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3 years ago

8 1/7 = 2 2/5 + m M=

Mathematics

1 answer:

DIA [1.3K]3 years ago

3 0

Step-by-step explanation:

multiply them

take m on one side

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Step-by-step explanation:

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2 years ago

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To find the equation of a line, we need the slope of the line and a point on the line. Since we are requested to find the equati

Ad libitum [116K]

Answer:

$m = \frac{1}{12}$

Step-by-step explanation:

Given

$(x,y) = (36,6)$

$f(x) = \sqrt x$ ----- the equation of the curve

Required

The slope of f(x)

The slope (m) is calculated using:

$m = \lim_{h \to 0} \frac{f(a + h) - f(a)}{h}$

$(x,y) = (36,6)$ implies that:

$a = 36; f(a) = 6$

So, we have:

$m = \lim_{h \to 0} \frac{f(a + h) - f(a)}{h}$

$m = \lim_{h \to 0} \frac{f(36 + h) - 6}{h}$

If $f(x) = \sqrt x$ ; then:

$f(36 + h) = \sqrt{36 + h}$

So, we have:

$m = \lim_{h \to 0} \frac{\sqrt{36 + h} - 6}{h}$

Multiply by: $\sqrt{36 + h} + 6$

$m = \lim_{h \to 0} \frac{(\sqrt{36 + h} - 6)(\sqrt{36 + h} + 6)}{h(\sqrt{36 + h} + 6)}$

Expand the numerator

$m = \lim_{h \to 0} \frac{36 + h - 36}{h(\sqrt{36 + h} + 6)}$

Collect like terms

$m = \lim_{h \to 0} \frac{36 - 36+ h }{h(\sqrt{36 + h} + 6)}$

$m = \lim_{h \to 0} \frac{h }{h(\sqrt{36 + h} + 6)}$

Cancel out h

$m = \lim_{h \to 0} \frac{1}{\sqrt{36 + h} + 6}$

$h \to 0$ implies that we substitute 0 for h;

So, we have:

$m = \frac{1}{\sqrt{36 + 0} + 6}$

$m = \frac{1}{\sqrt{36} + 6}$

$m = \frac{1}{6 + 6}$

$m = \frac{1}{12}$

<em>Hence, the slope is 1/12</em>

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3 years ago

What is the 10th 53rd & 200th Figure if the constant difference is +4

natulia [17]

Answer:

Step-by-step explanation:

Assuming we are doing an arithmetic sequence

It depends on where you start

If the first term is a₁ and the term number is n

each subsequent term is

aₙ = a₁ + 4(n - 1)

a₁₀ = a₁ + 4(10 - 1) = a₁ + 36

a₅₃ = a₁ + 4(53 - 1) = a₁ + 208

a₂₀₀ = a₁ + 4(200 - 1) = a₁ + 796

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3 years ago

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Which polynomial function is graphed below?

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Answer:

Step-by-step explanation:

We know that the graph passes through the point $(1, 0)$ , since the line crosses the X-axis at $x = 1$ , so that means our equation has $(x - 1)$ in it. The graph also passes through the point $(-3, 0)$ , so the equation also has $(x + 3)$ in it.

This narrows it down to options C and D.

When a graph "bounces" off the X-axis at a point, like it does at $x = -3$ , this is called a "double zero", meaning two factors of the equation equal zero. This means that the graph has $(x + 3)^{2}$ , leaving us with option D.

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4 years ago