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2 years ago

How write 632.745 word form

Mathematics

2 answers:

Vladimir79 [104]2 years ago

6 0

Six hundred thirty two and seven hundred forty five thousandths

Thepotemich [5.8K]2 years ago

4 0

Six hundred thirty two and seven hundred forty five thousandths.

600 + 30 + 2 + .7 + .04 + .005

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the area of a trapezoid is 152. the height is 16 and the length of one of the sides is 6. find the length of the second parallel

kari74 [83]

Formula for trapezoid
A=h*(b1+b2)/2,

b1 an db2 are the two lengths of the sides

152 = 16*(b1+6)/2 = 8*(b1+6)=8*b1+48--> 8*b1=104, b1 = 104/8=13

A = 16*(6+13)/2=16*9.5=152

4 0

3 years ago

H(t)=(t+3)^2+5

sineoko [7]

Answer:

Option C. −4≤t≤−3

Step-by-step explanation:

we have

$H(t)=(t+3)^2+5$

This is the equation of a vertical parabola written in vertex form

The parabola open upward (the leading coefficient is positive)

The vertex is a minimum

The vertex is the point (-3,5)

The function is increasing in the interval [-3,∞)

The function is decreasing in the interval (-∞,-3]

The function will have a negative average rate when the function will be decreasing

therefore

the answer is option C

7 0

4 years ago

Evaluate the expression for x = 2 and y = 4 <br> 16x0 + 2x2 • y –1

borishaifa [10]

Answer:

Step-by-step explanation:

$16x^0+2x^2*y-1\\$

$16+2(2)^2*4-1\\16+2*4*4-1\\16+32-1=47$

7 0

4 years ago

Kyra exercises at a gym for 1 2/3 hours. she stretches,lifts weights, runs, and showers. kyra spends 1/4 hour stretching 1/3 hou

topjm [15]

Total time (1 2/3 hours) minus the time spent stretching (1/4 hour) minus the time spent showering (1/3 hour) equals the time spent running.

x= time spent running

x= 1 2/3 - (1/4 + 1/3)
change 1 2/3 to improper fraction
x= 5/3 - (1/4 + 1/3)
need to multiply by a common factor
3 and 4 go into 12 evenly
5/3= 20/12
1/4= 3/12
1/3= 4/12

x= 20/12 - (3/12 + 4/12)
x= 20/12 - 7/12
x= 13/12
change back to mixed fraction

x= 1 1/12 hours spent running

Hope this helps! :)

5 0

4 years ago

The annual yield per walnut tree is fairly constant at 140 pounds per tree when the number of trees per acre is 20 or fewer. For

Vlad [161]

Answer:

a. To maximize the annual yield for the acre the number of walnut to be planted per acre is n = 45

b. the maximum number of pounds of walnuts per acre is T = 4,050

Step-by-step explanation:

Let T represent the total number of pounds of walnuts per acre and n represent the number of trees per acre.

From the question;

For each additional tree over 20, the annual yield per tree for all trees on the acre decreases by 2 pounds due to overcrowding

The number of walnut yield per tree can be represented as;

x = 140 - (n-20)2

the total number of pounds of walnuts per acre T is;

T = n(x)

T = n(140 - (n-20)2) = 140n - 2n(n-20) = 140n - 2n^2 + 40n

T = 180n - 2n^2

Maximizing T, we will differentiate T.

T is maximum at dT/dn = 0

dT/dn = 180 - 4n = 0

4n = 180

n = 180/4 = 45

Substituting the value of into the equation of T;

T = 180(45) - 2(45^2)

T = 4,050

Therefore,

a. To maximize the annual yield for the acre the number of walnut to be planted per acre is n = 45

b. the maximum number of pounds of walnuts per acre is T = 4,050

4 0

3 years ago