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2 years ago

Whoever answers this first gets brainliest Pls help ASAP

Mathematics

2 answers:

Vaselesa [24]2 years ago

8 0

Answer:

1. Rational Number

2. Irrational Number

3. Terminating Decimal

4. Repeating Decimal

Step-by-step explanation:

A Rational Number can be written as a ratio of two integers. For example, as a simple fraction. That leaves us for the answer to the second one as irrational. To terminate means to end, so a terminating decimal is a decimal that does not go on forever. Whereas, a repeating decimal continues repeating forever (or indefinitely).

I would like to apologize in advance if any of these were incorrect.

Mandarinka [93]2 years ago

7 0

Answer:

Terminating goes to 3

Repeating goes to 4

Rational goes to 1

Irrational goes to 2

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3 years ago

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Irina-Kira [14]

I think it’s 0.85d, but i’m not sure

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3 years ago

I’m confused on this one

GenaCL600 [577]

The slope of CD is

$m = \dfrac{6 - -4}{2 - 0} = 5$

The slope of the perpendicular bisector is the negative reciprocal,

$n = -1/m = - \dfrac 1 5$

The perpendicular bisector passes through M, the midpoint of CD

M =( (0+2)/2, (-4 + 6)/2) = (1, -1)

So point slope form for the perpendicular bisector is

$y - -1 = - \frac 1 5(x - 1)$

Solving for y gives slope intercept form.

$y = - \frac 1 5 x - \frac 4 5$

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Answer: y=(-1/5)x-4/5

3 0

3 years ago

Find all roots: x^3 + 7x^2 + 12x = 0 Show all work and check your answer.

Aliun [14]

The three roots of x^3 + 7x^2 + 12x = 0 is 0,-3 and -4

Solution:

We have been given a cubic polynomial.

$x^{3}+7 x^{2}+12 x=0$

We need to find the three roots of the given polynomial.

Since it is a cubic polynomial, we can start by taking ‘x’ common from the equation.

This gives us:

$x^{3}+7 x^{2}+12 x=0$

$x\left(x^{2}+7 x+12\right)=0$ ----- eqn 1

So, from the above eq1 we can find the first root of the polynomial, which will be:

x = 0

Now, we need to find the remaining two roots which are taken from the remaining part of the equation which is:

$x^{2}+7 x+12=0$

we have to use the quadratic equation to solve this polynomial. The quadratic formula is:

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

Now, a = 1, b = 7 and c = 12

By substituting the values of a,b and c in the quadratic equation we get;

$\begin{array}{l}{x=\frac{-7 \pm \sqrt{7^{2}-4 \times 1 \times 12}}{2 \times 1}} \\\\{x=\frac{-7 \pm \sqrt{1}}{2}}\end{array}$

Therefore, the two roots are:

$\begin{array}{l}{x=\frac{-7+\sqrt{1}}{2}=\frac{-7+1}{2}=\frac{-6}{2}} \\\\ {x=-3}\end{array}$

And,

$\begin{array}{c}{x=\frac{-7-\sqrt{1}}{2}} \\\\ {x=-4}\end{array}$

Hence, the three roots of the given cubic polynomial is 0, -3 and -4

4 0

3 years ago

1 fourth B minus 10 equals 7

creativ13 [48]

$\frac{1}{4}$ b - 10 = 7
Take 10 to the other side.

$\frac{1}{4}$ b = 17

Multiply by 4 to isolate b
$\frac{1}{4}$ b × 4 = 17 × 4

4 and 4 cancels out

b = 68

3 0

3 years ago