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marishachu [46]
3 years ago
15

When labeling a pre-image and it's image, we use a small tick mark. It's called a(n):

Mathematics
2 answers:
Dmitriy789 [7]3 years ago
7 0

Answer:

It’s a prime

Step-by-step explanation:

IgorLugansk [536]3 years ago
4 0

Answer:

Prime

Step-by-step explanation:

If we have a point on a pre-image, we can have it labelled as A(x,y)

Now, we can write this as the image

To write this as the image, we simply have to add a mark on the A

This mark is at the upper right

We have it as A’ (x,y)

The part added is called a prime

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The length of the base of the plot is 12 feet, and the length of one side is 5 feet, as shown in the diagram. Tammy makes a scal
Musya8 [376]
First, find the area of the plot.

Area = length x width
Area = 12 x 5
Area = 60 square feet.

The ratio of the scale to the plot is 1 inch : 2 feet

Square each of those:

1 square inch : 4 square feet

The plot is 60 square feet. 60 sq ft / 4 sq ft = 15

The area of the plot in the scale drawing is 15 square inches.
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4 years ago
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PLEASE HELP! A company makes triangular plates for individual slices of pizza. For each plate, the base is 7 inches and the heig
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3 years ago
A(t)=.892t^3-13.5t^2+22.3t+579 how to solve this
Minchanka [31]

Answer:

t = (5 ((446 sqrt(3188516012553) - 827891226)^(1/3) - 204292 (-1)^(2/3) (3/(413945613 - 223 sqrt(3188516012553)))^(1/3)))/(223 6^(2/3)) + 1125/223 or t = 1125/223 - (5 ((-2)^(1/3) (223 sqrt(3188516012553) - 413945613)^(1/3) - 204292 (-3/(413945613 - 223 sqrt(3188516012553)))^(1/3)))/(223 6^(2/3)) or t = 1125/223 - (5 ((827891226 - 446 sqrt(3188516012553))^(1/3) + 204292 (3/(413945613 - 223 sqrt(3188516012553)))^(1/3)))/(223 6^(2/3))

Step-by-step explanation:

Solve for t over the real numbers:

0.892 t^3 - 13.5 t^2 + 22.3 t + 579 = 0

0.892 t^3 - 13.5 t^2 + 22.3 t + 579 = (223 t^3)/250 - (27 t^2)/2 + (223 t)/10 + 579:

(223 t^3)/250 - (27 t^2)/2 + (223 t)/10 + 579 = 0

Bring (223 t^3)/250 - (27 t^2)/2 + (223 t)/10 + 579 together using the common denominator 250:

1/250 (223 t^3 - 3375 t^2 + 5575 t + 144750) = 0

Multiply both sides by 250:

223 t^3 - 3375 t^2 + 5575 t + 144750 = 0

Eliminate the quadratic term by substituting x = t - 1125/223:

144750 + 5575 (x + 1125/223) - 3375 (x + 1125/223)^2 + 223 (x + 1125/223)^3 = 0

Expand out terms of the left hand side:

223 x^3 - (2553650 x)/223 + 5749244625/49729 = 0

Divide both sides by 223:

x^3 - (2553650 x)/49729 + 5749244625/11089567 = 0

Change coordinates by substituting x = y + λ/y, where λ is a constant value that will be determined later:

5749244625/11089567 - (2553650 (y + λ/y))/49729 + (y + λ/y)^3 = 0

Multiply both sides by y^3 and collect in terms of y:

y^6 + y^4 (3 λ - 2553650/49729) + (5749244625 y^3)/11089567 + y^2 (3 λ^2 - (2553650 λ)/49729) + λ^3 = 0

