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borishaifa [10]
3 years ago
10

Is 3/10 greater than 5/10

Mathematics
2 answers:
allochka39001 [22]3 years ago
4 0
3 is always less than 5 in fractions and whole numbers. Think of it as 3 parts of ten squares are shaded and 5 out of ten squares shades, which it more. The answer is 3/10 is greater than 5/10
storchak [24]3 years ago
3 0

3 of anything is almost always less than 5 of the same thing.
That's true for cows, rocks, trees, fish, salt-shakers, and babies.
I can't think of any object where it wouldn't be true.
Why wouldn't it be true for tenths ?

Let me say it another way:

No.  3/10  is <em>less than</em>  5/10 .


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alexdok [17]
N+14=17. n=3. So the missing number is 3.
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Match each sentence with the inequality that could represent the situation
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Make <br> x<br> the subject of the formula<br> x<br> −<br> 3<br> =<br> q
lakkis [162]

Answer:

x = q + 3

General Formulas and Concepts:

<u>Pre-Algebra</u>

Order of Operations: BPEMDAS

  1. Brackets
  2. Parenthesis
  3. Exponents
  4. Multiplication
  5. Division
  6. Addition
  7. Subtraction
  • Left to Right

Equality Properties

Step-by-step explanation:

<u>Step 1: Define Equation</u>

x - 3 = q

<u>Step 2: Solve for </u><em><u>x</u></em>

  1. Add 3 to both sides:          x = q + 3
8 0
2 years ago
Factoring by grouping how do i solve it 8r^3-64^2+r-8
Lelu [443]
Squaring a number is the same as multiplying the number by itself <span>(<span><span>64⋅64</span>).</span></span><span> In this case, </span>64<span> squared is </span><span>4096.
</span><span>8<span>r^3</span>−4096+r−8

</span>Subtract 8<span> from </span><span>−4096</span><span> to get </span><span><span>−4104</span>.
</span><span>8<span>r^3</span>−4104+r

</span>Factor<span> the </span>polynomial<span> using the rational </span>roots<span> theorem.
</span><span>(r−8)(8<span>r^2</span>+64r+513)</span>
4 0
2 years ago
Solving a trigonometric equation involving an angle multiplied by a constant
PIT_PIT [208]

In these questions, we need to follow the steps:

1 - solve for the trigonometric function

2 - Use the unit circle or a calculator to find which angles between 0 and 2π gives that results.

3 - Complete these angles with the complete round repetition, by adding

2k\pi,k\in\Z

4 - these solutions are equal to the part inside the trigonometric function, so equalize the part inside with the expression and solve for <em>x</em> to get the solutions.

1 - To solve, we just use algebraic operations:

\begin{gathered} \sqrt[]{3}\tan (3x)+1=0 \\ \sqrt[]{3}\tan (3x)=-1 \\ \tan (3x)=-\frac{1}{\sqrt[]{3}} \\ \tan (3x)=-\frac{\sqrt[]{3}}{3} \end{gathered}

2 - From the unit circle, we can see that we will have one solution from the 2nd quadrant and one from the 4th quadrant:

The value for the angle that give positive

+\frac{\sqrt[]{3}}{3}

is known to be 30°, which is the same as π/6, so by symmetry, we can see that the angles that have a tangent of

-\frac{\sqrt[]{3}}{3}

Are:

\begin{gathered} \theta_1=\pi-\frac{\pi}{6}=\frac{5\pi}{6} \\ \theta_2=2\pi-\frac{\pi}{6}=\frac{11\pi}{6} \end{gathered}

3 - to consider all the solutions, we need to consider the possibility of more turn around the unit circle, so:

\begin{gathered} \theta=\frac{5\pi}{6}+2k\pi,k\in\Z \\ or \\ \theta=\frac{11\pi}{6}+2k\pi,k\in\Z \end{gathered}

Since 5π/6 and 11π/6 are π radians apart, we can put them together into one expression:

\theta=\frac{5\pi}{6}+k\pi,k\in\Z

4 - Now, we need to solve for <em>x</em>, because these solutions are for all the interior of the tangent function, so:

\begin{gathered} 3x=\theta \\ 3x=\frac{5\pi}{6}+k\pi,k\in\Z \\ x=\frac{5\pi}{18}+\frac{k\pi}{3},k\in\Z \end{gathered}

So, the solutions are:

x=\frac{5\pi}{18}+\frac{k\pi}{3},k\in\Z

4 0
1 year ago
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