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3 years ago

The difference of 54 and a number

Mathematics

2 answers:

jeka943 years ago

4 0

Answer would be 54 - x

melisa1 [442]3 years ago

3 0

Since there's nothing to subtract from, we have an UNKNOWN.

An UNKNOWN is represented by an "x"

So, the answer would be: 54 - x

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Please show me how you find x and determine what x is Thank you I appreciate it :)

AfilCa [17]

well, let's notice something, we have two 45° angles, now each of those will give use an equal opposite side, namely if that one side is "x", the other is her twin, "x" as well.

is a right-triangle, so we can use the pythagorean theorem, and we know the sides are x,x,6.

$\bf \textit{using the pythagorean theorem} \\\\ c^2=a^2+b^2\implies 6^2=x^2+x^2 \qquad \begin{cases} c=\stackrel{hypotenuse}{6}\\ a=\stackrel{adjacent}{x}\\ b=\stackrel{opposite}{x}\\ \end{cases} \\\\\\ 36=2x^2\implies \cfrac{36}{2}=x^2\implies 18=x^2\implies \sqrt{18}=x\implies \sqrt{9\cdot 2}=x \\\\\\ \sqrt{3^2\cdot 2}=x\implies \boxed{3\sqrt{2}=x}$

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3 years ago

The equations y = 1/2x + 1 and y= √2x+4 are graphed on the coordinate grid.

lana [24]

Answer:

Step-by-step explanation:

The graphs intersect at 2 points.

The 2 solutions are x = -2 and x = 6.

3 0

2 years ago

Find b, given that a = 20, angle A = 30°, and angle B = 45° in triangle ABC

Kamila [148]

The side b is opposite to the angle B, applying the law of the sines, we have:

 $\frac{a}{sinA} = \frac{b}{sinB}$
$\frac{20}{sin30^0} = \frac{b}{sin45^0}$
$\frac{20}{ \frac{1}{2} } = \frac{b}{ \frac{ \sqrt{2} }{2} }$
$20* \frac{ \sqrt{2} }{2} = b* \frac{1}{2}$
$\frac{20 \sqrt{2} }{2} = \frac{b}{2}$
$2*b =2*20 \sqrt{2}$
$2b = 40 \sqrt{2}$
$b = \frac{40 \sqrt{2} }{2}$
$\boxed{b = 20 \sqrt{2} }$

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3 years ago

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Help if x ≠ 3, how would I write x^2 - 9/ x^2 - 3x?

maria [59]

Do you want me to write it for you in standard form?

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3 years ago

Is 20x9x5 a associative or distributive property?

Oduvanchick [21]

It is neither. Distributive is like 20x9 20x5 then add those products. Associative is like (20x9) x5. You can however use both for this equation

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3 years ago

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