Use the digits in the number 531,642 to write a number that is greater than and a number that is less than 531,642.

Six decreased by 9 times a number is -39

Andrew [12]

The translation of the mathematical statement given as six decreased by 9 times a number is -39 is 9x - 6 = -39

<h3>How to translate the mathematical statement?</h3>

The mathematical statement is given as:

six decreased by 9 times a number is -39

9 times a number means

9 * x (assume x is the number)

So, we have:

six decreased by 9 times a number is -39 ⇒ six decreased by 9x is -39

Decreased by means we subtract 6 from 9x.

So, we have:

six decreased by 9 times a number is -39 ⇒ 9x - 6 = -39

Hence, the translation of the mathematical statement given as six decreased by 9 times a number is -39 is 9x - 6 = -39

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5 0

2 years ago

K/2 + 9 = 30 What is k

Tcecarenko [31]

K/2+9=30 1. Start by getting rid of the number farthest from K
-9 -9      2. The nines cancel out, 30-9=21
K/2 = 21   3. Next, get the 2 away from K
/-2 /-2 4. The 2 cancel out when divided, 21 divided by -2= -10.5
K = -10.5
K=-10.5

4 0

4 years ago

Use the unit circle to find the value of tan 30

Harlamova29_29 [7]

Answer:

$\frac{\sqrt{3}}{3}$

Explanation:

$\sin(30^{\circ}) = \frac{1 }{2 }$

$\cos(30^{\circ}) = \frac{\sqrt{3} }{2 }$

$\tan(30^{\circ}) = \frac{\text{opposite} }{\text{adjacent} }$

$\tan(30^{\circ}) = \frac{\frac{1 }{2 }} {\frac{\sqrt{3} }{ 2}}$

$\tan(30^{\circ}) = \frac{1 }{2 } \times{\frac{2}{\sqrt{3} }}$

$\tan(30^{\circ}) = \frac{1}{\sqrt{3}}$

$\tan(30^{\circ}) = \frac{\sqrt{3}}{3}$

7 0

3 years ago

When completing the square on the equation c2 + 110 = 12, the resulting solution is:<br> It

Delvig [45]

Answer:

$c=1$ and $c=-12$

Step-by-step explanation:

The correct equation is:

$c^2+11c=12$

To solve by completing the square method.

Solution:

We have:

$c^2+11c=12$

In order to solve by completing the square method we will carry out the following operations to the given equation to get a perfect square binomial.

We can write as:

$c^2+2.\frac{11}{2}c=12$

$c^2+2.\frac{11}{2}c+(\frac{11}{2})^2-(\frac{11}{2})^2=12$

$(c+\frac{11}{2})^2-(\frac{11}{2})^2=12$ [As $c^2+2.\frac{11}{2}c+(\frac{11}{2})^2=(c+\frac{11}{2})^2$ ]

Adding both sides by $(\frac{11}{2})^2$

$(c+\frac{11}{2})^2-(\frac{11}{2})^2+(\frac{11}{2})^2=12+(\frac{11}{2})^2$

$(c+\frac{11}{2})^2=12+\frac{121}{4}$

Taking LCD to add fraction.

$(c+\frac{11}{2})^2=\frac{48}{4}+\frac{121}{4}$

$(c+\frac{11}{2})^2=\frac{169}{4}$

Taking square root both sides.

$\sqrt{(c+\frac{11}{2})^2}=\sqrt{\frac{169}{4}}$

$c+\frac{11}{2}=\pm\frac{13}{2}$

Subtracting both sides by $\frac{11}{2}$ :

$c+\frac{11}{2}-\frac{11}{2}=\pm\frac{13}{2}-\frac{11}{2}$

$c=\pm\frac{13}{2}-\frac{11}{2}$

So, we have:

$c=\frac{13}{2}-\frac{11}{2}$ and $c=-\frac{13}{2}-\frac{11}{2}$

$c=\frac{2}{2}$ and $c=\frac{-24}{2}$

$c=1$ and $c=-12$ (Answer)

6 0

4 years ago

the sum of the first n terms of a sequence is 20 - 10 / 2 ^ n - 1 find the sum of the first five terms

Travka [436]

Answer:

$S_5=\dfrac{170}{9}$

Step-by-step explanation:

Given that,

The sum of first n term of a sequence is $S_n=20-\dfrac{10}{2n-1}$

We need to find the sum of the first 5 terms.

Put n = 5,

$S_5=20-\dfrac{10}{2(5)-1}\\\\=20-\dfrac{10}{10-1}\\\\=20-\dfrac{10}{9}\\\\=\dfrac{170}{9}$

So, the sum of first 5 terms is equal to $\dfrac{170}{9}$ .

6 0

3 years ago