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4 years ago

What are the factors of x2 – 100?

Mathematics

1 answer:

I am Lyosha [343]4 years ago

8 0

Answer:

(x-10)(x+10)

Step-by-step explanation:

Here, the given expression,

$x^2-100$

$=(x)^2-(10)^2$

By using the identity $a^2-b^2=(a+b)(a-b)$

$=(x+10)(x-10)$

$=(x-10)(x+10)$ ( By the commutative property )

Since, further factorization is not possible,

Hence, the factors of $x^2-100$ are,

$(x-10)(x+10)$

Second option is correct.

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What is the length of the hypotenuse, c? Round your answer to the nearest hundredth.

Afina-wow [57]

Answer:

Step-by-step explanation:

By using the Pythagorean theorem:

b^2+3^2=5^2
b^2+9=25

b=sqrt (25-9)

b=sqrt(16)

b=4

=> 4^2+7^2=c^2
16+49=c^2
65=c^2
=>c=sqrt(65)

c = (approximately) 8

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2 years ago

Write 0.31 Repeating<br> as a fraction in simplest form.

KengaRu [80]

Answer:

The answer is 31/100

Step-by-step explanation:

Convert the decimal number to a fraction by placing the decimal number over a power of ten. Since there are 2 numbers to the right of the decimal point, place the decimal number over 10 2 ( 100 ) . Next, add the whole number to the left of the decimal.

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3 years ago

What is the rule for the following sequence of numbers: 0,7,14,21_...?

Zielflug [23.3K]

You're adding 7 each time.

3 0

3 years ago

The bottom of a ladder must be placed 2 1/2 feet from the wall. The ladder is 9 feet long How far above ground is the ladder tou

Gnoma [55]

Answer:

Step-by-step explanation:

$h=\sqrt{9^2-(\frac{5}{2} )^2}\\=\sqrt{81-\frac{25}{4} } \\=\sqrt{\frac{299}{4} } \\=\frac{\sqrt{299} }{2} \\\approx 8.6458\\\approx 8.65 ~ft$

7 0

3 years ago

Read 2 more answers

Can i pls get some help here with how to even solve it using clear steps?

kirill [66]

first off, let's split the triplet into two equations, then from there on we'll do substitution.

$\cfrac{y}{x-z}=\cfrac{x}{y}=\cfrac{x+y}{z}\implies \begin{cases} \cfrac{y}{x-z}=\cfrac{x}{y}\\[2em] \cfrac{x}{y}=\cfrac{x+y}{z} \end{cases} \\\\[-0.35em] ~\dotfill\\\\ \stackrel{\textit{using the 1st equation}}{\cfrac{y}{x-z}=\cfrac{x}{y}\implies }y^2=\underline{x^2-xz} \\\\[-0.35em] ~\dotfill\\\\ \stackrel{\textit{using the 2nd equation}}{\cfrac{x}{y}=\cfrac{x+y}{z}\implies }xz=xy+y^2\implies \stackrel{\textit{substituting for }y^2}{xz=xy+(\underline{x^2-xz})}$

$2xz=xy+x^2\implies 2xz=x(y+x)\implies \cfrac{2xz}{x}=y+x \\\\\\ 2z=y+x\implies 2=\cfrac{y+x}{z}\implies 2=\cfrac{x+y}{z} \\\\[-0.35em] ~\dotfill\\\\ \stackrel{}{ \begin{cases} \cfrac{y}{x-z}=\cfrac{x}{y}\\[2em] \cfrac{x}{y}=\cfrac{x+y}{z} \end{cases}}\implies \begin{cases} \cfrac{y}{x-z}=\cfrac{x}{y}\\[2em] \cfrac{x}{y}=2 \end{cases}\implies \begin{cases} \cfrac{y}{x-z}=2\\[2em] \cfrac{x}{y}=2 \end{cases}$

that of course, is only true if x + y, or our numerator doesn't turn into 0, if it does then our fraction becomes 0 and our equation goes south. Keeping in mind that x,y and z are numeric values that correlate like so.

4 0

2 years ago