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3 years ago

5

Algebra help please?

2 answers:

Strike441 [17]3 years ago

6 0

Answer:

15 out 45 is the probability.

So out of 45 times he will hit 15 times.

Mashutka [201]3 years ago

6 0

If the player bats 45 times, and only hits 15, this can be represented by 15/45 (hits/bats) simplified, this is 1/3. So, he has about a 1/3 chance of getting a hit. :)

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What is the difference between (-58) and (+35) ?

romanna [79]

To do this question, you are best splitting it into two halves: getting to 0, and then 35
Starting at -58, we want to get to 0
-58 + 58 = 0
Then, we get to 35 from 0
0 + 35 = 35
Now, add the two values together
58 + 35 = 93
The difference is 93

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3 years ago

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Find the missing angle in a triangle with the given sides.

d1i1m1o1n [39]

Answer:

1

Step-by-step explanation:

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3y-(4y+6x) y=3 x=-2<br> Evaluate each expression for the given values of the variables

STALIN [3.7K]

Answer:

9

Step-by-step explanation:

3y-(4y+6x)

=3*3-(4*3+6*-2)

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3 years ago

James is solving a number puzzle that involves three integers, a, b, and c, where c is a positive integer. The product of a and

m_a_m_a [10]

Given:

Three integers, a, b, and c, where c is a positive integer.

The product of a and b is 6.

The product of a and c is -4.

The product of b and c is -6.

To find:

The values of a,b and c.

Solution:

According to the given information:

$ab=6$ ...(i)

$ac=-4$ ...(ii)

$bc=-6$ ...(iii)

From (ii), we get

$a=-\dfrsc{4}{c}$ ...(iv)

From (iii), we get

$b=-\dfrsc{6}{c}$ ...(v)

Putting $a=-\dfrsc{4}{c}$ and $b=-\dfrsc{6}{c}$ in (i), we get

$\dfrac{-4}{c}\times \dfrac{-6}{c}=6$

$\dfrac{24}{c^2}=6$

$\dfrac{24}{6}=c^2$

$4=c^2$

Taking square root on both sides, we get

$\pm \sqrt{4}=c$

$\pm 2=c$

It is given that c is a positive integer. So, it cannot be negative and the only value of c is $c=2$ .

Putting $c=2$ in (iv), we get

$a=-\dfrsc{4}{2}$

$a=-2$

Putting $c=2$ in (v), we get

$b=-\dfrsc{6}{2}$

$b=-3$

Therefore, the values of a,b,c are $a=-2,b=-3,c=2$ .

8 0

3 years ago

<img src="https://tex.z-dn.net/?f=Evaluate%3A%20%5Csqrt%7B%20%5Cfrac%20%7B1%20-%20sin%28x%29%7D%7B1%20%2B%20sin%28x%29%E2%80%8B%

stealth61 [152]

$\large\underline{\sf{Solution-}}$

We have to <u>evaluate</u> the given <u>expression</u>.

$\rm = \sqrt{ \dfrac{1 - \sin(x) }{1 + \sin(x) } }$

If we multiple both numerator and denominator by 1 - sin(x), then the value remains same. Let's do that.

$\rm = \sqrt{ \dfrac{[1 - \sin(x)][1 - \sin(x) ]}{[1 + \sin(x)][1 - \sin(x) ]} }$

$\rm = \sqrt{ \dfrac{[1 - \sin(x)]^{2}}{1- \sin^{2} (x) } }$

<u>We know that:</u>

$\rm \longmapsto { \sin}^{2}(x) + \cos^{2}(x) = 1$

$\rm \longmapsto \cos^{2}(x) = 1 - { \sin}^{2}(x)$

Therefore, <u>the expression becomes:</u>

$\rm = \sqrt{ \dfrac{[1 - \sin(x)]^{2}}{\cos^{2} (x)}}$

$\rm = \dfrac{1 - \sin(x)}{\cos(x)}$

$\rm = \dfrac{1}{\cos(x)} - \dfrac{ \sin(x) }{ \cos(x) }$

$\rm = \sec(x) - \tan(x)$

7 0

3 years ago

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