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3 years ago

Please show full solutions. Will mark brainliest for THE BEST ANSWER. THANK YOU.

Mathematics

1 answer:

aivan3 [116]3 years ago

7 0

Answer:

Step-by-step explanation:

12x² + 17x -5 = 12x² + 20x - 3x - 5

= 4x(3x + 5) -(3x + 5)

= (3x + 5)(4x - 1)

6x² + 7x - 5 = 6x² + 10x - 3x - 5

= 2x(3x + 5) -(3x+5)

= (3x + 5)(2x - 1)

$Area =\dfrac{1}{2}*b*h$

$= \dfrac{1}{2}*(4x -1) *\dfrac{6x^{2}+7x-5}{12x^{2}+17x-5}\\\\=\dfrac{1}{2}*(4x-1)*\dfrac{(3x+5)(2x-1)}{(3x+5)(4x-1)}\\\\=\dfrac{1}{2}*1*\dfrac{ 2x-1}{1}\\\\=\dfrac{2x-1}{2}$

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alukav5142 [94]

Answer:

x = 4

Step-by-step explanation:

$\frac{12}{16} =\frac{5x-2}{6x}$

$6x(\frac{12}{16} )=5x-2$

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$72x = 16(5x-2)$

$72x=80x-32$

$8x=32$

$x=32/8=4$

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2 years ago

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Answer:

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Step-by-step explanation:

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3 years ago

Read 2 more answers

4. What expression represents the width of the room?

Irina18 [472]

Answer:

35-4

Step-by-step explanation:

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3 years ago

Read 2 more answers

Please help me out!!!!!!! 54 points!

statuscvo [17]

Answer:

1. (x+5)(x-1)

6. (3x+4)(x-6)

9. (x+5)(x-5)

Step-by-step explanation:

To factor an equation, break the polynomial into two binomials which multiply to make it. There are three main ways to factor:

Regular: Find two numbers that multiply to C and add to B from $ax^2 + bx + c$ .

$x^2+4x-5 = (x+5)(x-1)$

Grouping: Find two numbers that multiply to A*C and add to B. Then split B into the two factors and factor by GCF.

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Difference of Squares: Factor by taking the square root of each term. $a^2-b^2 = (a-b)(a+b)$

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3 years ago

Select the functions that have a value of -1. sin180° cos180° tan180° csc180° sec180° cot180°

kupik [55]

We have to break each degree in terms of 90

A) $sin180^\circ=sin(90\times2+0)$

Which is in third quadrant, therefore sine is negative hence

$sin(90\times2+0)= -sin0 ^\circ = 0$

B) $cos180^\circ =cos(90\times2+0)$

Which is in third quadrant, therefore cosine is negative hence

$cos(90\times2+0)= -cos0^\circ = -1$

C) $tan180^\circ=tan(90\times2+0)$

Which is in third quadrant, therefore tangent is positive hence

$tan(90\times2+0)= tan0^\circ = 0$

D) $csc180^\circ=csc(90\times2+0)$

Which is in third quadrant, therefore cosec is negative hence

$cosec(90\times2+0)= -csc0^\circ =$ not defined

E) $sec180^\circ=sec(90\times2+0)$

Which is in third quadrant, therefore secant is negative hence

$sec(90\times2+0)= -sec0^\circ = -1$

F) $cot180^\circ=cot(90\times2+0)$

Which is in third quadrant, therefore tangent is positive hence

$cot(90\times2+0)= cot0^\circ =$ not defined

Hence only $cos 180^\circ$ and

$sec180^\circ$ have value -1

Hope this will help

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3 years ago