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3 years ago

Negative three times a number is at least 57.

Mathematics

1 answer:

EastWind [94]3 years ago

7 0

Answer:

given a number x, negative 3 times that number is -3x. If this has to be at least 57, it means that -3x must be greater than or equal to 57:

$-3\geq 57$ Dividing both sides by -3 solves the equation, but we have to switch the inequality sign because we're dividing by a negative number:

$x \leq \frac{57}{3} =19$

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1. Sarah applies the pearl guitar fret markers to her fret board as shown. Each fret marker is in the shape of a parallelogram.

Flura [38]

Answer:

315

Step-by-step explanation:

We are told in the question that each fret marker is in the shape of a parallelogram

The area of a parallelogram is given as

Base × Height.

Each of the bottoms of the fret markers has the following values

Base = 52.5mm

Height =6 mm

The area of each of the bottom fret markers

= Base × Height

= 52.5 × 6

= 315

Therefore, the area of each of the bottom fret markers without unit is 315

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3 years ago

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Alex17521 [72]

Answer:

B) y = -3x - 5

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3 years ago

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RoseWind [281]

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3 years ago

Read 2 more answers

The Product of a number m and 23

damaskus [11]

Answer:

23m

Step-by-step explanation:

The product of a number m and the numeral 23 is 23m.

6 0

3 years ago

What are the real zeroes of x3 + 6 x2 – 9x - 54?

o-na [289]

Answer:

Option B 3,-3,-6 is correct.

Step-by-step explanation:

We need to find real zeroes of $x^3+6x^2-9x-54$

Solving

$x^3+6x^2-9x-54\\=(x^3+6x^2)+(-9x-54)$

Taking x^2 common from first 2 terms and -9 from last two terms we get

$=(x^3+6x^2)+(-9x-54)\\=x^2(x+6)-9(x+6)\\$

Taking (x+6) common

$(x+6)(x^2-9)\\$

x^2-9 can be solved using formula a^2-b^2 = (a+b)(a-b)

$=(x+6)((x)^2-(3)^2)\\=(x+6)(x+3)(x-3)$

Putting it equal to zero,

$(x+6)(x+3)(x-3) =0\\x+6 =0, x+3=0\,\, and\,\, x-3=0\\x=-6, x=-3\,\, and\,\, x=3$

So, Option B 3,-3,-6 is correct.

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2 years ago