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3 years ago

Sqrt(0.32)*(2sqrt(800)-3sqrt(1800))

Mathematics

1 answer:

Iteru [2.4K]3 years ago

4 0

Answer:

paki ayos pohhh salamat or patingin Ng example para ma answer kupo salamat

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Nataly_w [17]

$\textit{circumference of a circle}\\\\ C=\pi d~~ \begin{cases} d=diameter\\[-0.5em] \hrulefill\\ d=16 \end{cases}\implies C=16\pi \implies \stackrel{\textit{3 times that much}}{C=48\pi \implies} C\approx \underset{inches}{150.8}$

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3 years ago

the three median of a triangle intersect as a point. which measurements could represent the segments of one of the medians? 2 an

topjm [15]

2 & 3 is your answer

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3 years ago

Help! First correct answer gets brainliest

lozanna [386]

All you have to do is plug in.

x = 1 , y= 8

3x+5=y

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3 years ago

Please solve this law of indices

PilotLPTM [1.2K]

4√36^-1 = 36^-1/4

36^3/4 • 36^-1/4

=36^2/4

or √36 = 6

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2 years ago

Read 2 more answers

Factor the polynomial, x2 + 5x + 6

patriot [66]

Answer:

Choice b.

$x^{2} + 5\, x + 6 = (x + 3)\, (x + 2)$ .

Step-by-step explanation:

The highest power of the variable $x$ in this polynomial is $2$ . In other words, this polynomial is quadratic.

It is thus possible to apply the quadratic formula to find the "roots" of this polynomial. (A root of a polynomial is a value of the variable that would set the polynomial to $0$ .)

After finding these roots, it would be possible to factorize this polynomial using the Factor Theorem.

Apply the quadratic formula to find the two roots that would set this quadratic polynomial to $0$ . The discriminant of this polynomial is $(5^{2} - 4 \times 1 \times 6) = 1$ .

$\begin{aligned}x_{1} &= \frac{-5 + \sqrt{1}}{2\times 1} \\ &= \frac{-5 + 1}{2} \\ &= -2\end{aligned}$ .

Similarly:

$\begin{aligned}x_{2} &= \frac{-5 - \sqrt{1}}{2\times 1} \\ &= \frac{-5 - 1}{2} \\ &= -3\end{aligned}$ .

By the Factor Theorem, if $x = x_{0}$ is a root of a polynomial, then $(x - x_0)$ would be a factor of that polynomial. Note the minus sign between $x$ and $x_{0}$ .

The root $x = -2$ corresponds to the factor $(x - (-2))$ , which simplifies to $(x + 2)$ .
The root $x = -3$ corresponds to the factor $(x - (-3))$ , which simplifies to $(x + 3)$ .

Verify that $(x + 2)\, (x + 3)$ indeed expands to the original polynomial:

$\begin{aligned}& (x + 2)\, (x + 3) \\ =\; & x^{2} + 2\, x + 3\, x + 6 \\ =\; & x^{2} + 5\, x + 6\end{aligned}$ .

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3 years ago