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2 years ago

A batting machine uses an automatic baseball feeder. During baseball

Mathematics

1 answer:

Inessa05 [86]2 years ago

5 0

The feeders in battling machine are represented in proportions and fractions.

The equation that represents the problem is: $\mathbf{\frac 16x + 15 = \frac 23x}$
The feeder can hold 30 baseballs, when full

The given parameters are:

$\mathbf{Initial = \frac 16x}$ ------ 1/6 full

$\mathbf{Additional = 15}$ --- baseballs added

$\mathbf{Final = \frac 23x}$ ---- 2/3 full

So, the equation that represents the problem is:

$\mathbf{Initial + Additional = Final}$

So, we have:

$\mathbf{\frac 16x + 15 = \frac 23x}$

The number of baseballs it can hold is calculated as follows:

$\mathbf{\frac 16x + 15 = \frac 23x}$

Multiply through by 6

$\mathbf{x + 90 = 4x}$

Collect like terms

$\mathbf{4x - x = 90 }$

$\mathbf{3x = 90 }$

Divide through by 3

$\mathbf{x = 30 }$

Hence, the feeder can hold 30 baseballs, when full

Read more about proportions and fractions at:

brainly.com/question/20337104

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Answer:

Step-by-step explanation:

6x5=30

30x2/3=20

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-12p^2 - 12pq + 2p -3q^2 + q

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3 years ago

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We have two varieties of concrete that we can mix together. One is 5% water and 95% concrete mix, and the other is 20% water and

Orlov [11]

Answers:

20 pounds of the 95% concrete mix
10 pounds of the 80% concrete mix

=================================================

Explanation:

x = amount of the 95% concrete mix

y = amount of the 80% concrete mix

Type A mix has 95% concrete. If we have x pounds of this, then 0.95x pounds is pure concrete. Type B mix has 80% concrete. If there are y pounds of this, then 0.80y pounds is pure concrete.

So far we have 0.95x+0.80y pounds of pure concrete from the two mixes

Divide this over 30 to get (0.95x+0.80y)/30

Setting this equal to 0.90 means we want the final mix to be 90% concrete.

Since we want 30 pounds total, we can say x+y = 30 which solves to y = 30-x. We'll use this later

-------------------------------

(0.95x+0.80y)/30 = 0.90

0.95x+0.80y = 30*0.90 .... multiply both sides by 30

0.95x+0.80y = 27

0.95x+0.80(30-x) = 27 ..... plug in y = 30-x

0.95x+0.80*30-0.80*x = 27

0.95x+24-0.80x = 27

0.15x+24 = 27

0.15x = 27-24 ..... subtracting 24 from both sides

0.15x = 3

x = 3/0.15 ........... dividing both sides by 0.15

x = 20

y = 30-x

y = 30-20

y = 10

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3 years ago

Need help with #6 Find the missing angles

puteri [66]

Answer:

p = 95

q = 95

r = 85

s = 85

Step-by-step explanation:

85 deg & p are same-side int angles & are supplementary.

p = 180 - 85 = 95

p & q are vertical & congruent.

q = 95

p & s are supplenmentary.

s = 85

s & r are alt int angles & congruent.

r = 85

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3 years ago

Calculus 2 master needed; stuck on evaluating the integral please show steps <img src="https://tex.z-dn.net/?f=%5Cint%20%7Bsec%2

kakasveta [241]

Answer:

$\int \sec(\frac{x}{2})\tan^5({\frac{x}{2}})dx=\frac{2\sec^5(\frac{1}{2}x)}{5}-\frac{4\sec^3(\frac{1}{2}x)}{3}+2\sec(\frac{1}{2}x)+C$

Step-by-step explanation:

So we have the integral:

$\int \sec(\frac{x}{2})\tan^5({\frac{x}{2}})dx$

First, let's use substitution to get rid of the x/2. I'm going to use the variable y. So, let y be x/2. Thus:

$y=\frac{x}{2}\\dy=\frac{1}{2}dx\\2dy=dx$

Therefore, the integral is:

$=2\int \sec(y)\tan^5(y)dy$

Now, as you had done, let's expand the tangent term. However, let's do it to the fourth. Thus:

$=2\int \sec(y)\tan^4(y)\tan(y)dy$

Now, we can use a variation of the trigonometric identity:

$\tan^2(y)+1=\sec^2(y)$

So:

$\tan^2(y)=\sec^2(y)-1$

Substitute this into the integral. Note that tan^4(x) is the same as (tan^2(x))^2. Thus:

$=2\int \sec(y)(\tan^2(y))^2\tan(y)dy\\=2\int \sec(y)(\sec^2(y)-1)^2\tan(y)dy$

Now, we can use substitution. We will use it for sec(x). Recall what the derivative of secant is. Thus:

$u=\sec(y)\\du=\sec(y)\tan(y)dy$

Substitute:

$2\int (\sec^2(y)-1)^2(\sec(y)\tan(y))dy\\=2\int(u^2-1)^2 du$

Expand the binomial:

$=2\int u^4-2u^2+1du$

Spilt the integral:

$=2(\int u^4 du+\int-2u^2du+\int +1du)$

Factor out the constant multiple:

$=2(\int u^4du-2\int u^2du+\int(1)du)$

Reverse Power Rule:

$=2(\frac{u^{4+1}}{4+1}-2(\frac{u^{2+1}}{2+1})+(\frac{u^{0+1}}{0+1}}))$

Simplify:

$=2(\frac{u^5}{5}-\frac{2u^3}{3}+u)$

Distribute the 2:

$=\frac{2u^5}{5}-\frac{4u^3}{3}+2u$

Substitute back secant for u:

$=\frac{2\sec^5(y)}{5}-\frac{4\sec^3(y)}{3}+2\sec(y)$

And substitute back 1/2x for y. Therefore:

$=\frac{2\sec^5(\frac{1}{2}x)}{5}-\frac{4\sec^3(\frac{1}{2}x)}{3}+2\sec(\frac{1}{2}x)$

And, finally, C:

$=\frac{2\sec^5(\frac{1}{2}x)}{5}-\frac{4\sec^3(\frac{1}{2}x)}{3}+2\sec(\frac{1}{2}x)+C$

And we're done!

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3 years ago

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