Bruce wants to make 50 ml of an alcohol solution with a 12% concentration. He has a 10% alcohol solution and a 15% alcohol solution. The equation 0.10x + 0.15(50 – x) = 0.12(50) can be used to find the amount of 10% alcohol solution Bruce should use. How much of the 15% alcohol solution should Bruce use?

Answer:

[tex]d=22t+10...(1)[/tex]

[tex]d=28t...(2)[/tex]

Step-by-step explanation:

Let t represent the number of hours and d represent the total distance after t hours.

We have been given that John bikes 22 km per hour and starts at mile 10. This means that slope of line will be 22 and y-intercept will be 10. So total distance covered by John in t hours will be:

[tex]d=22t+10...(1)[/tex]

We are also told that Gwynn bikes 28 km per hour and starts at mile 0. This means that slope of line will be 28 and y-intercept will be 0. So total distance covered by Gwynn in t hours will be:

[tex]d=28t+0[/tex]

[tex]d=28t...(2)[/tex]

Therefore, our required system of linear equations will be:

[tex]d=22t+10...(1)[/tex]

[tex]d=28t...(2)[/tex]

The equation: [tex]\( 21x^2 - 75x + 36 = 0 \)[/tex] represents "75 times an integer minus 36 equals 21 times the square of the integer."

To solve this problem, let's represent the unknown integer with the variable [tex]\( x \)[/tex]. Then we can set up the equation based on the given information.

The problem states: "Seventy-five times an integer, minus 36, equals 21 times the square of the integer."

Mathematically, this can be represented as:

[tex]\[ 75x - 36 = 21x^2 \][/tex]

So, the equation that could be used to solve for the unknown integer is:

[tex]\[ 21x^2 - 75x + 36 = 0 \][/tex]

Answer:

2.86 quarts ( approx )

Step-by-step explanation:

Given,

The initial quantity of the Mr. Gittleboro's radiator that contains 30% antifreeze = 10 quarts,

Let x quarts of pure antifreeze replaced x quarts of Mr. Gittleboro's radiator to bring it up to a required 50% antifreeze,

So, the quantity of 30% antifreeze radiator after drained off x quarts = (10-x) quarts

Also, the quantity of final antifreeze radiator 50% antifreeze = 10 quarts

Thus, we can write,

30% of (10-x) + 100% of x = 50% of 10

30(10-x) + 100x = 500

300 - 30x + 100x = 500

300 + 70x = 500

70x = 200

x = 2.85714285714 quarts ≈ 2.86 quarts

[tex]\boxed{{\mathbf{2}}{\mathbf{.86 quartz}}}[/tex]  of [tex]30\%[/tex]  antifreeze replaced by pure antifreeze to maintain [tex]50\%[/tex]  antifreeze in the radiator.

Further explanation:

Given:

It is given that radiator contains [tex]30\%[/tex]  antifreeze and the radiator holds 10 quartz.

Step by step explanation:

Step 1:

Consider [tex]x[/tex]  as the number of quartz of pure antifreeze that is replaced by [tex]x[/tex]  number of quartz of Mr. Gittleboro’s radiator to maintain the level of [tex]50\%[/tex] antifreeze in the radiator.

It is given that radiator contains [tex]30\%[/tex]  antifreeze and the radiator holds 10 quartz.

The amount of [tex]30\%[/tex]  antifreeze radiator after drained off [tex]x{\text{ quartz}}[/tex]  is [tex]\left({10-x}\right){\text{quartz}}[/tex] .

The radiator only holds [tex]{\text{10 quartz}}[/tex] .

The amount of final antifreeze radiator [tex]50\%[/tex]  antifreeze is [tex]{\text{10 quartz}}[/tex] .

