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Westkost [7]
3 years ago
9

The area of a square field is 1 17/64 m2. What is the perimeter of the square field? Can some1 say this ans fast pls

Mathematics
1 answer:
AURORKA [14]3 years ago
4 0

Answer:

5.41 m

Step-by-step explanation:

First, let's find a side length by taking the square root of the area.

117/64 ^ 1/2 = 1.352...

Next, we need to multiply by 4.

1.352... x 4 = 5.408...

= approximately 5.41

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Blababa [14]

Answer:

d) 500 scoops

Step-by-step explanation:

looking at the trend of scoops sold per year, 500 seems to be the correct answer. let me know if this is wrong.

8 0
2 years ago
Read 2 more answers
Please HELP!!!!! I need to know the missing value. PLease!!!!!!!!!!!!!!
GuDViN [60]

Answer:

w = 21.4 ft

Step-by-step explanation:

Area of Trapezoid = Area = (a+b)/2 * h

Where a and b are the bases of trapezoid, and h is the height.

In the figure we are given:

a= 32.3 ft

b=w=?

h= 22.9 ft

Area = 614.865 ft²

Putting values in formula

Area = (a+b)/2 * h

614.865 = (32.3 + w) / 2 * 22.9

1229.73 = 32.3 + w * 22.9

1229.73 / 22.9 = 32.3 +w

53.7 -32.3 = w

=> w = 21.4 ft

7 0
3 years ago
A cruise ship is traveling south going approximately 22 mph when it hits the Gulf Stream flowing east at 4 mph. Which expression
alexira [117]
<h2>Hello!</h2>

The answer is:

The correct option is

a) Resultant=\sqrt{22^{2} +4^{2} }

<h2>Why?</h2>

Since a right triangle is formed, we can calculate the resultant force using the Pythagorean Theorem which states that:

c=\sqrt{a^{2}+b^{2}}

Where,

c, is the hypothenuse.

a, is one triangle leg.

b, is the other triangle leg.

So, we are given the information:

a=22mph

b=4mph

So, calculating the resultant force (hypothenuse), we have:

Resultant=\sqrt{22^{2} +4^{2} }

Hence, the expression that would find the result velocity is:

a) [tex]Resultant=\sqrt{22^{2} +4^{2} }[/tex]

Have a nice day!

4 0
3 years ago
Y=1/2x-1
Alinara [238K]

Answer:

<h2>Y=1/2x-1</h2>

Find where the expression

1/2x−1 is undefined.

x=1/2

Consider the rational function

R(x)=ax^n/bx^m where n is the degree of the numerator and m is the degree of the denominator.

1. If n<m, then the x-axis, y=0, is the horizontal asymptote.

2. If n=m, then the horizontal asymptote is the line y = ab.

3. If n>m, then there is no horizontal asymptote (there is an oblique asymptote).

Find n and m.n=0m=1

Since n<m, the x-axis, y=0, is the horizontal asymptote.y=0

There is no oblique asymptote because the degree of the numerator is less than or equal to the degree of the denominator.

No Oblique Asymptotes

This is the set of all asymptotes.

Vertical Asymptotes:

x=1/2

Horizontal Asymptotes:

y=0

No Oblique Asymptotes

<h2>2x+y=4</h2>

Subtract 2x from both sides of the equation.

y=4−2x

Rewrite in slope-intercept form.

The slope-intercept form is

y=mx+b, where m is the slope and b is the y-intercept.

y=mx+b

Reorder 4 and −2x.

y=−2x+4

Use the slope-intercept form to find the slope and y-intercept.

Find the values of m and b using the form

y=mx+b.

m= - 2b=4

The slope of the line is the value of m, and the y-intercept is the value of b.

Slope: −2

y-intercept: (0,4)

Any line can be graphed using two points. Select two x values, and plug them into the equation to find the corresponding y

values.

Find the x-intercept.

x-intercept(s):

(2,0)

Find the y-intercept.

y-intercept(s):

(0,4)

Create a table of the x and y values.

x y

0 4

2 0

Graph the line using the slope and the y-intercept, or the points.

Slope: −2

y-intercept:

(0,4)

Step-by-step explanation:

Hope it is helpful...

8 0
3 years ago
Please read the attachment that is there.
Firlakuza [10]

A is equivalent to -(8/3)

8 0
3 years ago
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