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garri49 [273]
1 year ago
7

96.5 times 2.54 i know the answer i just need help showing the work

Mathematics
2 answers:
Ivanshal [37]1 year ago
6 0
At first just take out the decimals and multply numbers normally.

Then get your answer. Then check how many decimal places there are in 

96.5 * 2.54 which is 3

then in the answer count three down from the end at 0  and place your point at 3 decimal places!

kogti [31]1 year ago
3 0
You just solve it like a normal multiplication problem and then start putting the decimal begin the answer and place it 3 numbers forward
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Given the general identity tan X =sin X/cos X , which equation relating the acute angles, A and C, of a right ∆ABC is true?
irakobra [83]

First, note that m\angle A+m\angle C=90^{\circ}. Then

m\angle A=90^{\circ}-m\angle C \text{ and } m\angle C=90^{\circ}-m\angle A.

Consider all options:

A.

\tan A=\dfrac{\sin A}{\sin C}

By the definition,

\tan A=\dfrac{BC}{AB},\\ \\\sin A=\dfrac{BC}{AC},\\ \\\sin C=\dfrac{AB}{AC}.

Now

\dfrac{\sin A}{\sin C}=\dfrac{\dfrac{BC}{AC}}{\dfrac{AB}{AC}}=\dfrac{BC}{AB}=\tan A.

Option A is true.

B.

\cos A=\dfrac{\tan (90^{\circ}-A)}{\sin (90^{\circ}-C)}.

By the definition,

\cos A=\dfrac{AB}{AC},\\ \\\tan (90^{\circ}-A)=\dfrac{\sin(90^{\circ}-A)}{\cos(90^{\circ}-A)}=\dfrac{\sin C}{\cos C}=\dfrac{\dfrac{AB}{AC}}{\dfrac{BC}{AC}}=\dfrac{AB}{BC},\\ \\\sin (90^{\circ}-C)=\sin A=\dfrac{BC}{AC}.

Then

\dfrac{\tan (90^{\circ}-A)}{\sin (90^{\circ}-C)}=\dfrac{\dfrac{AB}{BC}}{\dfrac{BC}{AC}}=\dfrac{AB\cdot AC}{BC^2}\neq \dfrac{AB}{AC}.

Option B is false.

3.

\sin C = \dfrac{\cos A}{\tan C}.

By the definition,

\sin C=\dfrac{AB}{AC},\\ \\\cos A=\dfrac{AB}{AC},\\ \\\tan C=\dfrac{AB}{BC}.

Now

\dfrac{\cos A}{\tan C}=\dfrac{\dfrac{AB}{AC}}{\dfrac{AB}{BC}}=\dfrac{BC}{AC}\neq \sin C.

Option C is false.

D.

\cos A=\tan C.

By the definition,

\cos A=\dfrac{AB}{AC},\\ \\\tan C=\dfrac{AB}{BC}.

As you can see \cos A\neq \tan C and option D is not true.

E.

\sin C = \dfrac{\cos(90^{\circ}-C)}{\tan A}.

By the definition,

\sin C=\dfrac{AB}{AC},\\ \\\cos (90^{\circ}-C)=\cos A=\dfrac{AB}{AC},\\ \\\tan A=\dfrac{BC}{AB}.

Then

\dfrac{\cos(90^{\circ}-C)}{\tan A}=\dfrac{\dfrac{AB}{AC}}{\dfrac{BC}{AB}}=\dfrac{AB^2}{AC\cdot BC}\neq \sin C.

This option is false.

8 0
1 year ago
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A motorcycle has an initial speed of u m/s. It accelerates to a speed of 1.2u in 10 seconds. The motorcycle then travels at a co
kolbaska11 [484]

Answer:

Step-by-step explanation:

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According to the Rational Root Theorem, the following are potential roots of f(x) = 6x4 + 5x3 – 33x2 – 12x + 20. -5/2,-2, 1, 10/
stiks02 [169]
-5/2 is the answer. I just took the test and got it right.
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1 year ago
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5. When looking at a map, a student realizes that Birmingham is nearly due west of Atlanta, and Nashville is nearly due north of
leonid [27]

Answer: 250 mi

Step-by-step explanation:

Here we can think in a triangle rectangle:

The distance from Birmingham to Atlanta is roughly 150 mi, and this is one of the cathetus.

And the distance from Birmingham to Nashville is roughly 200 mi, this is the other cathetus of the triangle.

Now, the distance from Atlanta to Nashville will be the hypotenuse of this triangle rectangle.

Now we can apply the Pythagorean's theorem:

A^2 + B^2 = H^2

Where A and B are the cathetus, and H is the hypotenuse:

Then:

H = √(A^2 + B^2)

H = √(150^2 + 200^2) mi = √(62,500) mi = 250 mi

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Tim was given a large bag of sweets and ate one third of the sweets before stopping as he was feeling sick. The next day he ate
mars1129 [50]

Answer: In the beginning he was given 27 sweets.

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On the third day he ate one-third and had 8 sweets left over. To determine how many he started with on the third day, let the total on day three be called a. If one-third of a is eaten, then the left over which is two-thirds is 8. That is;

8/a = 2/3

By cross multiplication we now have

8 x 3 = 2a

24/2 = a

a = 12

Let the number of sweets he had on day two be called b. If he ate one-third of b and he had 12 left over, then the two-thirds left over is 12 and we now have;

12/b = 2/3

By cross multiplication we now have

12 x 3 = 2b

36 = 2b

36/2 = b

b = 18

Let the number of sweets he had on day one be called x. If he ate one-third of x and he had 18 left over, then the two-thirds left over is 18, and we now have;

18/x = 2/3

By cross multiplication we now have

18 x 3 = 2x

54 = 2x

x = 27

Therefore Tim was given 27 sweets at the beginning.

3 0
2 years ago
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