Step 2:

The equation for maintain the level of [tex]50\%[/tex]  antifreeze in the radiator by replacing the [tex]x[/tex]  number of quartz of [tex]30\%[/tex]  antifreeze by pure antifreeze as,

[tex]\begin{aligned}30\%{\text{ of}}\left({10-x}\right)+100\% {\text{ of }}x&=50\% {\text{ of 10}}\\\frac{{30}}{{100}}\times\left({10-x}\right)+\frac{{100}}{{100}}{\text{}}\times{\text{}}x&=\frac{{50{\text{ }}}}{{100}}\times{\text{10}}\\{\text{30}}\left({10-x}\right)+100x&=500\\\end{aligned}[/tex]

Now simplify the further equation using the distributive property.

[tex]\begin{aligned}{\text{30}}\left({10-x}\right)+100x&=500\\300-30x+100x&=500\\300+70x&=500\\70x&=500-300\\\end{aligned}[/tex]

Now simplify the further equation to obtain the value of [tex]x[/tex] .

[tex]\begin{aligned}70x&=500-300\hfill\\70x&=200\hfill\\x&=\frac{{200}}{{70}}\hfill\\x&=2.85714\approx 2.86{\text{ quartz}}\hfill\\\end{aligned}[/tex]

Therefore, [tex]2.86{\text{ quartz}}[/tex]  of [tex]30\%[/tex]  antifreeze replaced by pure antifreeze to maintain [tex]50\%[/tex]  antifreeze in the radiator.

Learn more:  

Answer details:

Grade: High school

Subject: Mathematics

Chapter: Linear equation

Keywords: radiator, amount, replaced, Gittlebro, antifreeze, drained, maintain, quartz, number, linear equation, percentage, fraction, final antifreeze.

There are 11.3 feet far above the ground does the ladder touch the wall.

The Pythagoras theorem states that in a right triangle, the square of the hypotenuse is equal to the sum of the square of the other two sides.

Given that;

The bottom of a ladder must be placed 4 feet from a wall. the ladder is 12 feet long.

Now, Let the length of the ladder above the ground does the ladder touch the wall = x

Hence, We get;

x² = 12² - 4²

x² = 144 - 16

x² = 128

x = √128

x = 11.3 feet

Thus, the length of the ladder above the ground does the ladder touch the wall = 1.3 feet

Learn more about the Pythagoras theorem visit:

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To find the number of black marbles, calculate the percentage of white marbles first and then subtract it from the total marbles, there are 4,394 black marbles in the jar.

To find the number of black marbles:

Therefore, there are 4,394 black marbles in the jar.

Answer: The answer is the number of hours of daylight each day.

Step-by-step explanation:  We are to give the best  example of a natural phenomenon of periodic behavior.

A periodic behavior means the behavior which repeats itself in equal intervals of time or periods. Since we are talking about a natural phenomenon, so the rotation of earth around the sun can be an example.

In other words, we can say the number of hours of daylight each day will be the proper example, because the sun gives light for almost same number of hours each day.

Thus, the natural phenomenon is the number of hours of daylight each day.

Answer: 2x-3=x+2

. -x -x

1x -3=2

+3 +3

1x=5

1x divide by 1x, 5 divide by 1x

Which will be:

X=5

The Boy: 4.4 x 60 = 264ft per minute.  264 / 3 = 88yrds Per minute

The Fish: 12 x 5280 = 63360ft hr. 5280 / 60 = 1056ft minute 1056 / 3 = 352yrds

352 - 88 = 44 yards quicker then the boy.

44 x 3 = 132 feet quicker then the boy.

To find the value of k in the equation 2k+9=7−3k, we rearrange and simplify to isolate k, resulting in k = -2/5.

To solve for the value of k in the equation 2k+9=7−3k, we must isolate the variable on one side of the equation. Let's start by adding 3k to both sides to get all the k terms on one side:

2k + 3k + 9 = 7 − 3k + 3k
5k + 9 = 7

Next, subtract 9 from both sides to get:

5k = 7 − 9
5k = −2

Now, to find the value of k, divide both sides by 5:

k = −2/5

Therefore, the value of k is −2/5.

