqrt{13}}{6} \right )}\\=\pm \sqrt{\left ( \frac{-1+ 3.61}{6} \right )}\\=\pm  \sqrt{\left ( \frac{2.61}{6} \right )} \\=\pm \sqrt{0.44}\\=\pm 0.7

5 0
3 years ago
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