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N76 [4]
2 years ago
5

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 <em>30 baseballs</em>, when full

The given parameters are:

<em />\mathbf{Initial = \frac 16x}<em> ------ 1/6 full</em>

<em />\mathbf{Additional = 15}<em> --- baseballs added</em>

<em />\mathbf{Final = \frac 23x}<em> ---- 2/3 full</em>

<em />

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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What is the product of 2p + q and –3q – 6p + 1?
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Orlov [11]

Answers:

  • <u>20 pounds</u> of the 95% concrete mix
  • <u>10 pounds</u> 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
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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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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:

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Spilt the integral:

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Factor out the constant multiple:

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Reverse Power Rule:

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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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