Plug in -0.5 for c

a = 3 - 5(-0.5)

Multiply -5 with (-0.5)

a = 3 + 2.5

Simplify

a = 5.5

a = 5.5 is your answer

hope this helps

In the formula 'a = 3-5c', 'a' becomes 5.5 when c=-0.5. This is derived by substituting -0.5 as 'c' in the given formula yielding the result of 5.5 after the calculation.

The subject of your question is mathematics and it's about finding the value of 'a' in the formula: a = 3-5c when c=-0.5.

To answer your question, we'll substitute the value of 'c' into the formula:a = 3 - 5*(-0.5). When you multiply -5 with -0.5, the answer is 2.5.

Therefore, when you add 3 and 2.5, the answer is 5.5. So, when c = -0.5, the value of 'a' will be 5.5.

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4 total quizzes,

 for a 40 average on 4 quizzes she needs 40*4 = 160 total points

for 3 quizzes she has 45 +32 +37 = 114 points

160-114 = 46, she needs at least a 46 on the 4th quiz

Answer:

ΔKML                                               Option C

Step-by-step explanation:

Given right triangle is ΔJKL

Where ∠JKL = 90°

Here KM is perpendicular to LJ.

In triangle ΔJKL & ΔKML

∠JKL ≅ ∠KML  (both are right angle)

∠JLK ≅ ∠KLM   (common angle in both the triangles)

So,

ΔJKL is similar to ΔKML.

That's the final answer.

I hope it will help you.

To find the rate of change of the x-coordinate at the point (1, 5) with a given y-coordinate rate of change, we use the chain rule to differentiate the equation y = 24 + [tex]x^3[/tex] with respect to time, and solve for dx/dt.

The student's question involves the relationship between the rates of change of the x-coordinate and y-coordinate of a particle moving along a curve. Given the equation y = 24 +[tex]x^3[/tex], we need to find how fast the x-coordinate is changing at the point (1, 5), given that the y-coordinate is increasing at a rate of 4 cm/s. To find this, we can use the chain rule from calculus to relate the rates of change. Differentiating both sides with respect to time t, we get dy/dt = 3[tex]x^2[/tex] dx/dt. Solving for dx/dt, we find the rate of change of the x-coordinate when x = 1 and dy/dt = 4 cm/s.

At the point (1, 5), [tex]\( \frac{{dx}}{{dt}} = \frac{4}{3} \)[/tex] cm/s when [tex]\( \frac{{dy}}{{dt}} = 4 \)[/tex] cm/s.

To find how fast the x-coordinate is changing [tex](\( \frac{{dx}}{{dt}} \))[/tex] when the y-coordinate is increasing [tex](\( \frac{{dy}}{{dt}} \)),[/tex] we'll use implicit differentiation.

Given [tex]\( y = 24 + x^3 \),[/tex] differentiate both sides with respect to time [tex](\( t \)):[/tex]

[tex]\[ \frac{{dy}}{{dt}} = \frac{{d}}{{dt}}(24 + x^3) \][/tex]

Given that [tex]\( \frac{{dy}}{{dt}} = 4 \)[/tex], and when [tex]\( x = 1 \) and \( y = 5 \),[/tex] we can find [tex]\( \frac{{dx}}{{dt}} \):[/tex]

[tex]\[ 4 = 0 + 3x^2 \cdot \frac{{dx}}{{dt}} \][/tex]

[tex]At \( x = 1 \):[/tex]

[tex]\[ 4 = 3(1)^2 \cdot \frac{{dx}}{{dt}} \][/tex]

[tex]\[ 4 = 3 \cdot \frac{{dx}}{{dt}} \][/tex]

[tex]\[ \frac{{dx}}{{dt}} = \frac{4}{3} \][/tex]

So, the x-coordinate of the point is changing at a rate of [tex]\( \frac{4}{3} \)[/tex] cm/s at that instant.