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loris [4]
3 years ago
15

on a baseball diamond , the  distance from home plate to the pitchers mound is 60.5 feet. from the pitchers mound to second base

 is 66.8.                                                                                                    A. how far is it from home plate to second base?                                                                            B.from home plate to first base is 90 feet.how much less is this than the distance from                  home plate to second base.                                                                                                                                                                                                                                                                                                                 show work please?               
Mathematics
1 answer:
IRINA_888 [86]3 years ago
8 0
Home plate -> pitchers mound = 60.5 feet
Pitchers mound -> second base = 66.8 feet
Home plate -> first base = 90 feet

A. Home plate -> second base = 60.5 + 66.8 = 127,3 feet
B. 90 feet is less than 127,3 feet by 127,3 - 90 = 37,3 feet.
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Please help me!!!!!​
Bas_tet [7]

Answer:  see proof below

<u>Step-by-step explanation:</u>

Use the Sum & Difference Identity: tan (A - B) = (tanA - tanB)/(1 + tanA tanB)

Use the Half-Angle Identity: tan (A/2) = (1 - cosA)/(sinA)

Use the Unit Circle to evaluate tan (π/4) = 1

Use Pythagorean Identity:     cos²A + sin²A = 1

<u>Proof LHS → RHS</u>

\text{Given:}\qquad \qquad \qquad\dfrac{2\tan\bigg(\dfrac{\pi}{4}-\dfrac{A}{2}\bigg)}{1+\tan^2\bigg(\dfrac{\pi}{4}-\dfrac{A}{2}\bigg)}

\text{Difference Identity:}\qquad \dfrac{2 \bigg( \frac{\tan\frac{\pi}{4}-\tan\frac{A}{2}}{1+\tan\frac{\pi}{4}\cdot \tan\frac{A}{2}}\bigg)}{1+ \bigg( \frac{\tan\frac{\pi}{4}-\tan\frac{A}{2}}{1+\tan\frac{\pi}{4}\cdot \tan\frac{A}{2}}\bigg)^2}

\text{Substitute:}\qquad \qquad \dfrac{2 \bigg( \frac{1-\tan\frac{A}{2}}{1+\tan\frac{A}{2}}\bigg)}{1+ \bigg( \frac{1-\tan\frac{A}{2}}{1+\tan\frac{A}{2}}\bigg)^2}

\text{Simplify:}\qquad \qquad \qquad \dfrac{1-\tan^2\frac{A}{2}}{1+\tan^2\frac{A}{2}}

\text{Half-Angle Identity:}\qquad \quad \dfrac{1-(\frac{1-\cos A}{\sin A})^2}{1+(\frac{1-\cos A}{\sin A})^2}

\text{Simplify:}\qquad \qquad \dfrac{\sin^2 A-1+2\cos A-\cos^2 A}{\sin^2 A+1-2\cos A+\cos^2 A}

\text{Pythagorean Identity:}\qquad \qquad \dfrac{1-\cos^2 A-1+2\cos A}{2-2\cos A}

\text{Simplify:}\qquad \qquad \qquad \dfrac{2\cos A-2\cos^2 A}{2(1-\cos A)}\\\\.\qquad \qquad \qquad \qquad =\dfrac{2\cos A(1-\cos A)}{2(1-\cos A)}

                               =  cos A

LHS = RHS:  cos A = cos A   \checkmark

4 0
3 years ago
An important factor in selling a residential property is the number of times real estate agents show a home. A sample of 23 home
liberstina [14]

Using the t-distribution, it is found that the 95% confidence interval for the mean number of people the houses were shown is (20.1, 27.9).

We have the <u>standard deviation for the sample</u>, hence the t-distribution is used to build the confidence interval. Important information are given by:

  • Sample mean of \overline{x} = 24.
  • Sample standard deviation of s = 9.
  • Sample size of n = 23

The confidence interval is:

\overline{x} \pm t\frac{s}{\sqrt{n}}

In which t is the critical value for a <u>95% confidence interval with 23 - 1 = 22 df</u>, thus, looking at a calculator or at the t-table, it is found that t = 2.0739.

Then:

\overline{x} - t\frac{s}{\sqrt{n}} = 24 - 2.0739\frac{9}{\sqrt{23}} = 20.1

\overline{x} + t\frac{s}{\sqrt{n}} = 24 + 2.0739\frac{9}{\sqrt{23}} = 27.9

The 95% confidence interval for the mean number of people the houses were shown is (20.1, 27.9).

A similar problem is given at brainly.com/question/15180581

6 0
2 years ago
30% of doctors on a hospital are female. if there are 780 doctors altogether, how many doctors are female ? round to the nearest
Archy [21]
30% as a decimal is 0.3

Multiply 780 by 0.3
780(0.3)=234 female doctors
4 0
3 years ago
Read 2 more answers
The man is pushing with a force of 75 N. The rock is not moving. This is an example of what kind of force?
Oksi-84 [34.3K]
Kinetic energy
Pushing is energy in motion and energy in motion is classified as (KE) kinetic energy
3 0
3 years ago
12.<br> 1/2(4x — 2) – 2/3(6x + 9) &lt;4
RSB [31]

Answer: x > -11/2

svdbhtdtgifubhonj

5 0
3 years ago
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