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2 years ago

Estimate the sum 10 1/9+ 5 14/15

Mathematics

2 answers:

mezya [45]2 years ago

7 0

Answer: $16\frac{2}{45}$

Turn 10 1/9 into an Improper Fraction

Multiply $10*9$ and get $90$ . Now add $1$ and get $91$ .

$10\frac{1}{9} =91/9$

Turn 5 14/15 into an Improper Fraction

Multiply $5*15$ and get $75$ . Now add $14$ and get $89$ .

$5\frac{14}{15} =89/15$

New Problem: $91/9+89/15$

Find Common Denominator

$9: 9,18,27,36,45\\15: 15,30,45,60,75\\CD=45$

New Problem: $91/45+89/45$

Add

$91/45+89/45=722/45$

Simplify

$722/45=16\frac{2}{45}$

cupoosta [38]2 years ago

6 0

Answer:

$16 \frac{2}{45}$

Step-by-step explanation:

$10 \frac{1}{9} + 5 \frac{14}{15} \\ \frac{91}{9} + \frac{89}{15} \\ \frac{91 \times 5}{9 \times 5} + \frac{89 \times 3}{15 \times 3} \\ \frac{455}{45} + \frac{267}{45} \\ = \frac{722}{45} \\ = 16 \frac{2}{45}$

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Divide x^4 + 7 by x - 3.

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Answer:

Step-by-step explanation:

Divide using synthetic division, remembering to place a zero to denote the terms x³, x² and x

Since division by (x - 3) then evaluate using h = 3

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Intersection point of Y=logx and y=1/2log(x+1)

GalinKa [24]

Answer:

The intersection is $(\frac{1+\sqrt{5}}{2},\log(\frac{1+\sqrt{5}}{2})$ .

The Problem:

What is the intersection point of $y=\log(x)$ and $y=\frac{1}{2}\log(x+1)$ ?

Step-by-step explanation:

To find the intersection of $y=\log(x)$ and $y=\frac{1}{2}\log(x+1)$ , we will need to find when they have a common point; when their $x$ and $y$ are the same.

Let's start with setting the $y$ 's equal to find those $x$ 's for which the $y$ 's are the same.

$\log(x)=\frac{1}{2}\log(x+1)$

By power rule:

$\log(x)=\log((x+1)^\frac{1}{2})$

Since $\log(u)=\log(v)$ implies $u=v$ :

$x=(x+1)^\frac{1}{2}$

Squaring both sides to get rid of the fraction exponent:

$x^2=x+1$

This is a quadratic equation.

Subtract $(x+1)$ on both sides:

$x^2-(x+1)=0$

$x^2-x-1=0$

Comparing this to $ax^2+bx+c=0$ we see the following:

$a=1$

$b=-1$

$c=-1$

Let's plug them into the quadratic formula:

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

$x=\frac{1 \pm \sqrt{(-1)^2-4(1)(-1)}}{2(1)}$

$x=\frac{1 \pm \sqrt{1+4}}{2}$

$x=\frac{1 \pm \sqrt{5}}{2}$

So we have the solutions to the quadratic equation are:

$x=\frac{1+\sqrt{5}}{2}$ or $x=\frac{1-\sqrt{5}}{2}$ .

The second solution definitely gives at least one of the logarithm equation problems.

Example: $\log(x)$ has problems when $x \le 0$ and so the second solution is a problem.

So the $x$ where the equations intersect is at $x=\frac{1+\sqrt{5}}{2}$ .

Let's find the $y$ -coordinate.

You may use either equation.

I choose $y=\log(x)$ .

$y=\log(\frac{1+\sqrt{5}}{2})$

The intersection is $(\frac{1+\sqrt{5}}{2},\log(\frac{1+\sqrt{5}}{2})$ .

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≈≈≈≈ Glad I Could Help, And Good Luck! ≈≈≈≈

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Step-by-step explanation:

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