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3 years ago

What is the distance from (−4, 0) to (2, 5)? Round your answer to the nearest hundredth. Explain your answer.

Mathematics

1 answer:

11Alexandr11 [23.1K]3 years ago

5 0

Answer:

A 7.81

Step-by-step explanation:

Use the distance formula to find the distance between coordinates.

$d = \sqrt{(2--4)^2 + (5-0)^2} \\d = \sqrt{(2+4)^2 + (5)^2} \\d = \sqrt{(6)^2 + (5)^2} \\d = \sqrt{36 + 25} \\d = \sqrt{61} \\d=7.81$

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1.5/4x+1=0.4/x+4 PLEASE HELP ITS PROPORTIONS

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Answer:

Proportion states that the two fractions or ratios are equal

Given the equation: $\frac{1.5}{4x+1} = \frac{0.4}{x+4}$

By cross multiply we get;

$1.5(x+4) = 0.4(4x+1)$

Using distributive property; $a\cdot (b+c) = a\cdot b+ a\cdot c$

$1.5x + 6= 1.6x + 0.4$

Subtract 0.4 from both sides we get;

$1.5x +5.6= 1.6x$

Subtract 1.5x from both sides we get;

$5.6= 0.1x$

Divide both sides by 0.1 we get;

$x = \frac{5.6}{0.1}$

Simplify:

x = 56

Therefore, the value of x that satisfy the equation $\frac{1.5}{4x+1} = \frac{0.4}{x+4}$ is, 56

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2 years ago

Write each word form in standard form. <br> Four thousand

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4,000 +000 +00 + 0 is stander form for four thousand

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3 years ago

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1) Determine the discriminant of the 2nd degree equation below:

Aleksandr-060686 [28]

$\LARGE{ \boxed{ \mathbb{ \color{purple}{SOLUTION:}}}}$

We have, Discriminant formula for finding roots:

$\large{ \boxed{ \rm{x = \frac{ - b \pm \: \sqrt{ {b}^{2} - 4ac} }{2a} }}}$

Here,

x is the root of the equation.
a is the coefficient of x^2
b is the coefficient of x
c is the constant term

1) Given,

3x^2 - 2x - 1

Finding the discriminant,

➝ D = b^2 - 4ac

➝ D = (-2)^2 - 4 × 3 × (-1)

➝ D = 4 - (-12)

➝ D = 4 + 12

➝ D = 16

2) Solving by using Bhaskar formula,

❒ p(x) = x^2 + 5x + 6 = 0

$\large{ \rm{ \longrightarrow \: x = \dfrac{ - 5\pm \sqrt{( - 5) {}^{2} - 4 \times 1 \times 6 }} {2 \times 1}}}$

$\large{ \rm{ \longrightarrow \: x = \dfrac{ - 5 \pm \sqrt{25 - 24} }{2 \times 1} }}$

$\large{ \rm{ \longrightarrow \: x = \dfrac{ - 5 \pm 1}{2} }}$

So here,

$\large{\boxed{ \rm{ \longrightarrow \: x = - 2 \: or - 3}}}$

❒ p(x) = x^2 + 2x + 1 = 0

$\large{ \rm{ \longrightarrow \: x = \dfrac{ - 2 \pm \sqrt{ {2}^{2} - 4 \times 1 \times 1} }{2 \times 1} }}$

$\large{ \rm{ \longrightarrow \: x = \dfrac{ - 2 \pm \sqrt{4 - 4} }{2} }}$

$\large{ \rm{ \longrightarrow \: x = \dfrac{ - 2 \pm 0}{2} }}$

So here,

$\large{\boxed{ \rm{ \longrightarrow \: x = - 1 \: or \: - 1}}}$

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$\large{ \rm{ \longrightarrow \: x = \dfrac{ - ( - 1) \pm \sqrt{( - 1) {}^{2} - 4 \times 1 \times ( - 20) } }{2 \times 1} }}$

$\large{ \rm{ \longrightarrow \: x = \dfrac{ 1 \pm \sqrt{1 + 80} }{2} }}$

$\large{ \rm{ \longrightarrow \: x = \dfrac{1 \pm 9}{2} }}$

So here,

$\large{\boxed{ \rm{ \longrightarrow \: x = 5 \: or \: - 4}}}$

❒ p(x) = x^2 - 3x - 4 = 0

$\large{ \rm{ \longrightarrow \: x = \dfrac{ - ( - 3) \pm \sqrt{( - 3) {}^{2} - 4 \times 1 \times ( - 4) } }{2 \times 1} }}$

$\large{ \rm{ \longrightarrow \: x = \dfrac{3 \pm \sqrt{9 + 16} }{2 \times 1} }}$

$\large{ \rm{ \longrightarrow \: x = \dfrac{3 \pm 5}{2} }}$

So here,

$\large{\boxed{ \rm{ \longrightarrow \: x = 4 \: or \: - 1}}}$

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Step-by-step explanation:

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Write the coordinate of G'after G(1, 2) was reflected over the line y=x.

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Step-by-step explanation:

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