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3 years ago

How can I estimate the square root of a non-perfect square

Mathematics

2 answers:

BlackZzzverrR [31]3 years ago

8 0

Hope this helped you :)

allsm [11]3 years ago

3 0

Use the two perfect squares it falls between to give you an accurate point of reference.

For example, \/```28 (square root of 28)
The square root of 28 falls between the square root of 25 and 36, so you know the value will be between 5 and 6.

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A line passes (-8,-2) and has a slope of 5/4. Write an equation in Ax + By=C

klemol [59]

Answer:

5x-4y = -32

Step-by-step explanation:

First write the equation in point slope form

y-y1 = m(x-x1)

y - -2 = 5/4 ( x- -8)

y+2 = 5/4 (x+8)

Multiply each side by 4 to clear the fraction

4( y+2 )= 4*5/4 (x+8)

4y +8 = 5(x+8)

4y+8 = 5x+40

Subtract 4y from each side

8 = 5x-4y +40

Subtract 40 from each side

-32 = 5x-4y

5x-4y = -32

6 0

4 years ago

Read 2 more answers

A negative number divided by a positive number is:

Verdich [7]

Answer:

When you divide a negative number by a positive number then the quotient is negative.

Step-by-step explanation:

hope this helped

7 0

3 years ago

What is the length of the line segment with end points (5, -7) and (5, 11)

Law Incorporation [45]

Answer:

Step-by-step explanation:

The formula of a distance between two points:

$d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

We have the points (5, -7) and (5, 11). Substitute:

$d=\sqrt{(5-5)^2+(11-(-7))^2}=\sqrt{0^2+18^2}=\sqrt{18^2}=18$

4 0

3 years ago

Graph the linear inequality: 2x + 3y = 12, by creating a t-chart to plot the points. Shade by

Crazy boy [7]

Answer:

See explanation

Step-by-step explanation:

Assuming the given inequality is $2x+3y\le12$

Then the corresponding linear equation is $2x+3y=12$

When x=0, we have $2*0+3y=12$

$\implies 3y=12\implies y=4$

When y=0, we have $2x+3*0=12$

$\implies 2x=12\implies x=6$

The T-table is:

0 | 4

6 | 12

We plot this points and draw a solid straight line as shown in the attachment.

Now let us test the origin: (0,0) by plugging x=0 and y=0 into the inequality.

$2*0+3*0\le12$

$0\le12$ ....This is true so we shade the lower half plane as shown in the attachment.

6 0

3 years ago

If a ball is thrown straight up into the air with an initial velocity of 70 ft/s, its height in feet after t seconds is given by

Vitek1552 [10]

Answer:

a) Average velocity at 0.1 s is 696 ft/s.

b) Average velocity at 0.01 s is 7536 ft/s.

c) Average velocity at 0.001 s is 75936 ft/s.

Step-by-step explanation:

Given : If a ball is thrown straight up into the air with an initial velocity of 70 ft/s, its height in feet after t seconds is given by $y = 70t-16t^2$ .

To find : The average velocity for the time period beginning when t = 2 and lasting. a. 0.1 s. , b. 0.01 s. , c. 0.001 s.

Solution :

a) The average velocity for the time period beginning when t = 2 and lasting 0.1 s.

$(\text{Average velocity})_{0.1\ s}=\frac{\text{Change in height}}{0.1}$

$(\text{Average velocity})_{0.1\ s}=\frac{h_{2.1}-h_2}{0.1}$

$(\text{Average velocity})_{0.1\ s}=\frac{(70(2.1)-16(2.1)^2)-(70(0.1)-16(0.1)^2)}{0.1}$

$(\text{Average velocity})_{0.1\ s}=\frac{69.6}{0.1}$

$(\text{Average velocity})_{0.1\ s}=696\ ft/s$

b) The average velocity for the time period beginning when t = 2 and lasting 0.01 s.

$(\text{Average velocity})_{0.01\ s}=\frac{\text{Change in height}}{0.01}$

$(\text{Average velocity})_{0.01\ s}=\frac{h_{2.01}-h_2}{0.01}$

$(\text{Average velocity})_{0.01\ s}=\frac{(70(2.01)-16(2.01)^2)-(70(0.01)-16(0.01)^2)}{0.01}$

$(\text{Average velocity})_{0.01\ s}=\frac{75.36}{0.01}$

$(\text{Average velocity})_{0.01\ s}=7536\ ft/s$

c) The average velocity for the time period beginning when t = 2 and lasting 0.001 s.

$(\text{Average velocity})_{0.001\ s}=\frac{\text{Change in height}}{0.001}$

$(\text{Average velocity})_{0.001\ s}=\frac{h_{2.001}-h_2}{0.001}$

$(\text{Average velocity})_{0.001\ s}=\frac{(70(2.001)-16(2.001)^2)-(70(0.001)-16(0.001)^2)}{0.001}$

$(\text{Average velocity})_{0.001\ s}=\frac{75.936}{0.001}$

$(\text{Average velocity})_{0.001\ s}=75936\ ft/s$

5 0

4 years ago