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2 years ago

Which statement describes the first step to solve the equation by completing the square?

Mathematics

2 answers:

Readme [11.4K]2 years ago

6 0

Answer:

It's Multiply both sides of the equation by 1/2, I literally just took the quiz.

mart [117]2 years ago

5 0

Answer:

Your answer should be B

Step-by-step explanation:

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Convert 0.9 cm into m

Whitepunk [10]

<em>Answer is 0.009 </em>

0.9 ( 0.01 m / cm ) = 0.009

4 0

2 years ago

Total number of faculty is 136. There are 6 professors to every 11 lecturers. How many professors are there

laila [671]

Answer:

The Total number of Professors = 48

Step-by-step explanation:

Given as,

The total number of Faculty = 136

And the Ratio of Professors to Lecturers = 6 Ratio 11

Now According to question

Professors : Lecturers = 6 : 11

So, Let the Number of Professors = 6x

And The Number of Lecturers = 11x

∴ 6x + 11x = 136

Or, 17x = 136

Or, x = $\frac{136}{17}$ = 8

Hence, the number of Professors = 6 × 8 = 48 Answer

3 0

2 years ago

What is 8 x 10 power of 0

Lapatulllka [165]

Question:- Find

$8 \times {10}^{0}$

Answer:- we know that

${10}^{0} = 1$

so by putting the value:-

$8 \times {10}^{0} \\ 8 \times 1 \\ 8 \: ans$

7 0

2 years ago

Read 2 more answers

In a G.P the difference between the 1st and 5th term is 150, and the difference between the

liubo4ka [24]

Answer:

Either $\displaystyle \frac{-1522}{\sqrt{41}}$ (approximately $-238$ ) or $\displaystyle \frac{1522}{\sqrt{41}}$ (approximately $238$ .)

Step-by-step explanation:

Let $a$ denote the first term of this geometric series, and let $r$ denote the common ratio of this geometric series.

The first five terms of this series would be:

$a$ ,
$a\cdot r$ ,
$a \cdot r^2$ ,
$a \cdot r^3$ ,
$a \cdot r^4$ .

First equation:

$a\, r^4 - a = 150$ .

Second equation:

$a\, r^3 - a\, r = 48$ .

Rewrite and simplify the first equation.

$\begin{aligned}& a\, r^4 - a \\ &= a\, \left(r^4 - 1\right)\\ &= a\, \left(r^2 - 1\right) \, \left(r^2 + 1\right) \end{aligned}$ .

Therefore, the first equation becomes:

$a\, \left(r^2 - 1\right) \, \left(r^2 + 1\right) = 150$ ..

Similarly, rewrite and simplify the second equation:

$\begin{aligned}&a\, r^3 - a\, r\\ &= a\, \left( r^3 - r\right) \\ &= a\, r\, \left(r^2 - 1\right) \end{aligned}$ .

Therefore, the second equation becomes:

$a\, r\, \left(r^2 - 1\right) = 48$ .

Take the quotient between these two equations:

$\begin{aligned}\frac{a\, \left(r^2 - 1\right) \, \left(r^2 + 1\right)}{a\cdot r\, \left(r^2 - 1\right)} = \frac{150}{48}\end{aligned}$ .

Simplify and solve for $r$ :

$\displaystyle \frac{r^2+ 1}{r} = \frac{25}{8}$ .

$8\, r^2 - 25\, r + 8 = 0$ .

Either $\displaystyle r = \frac{25 - 3\, \sqrt{41}}{16}$ or $\displaystyle r = \frac{25 + 3\, \sqrt{41}}{16}$ .

Assume that $\displaystyle r = \frac{25 - 3\, \sqrt{41}}{16}$ . Substitute back to either of the two original equations to show that $\displaystyle a = -\frac{497\, \sqrt{41}}{41} - 75$ .

Calculate the sum of the first five terms:

$\begin{aligned} &a + a\cdot r + a\cdot r^2 + a\cdot r^3 + a \cdot r^4\\ &= -\frac{1522\sqrt{41}}{41} \approx -238\end{aligned}$ .

Similarly, assume that $\displaystyle r = \frac{25 + 3\, \sqrt{41}}{16}$ . Substitute back to either of the two original equations to show that $\displaystyle a = \frac{497\, \sqrt{41}}{41} - 75$ .

Calculate the sum of the first five terms:

$\begin{aligned} &a + a\cdot r + a\cdot r^2 + a\cdot r^3 + a \cdot r^4\\ &= \frac{1522\sqrt{41}}{41} \approx 238\end{aligned}$ .

4 0

2 years ago

Write <, >, or = to make the statement true : .293 ___ .29

JulsSmile [24]

If you add a zero to the end of .29 it becomes .290
So, the answer is .293 > .29

7 0

2 years ago