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3 years ago

80 as a product of prime factors

Mathematics

2 answers:

Ilia_Sergeevich [38]3 years ago

5 0

2x2x2x2x5 and 2,2,2,2,5

anastassius [24]3 years ago

5 0

Here is our: 80
Multiply:
8 x 10
2 x 4 x 5 x 2
2 x 2 x 2 x 2 x 5
= 80
<span>~~~~~~~~~~~~~~~~~~
Hn, these are the steps and answer to your question.
Best regards~Sans</span>

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Solve for the equation 1+(11-7x)^1/4=4 for x.

Gwar [14]

-10.

(11-7x)^1/4 =3
(11-7x)=81
-7x=70
X=-10

4 0

3 years ago

Read 2 more answers

Explain the steps to finding the vertex of g(x) = 3x2 + 12x + 15

RUDIKE [14]

Answer:

The vertex of this parabola, $(-2, 3)$ , can be found by completing the square.

Step-by-step explanation:

The goal is to express this parabola in its vertex form:

$g(x) = a\, (x - h)^2 + k$ ,

where $a$ , $h$ , and $k$ are constants. Once these three constants were found, it can be concluded that the vertex of this parabola is at $(h,\, k)$ .

The vertex form can be expanded to obtain:

$\begin{aligned}g(x)&= a\, (x - h)^2 + k \\ &= a\, \left(x^2 - 2\, x\, h + h^2\right) + k = a\, x^2 - 2\, a\, h\, x + \left(a\,h^2 + k\right)\end{aligned}$ .

Compare that expression with the given equation of this parabola. The constant term, the coefficient for $x$ , and the coefficient for $x^2$ should all match accordingly. That is:

$\left\lbrace\begin{aligned}& a = 3 \\ & -2\,a\, h = 12 \\& a\, h^2 + k = 15\end{aligned}\right.$ .

The first equation implies that $a$ is equal to $3$ . Hence, replace the " $a\!$ " in the second equation with $3\!$ to eliminate $\! a$ :

$(-2\times 3)\, h = 12$ .

$h = -2$ .

Similarly, replace the " $a$ " and the " $h$ " in the third equation with $3$ and $(-2)$ , respectively:

$3 \times (-2)^2 + k = 15$ .

$k = 3$ .

Therefore, $g(x) = 3\, x^2 + 12\, x + 15$ would be equivalent to $g(x) = 3\, (x - (-2))^2 + 3$ . The vertex of this parabola would thus be:

$\begin{aligned}&(-2, \, 3)\\ &\phantom{(}\uparrow \phantom{,\,} \uparrow \phantom{)} \\ &\phantom{(}\; h \phantom{,\,} \;\;k\end{aligned}$ .

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3 years ago

Question number 8. Prove that 3 + √5 is an irrational number.

Marina86 [1]

Answer :

Let's assume the opposite of the statement i.e., 3 + √5 is a rational number.

$\\ \sf \: 3 + \sqrt{5} = \frac{a}{b} \\ \\ \qquad \: \tiny \sf{(where \: \: a \: \: and \: \: b \: \: are \: \: integers \: \: and \: \: b \: \neq \: 0)} \\$

$\\ \sf \: \sqrt{5} = \frac{a}{b} - 3 = \frac{a - 3b}{b} \\$

Since, a, b and 3 are integers. So,

$\\ \sf \: \frac{p - 3b}{b} \\ \\ \qquad \tiny \sf{ \: (is \: \: a \: \: rational \: \: number \:) } \\$

Here, it contradicts that √5 is an irrational number.

because of the wrong assumption that 3 + √5 is a rational number.

$\\$

Hence, 3 + √5 is an irrational number.

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3 years ago

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charle [14.2K]

Answer:

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b. $223,333.31

Step-by-step explanation:

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3 years ago

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worty [1.4K]

Answer:

2 kids plan to do cross country

Step-by-step explanation:

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3 years ago