08.03.2021

If GHI is 78 what is the measure of GHI after the dilation

. 4

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Mathematics
Step-by-step answer
P Answered by PhD

9514 1404 393

EF = 1.98AB = 16.25HI = 12.22

Step-by-step explanation:

The mnemonic SOH CAH TOA is useful for these. It reminds you of the relations between sides and trig functions in a right triangle.

__

1. The side of interest is opposite the given angle. The given side is the hypotenuse.

  Sin = Opposite/Hypotenuse

  sin(26°) = EF/DF = EF/4.5

  EF = 4.5·sin(26°) = 4.5·0.44 = 1.98 . . . units

__

2. The given side is adjacent to the given angles. The desired length is that of the hypotenuse.

  Cos = Adjacent/Hypotenuse

  cos(52°) = AC/AB = 10.01/c

  c = 10.01/cos(52°) = 10.01/0.616 ≈ 16.25 . . . units

__

3. The given side is opposite the given angle, and the desired side length is adjacent to the given angle.

  Tan = Opposite/Adjacent

  tan(42°) = 11/HI

  HI = 11/tan(42°) = 11/0.90 = 12.22 . . . units

Mathematics
Step-by-step answer
P Answered by PhD

9514 1404 393

EF = 1.98AB = 16.25HI = 12.22

Step-by-step explanation:

The mnemonic SOH CAH TOA is useful for these. It reminds you of the relations between sides and trig functions in a right triangle.

__

1. The side of interest is opposite the given angle. The given side is the hypotenuse.

  Sin = Opposite/Hypotenuse

  sin(26°) = EF/DF = EF/4.5

  EF = 4.5·sin(26°) = 4.5·0.44 = 1.98 . . . units

__

2. The given side is adjacent to the given angles. The desired length is that of the hypotenuse.

  Cos = Adjacent/Hypotenuse

  cos(52°) = AC/AB = 10.01/c

  c = 10.01/cos(52°) = 10.01/0.616 ≈ 16.25 . . . units

__

3. The given side is opposite the given angle, and the desired side length is adjacent to the given angle.

  Tan = Opposite/Adjacent

  tan(42°) = 11/HI

  HI = 11/tan(42°) = 11/0.90 = 12.22 . . . units

Mathematics
Step-by-step answer
P Answered by PhD

9514 1404 393

AB = 16.25HI = 12.22

Step-by-step explanation:

The mnemonic SOH CAH TOA reminds you of the relevant relationships.

__

1. Cos = Adjacent/Hypotenuse

  cos(52°) = AC/AB

  AB = AC/cos(52°) = 10.01/0.616

  AB = 16.25

__

2. Tan = Opposite/Adjacent

  tan(42°) = GH/HI = 11/HI

  HI = 11/tan(42°) = 11/0.90

  HI ≈ 12.22

Mathematics
Step-by-step answer
P Answered by PhD

9514 1404 393

AB = 16.25HI = 12.22

Step-by-step explanation:

The mnemonic SOH CAH TOA reminds you of the relevant relationships.

__

1. Cos = Adjacent/Hypotenuse

  cos(52°) = AC/AB

  AB = AC/cos(52°) = 10.01/0.616

  AB = 16.25

__

2. Tan = Opposite/Adjacent

  tan(42°) = GH/HI = 11/HI

  HI = 11/tan(42°) = 11/0.90

  HI ≈ 12.22

Mathematics
Step-by-step answer
P Answered by PhD

1- c. Two-way ANOVA

2- a. 1

3- c. 78.1%

4- 48.9

Step-by-step explanation:

ANOVA is a statistical technique designed to test mean of one or more quantitative populations.  In two-way ANOVA it equals the block mean. Column block means square is three-way ANOVA.

Oxygen intake rate varies for different people. It also depends on the age factor. Oxygen intake rate is indicator of normal function of cells. A 48 year old man breath slower than a younger man. The babies breathe faster than adults.

Mathematics
Step-by-step answer
P Answered by PhD

1- c. Two-way ANOVA

2- a. 1

3- c. 78.1%

4- 48.9

Step-by-step explanation:

ANOVA is a statistical technique designed to test mean of one or more quantitative populations.  In two-way ANOVA it equals the block mean. Column block means square is three-way ANOVA.

Oxygen intake rate varies for different people. It also depends on the age factor. Oxygen intake rate is indicator of normal function of cells. A 48 year old man breath slower than a younger man. The babies breathe faster than adults.

Chemistry
Step-by-step answer
P Answered by Specialist

The ppm carbon in the seawater is 104.01.

Explanation:

Mass of seawater = 6.234 g

Mass of carbon dioxide gas  produced = 2.378 mg = 0.002378 g

1 mg = 0.001 g

Moles of carbon dioxide gas = \frac{0.002378 g}{44.009 g/mol}=5.4034\times 10^{-5} mol

1 mol of carbon dioxde gas has 1 mole pf carbon atoms,then 5.4034\times 10^{-5} mol will have ;

1\times 5.4034\times 10^{-5} mol=5.4034\times 10^{-5} mol of carbon

Mass of 5.4034\times 10^{-5} mol of carbon ;

12 g/mol\times 5.4034\times 10^{-5} mol=0.0006484 g

To calculate the ppm of oxygen in sea water, we use the equation:

\text{ppm}=\frac{\text{Mass of solute}}{\text{Mass of solution}}\times 10^6

(Both the masses are in grams)

=\frac{0.0006484 g}{ 6.234 g}\times 10^6=104.01 ppm

The ppm carbon in the seawater is 104.01.

Chemistry
Step-by-step answer
P Answered by Master

The ppm carbon in the seawater is 104.01.

Explanation:

Mass of seawater = 6.234 g

Mass of carbon dioxide gas  produced = 2.378 mg = 0.002378 g

1 mg = 0.001 g

Moles of carbon dioxide gas = \frac{0.002378 g}{44.009 g/mol}=5.4034\times 10^{-5} mol

1 mol of carbon dioxde gas has 1 mole pf carbon atoms,then 5.4034\times 10^{-5} mol will have ;

1\times 5.4034\times 10^{-5} mol=5.4034\times 10^{-5} mol of carbon

Mass of 5.4034\times 10^{-5} mol of carbon ;

12 g/mol\times 5.4034\times 10^{-5} mol=0.0006484 g

To calculate the ppm of oxygen in sea water, we use the equation:

\text{ppm}=\frac{\text{Mass of solute}}{\text{Mass of solution}}\times 10^6

(Both the masses are in grams)

=\frac{0.0006484 g}{ 6.234 g}\times 10^6=104.01 ppm

The ppm carbon in the seawater is 104.01.

Mathematics
Step-by-step answer
P Answered by PhD

For 1 flavor there are 9 topping

Therefore, for 5 different flavors there will be 5*9 choices

No of choices= 5*9

=45 

Mathematics
Step-by-step answer
P Answered by PhD

The answer is in the image 

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