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3 years ago

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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Kryger [21]

Answer:

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

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

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

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

Step-by-step explanation:

Solving (a):

Given

$10x - 5$

Required

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$10x - 5 = 5 * 2x - 5$

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:

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

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3 years ago

What Is the.value of 5^3i^9

gregori [183]

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4 years ago

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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 $y=x^2$

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

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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:

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

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