Substitute λ = 2553650/149187 and then z = y^3, yielding a quadratic equation in the variable z:

z^2 + (5749244625 z)/11089567 + 16652679340752125000/3320419398682203 = 0

Find the positive solution to the quadratic equation:

z = (125 (223 sqrt(3188516012553) - 413945613))/199612206

Substitute back for z = y^3:

y^3 = (125 (223 sqrt(3188516012553) - 413945613))/199612206

Taking cube roots gives (5 (223 sqrt(3188516012553) - 413945613)^(1/3))/(223 2^(1/3) 3^(2/3)) times the third roots of unity:

y = (5 (223 sqrt(3188516012553) - 413945613)^(1/3))/(223 2^(1/3) 3^(2/3)) or y = -(5 (-1/2)^(1/3) (223 sqrt(3188516012553) - 413945613)^(1/3))/(223 3^(2/3)) or y = (5 (-1)^(2/3) (223 sqrt(3188516012553) - 413945613)^(1/3))/(223 2^(1/3) 3^(2/3))

Substitute each value of y into x = y + 2553650/(149187 y):

x = (5 ((223 sqrt(3188516012553) - 413945613)/2)^(1/3))/(223 3^(2/3)) - 510730/223 (-1)^(2/3) (2/(3 (413945613 - 223 sqrt(3188516012553))))^(1/3) or x = 510730/223 ((-2)/(3 (413945613 - 223 sqrt(3188516012553))))^(1/3) - (5 ((-1)/2)^(1/3) (223 sqrt(3188516012553) - 413945613)^(1/3))/(223 3^(2/3)) or x = (5 (-1)^(2/3) ((223 sqrt(3188516012553) - 413945613)/2)^(1/3))/(223 3^(2/3)) - 510730/223 (2/(3 (413945613 - 223 sqrt(3188516012553))))^(1/3)

Bring each solution to a common denominator and simplify:

x = (5 ((446 sqrt(3188516012553) - 827891226)^(1/3) - 204292 (-1)^(2/3) (3/(413945613 - 223 sqrt(3188516012553)))^(1/3)))/(223 6^(2/3)) or x = -(5 ((-2)^(1/3) (223 sqrt(3188516012553) - 413945613)^(1/3) - 204292 ((-3)/(413945613 - 223 sqrt(3188516012553)))^(1/3)))/(223 6^(2/3)) or x = -(5 ((827891226 - 446 sqrt(3188516012553))^(1/3) + 204292 (3/(413945613 - 223 sqrt(3188516012553)))^(1/3)))/(223 6^(2/3))

Substitute back for t = x + 1125/223:

Answer: t = (5 ((446 sqrt(3188516012553) - 827891226)^(1/3) - 204292 (-1)^(2/3) (3/(413945613 - 223 sqrt(3188516012553)))^(1/3)))/(223 6^(2/3)) + 1125/223 or t = 1125/223 - (5 ((-2)^(1/3) (223 sqrt(3188516012553) - 413945613)^(1/3) - 204292 (-3/(413945613 - 223 sqrt(3188516012553)))^(1/3)))/(223 6^(2/3)) or t = 1125/223 - (5 ((827891226 - 446 sqrt(3188516012553))^(1/3) + 204292 (3/(413945613 - 223 sqrt(3188516012553)))^(1/3)))/(223 6^(2/3))

6 0
4 years ago
Jordan is five years old. his mother says to him "if you have two apples and i give you two more apples, how many apples will yo
Mars2501 [29]
Answer should be 2 and he says 4
8 0
4 years ago
Assume the given distributions are normal. Cucumbers grown on a certain farm have weights with a standard deviation of 2 ounces.
forsale [732]

Answer:

13.92

Step-by-step explanation:

We have that the critical z-score associated with 85% to the left is 1.04, we know that by table.

So we have to:

m + z * (sd) = 16

where m is the mean, z is the critical z-scor and sd is the standard deviation, if we replace we are left with:

m + 1.04 * (2) = 16

m = 16 - 1.04 * (2)

m = 13.92

Therefore, the average weight if 85% of cucumbers weigh less than 16 ounces is 13.92

4 0
3 years ago
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