Logx - logx - 1^2 = x

What is the answer to 36÷9 using a number line?

Katarina [22]

To divide 36/9 using the number line you have to jump from zero with length of 9 until reach 36, and the result will be the number of jumps.

I do the jumps by steps, but you can draw in the number line:

0. First jump from 0 to 9.

,

1. Second jump from 9 to 9+9=18.

,

2. Third jump from 18 to 18+9=27.

,

3. Fourth jump from 27 to 27+9=36.

,

4. Great!! We already reach 36.

So, we need four jumps of 9 to reach 36 from 0.

So, the result is 36/9=4

4 0

1 year ago

The hypothesis test of cost parameters indicates whether: a.the total cost is similar to that of the prior periods. b.the parame

maksim [4K]

Answer:

B. The Parameter are different from Zero

Step-by-step explanation:

3 0

3 years ago

A box of chalk contains 10 pieces. how many pieces of chalk are in 40 boxes

Ivanshal [37]

10 x 40 = 400 so the answer is 400 pieces of chalk.

3 0

2 years ago

Read 2 more answers

X^2+4x+y^2-10y+20=30 find the center of the circle by completing the square

swat32

Answer:

a). Center of the circle = (-2, 5)

b). Equation of the line ⇒ y = $-\frac{4}{5}x+\frac{58}{5}$

Step-by-step explanation:

Equation of the circle is,

x² + 4x + y²- 10y + 20 = 30

a). [x² + 2(2)x + 4 - 4] + [y²- 2(5)y + 25] - 25 + 20 = 30

[x² + 2(2)x + 4] - 4 + [y² - 2(5)y + 25] - 25 + 20 = 30

(x + 2)² + (y - 5)²- 29 + 20 = 30

(x + 2)² + (y - 5)²- 9 = 30

(x + 2)² + (y - 5)² = 39

By comparing this equation with the standard equation of a circle,

Center of the circle is (-2, 5).

b). A point (2, 10) lies on this circle.

Slope of the line joining this point to the center (-2, 5),

$m_{1}=\frac{y_{2}-y_{1}}{x_{2}-x_{1}}$

= $\frac{10-5}{2+2}$

= $\frac{5}{4}$

Let the slope of the tangent which is perpendicular to this line is ' $m_{2}$ '

Then by the property of perpendicular lines,

$m_{1}\times m_{2}=-1$

$\frac{5}{4}\times m_{2}=-1$

$m_{2}=-\frac{4}{5}$

Now the equation of the line passing though (2, 10) having slope $m_{2}=-\frac{4}{5}$

y - y' = $m_{2}(x-x')$

y - 10 = $-\frac{4}{5}(x-2)$

y - 10 = $-\frac{4}{5}x+\frac{8}{5}$

y = $-\frac{4}{5}x+\frac{8}{5}+10$

y = $-\frac{4}{5}x+\frac{58}{5}$

Therefore, equation of the line will be, y = $-\frac{4}{5}x+\frac{58}{5}$

7 0

3 years ago

If it takes 6 hours for a plane to travel 720km with a tail wind and 8 hours to make the return trip with a head wind. Find the

denis-greek [22]

The speed of wind and plane are 105 kmph and 15 kmph respectively.

<u>Solution:</u>

Given, it takes 6 hours for a plane to travel 720 km with a tail wind and 8 hours to make the return trip with a head wind.

We have to find the air speed of the plane and speed of the wind.

Now, let the speed wind be "a" and speed of aeroplane be "b"

And, we know that, distance = speed x time.

$\text { Now, at tail wind } \rightarrow 720=(a+b) \times 6 \rightarrow a+b=\frac{720}{6} \rightarrow a+b=120 \rightarrow(1)$

Now at head wind → $720=(a-b) \times 8 \rightarrow a-b=\frac{720}{8} \rightarrow a-b=90 \rightarrow(2)$

So, solve (1) and (2) by addition

2a = 210

a = 105

substitute a value in (1) ⇒ 105 + b = 120

⇒ b = 120 – 105 ⇒ b = 15.

Here, relative speed of plane during tail wind is 120 kmph and during head wind is 90 kmph.

Hence, speed of wind and plane are 105 kmph and 15 kmph respectively.

6 0

3 years ago

Logx - logx - 1^2 = x - 4