The sum that represents the number of tickets sold if 35 tickets were sold on monday, half of the remaining tickets were sold on tuesday , and 14 ticket were sold on wenesday

10,000 tiles  * 0.25 = 2500

 3000 -2500 = 500

 so the delivery fee is 500 and the tiles cost 0.25 each

 so the equation would be y = 0.25x +500

Answer is C

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.

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.

The algebraic expression equivalent to cot(arctan x) is 1 / x, derived by recognizing that cot(θ) is the reciprocal of tan(θ), and tan(θ) = x for θ = tan-1(x)

To find an algebraic expression for cot(arctan x), let's follow a step-by-step transformation using trigonometric identities: cot(arctan x) is denoted as cot(tan-1(x))

Therefore, the algebraic expression equivalent to cot(tan-1(x)) is 1 / x.

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.

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.

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:

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.