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3 years ago

A person can pay $33 for a membership to the history museum and then go to the museum

Mathematics

2 answers:

Lemur [1.5K]3 years ago

4 0

Answer:

42$

Step-by-step explanation:

The person pays 33$ for a membership,and then visits the museum 9 times.So 33+(9*1)=42

hram777 [196]3 years ago

4 0

Answer:

$42

Step-by-step explanation:

$33+$9=$42

Add the membership cost to the number of visits, and you have the answer.

Have a great day, and mark me brainliest if I am most helpful!

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Construct a quadratic polynomial whose zeroes are negatives of the zeroes of the

sp2606 [1]

Given:

The given quadratic polynomial is :

$x^2-x-12$

To find:

The quadratic polynomial whose zeroes are negatives of the zeroes of the given polynomial.

Solution:

We have,

$x^2-x-12$

Equate the polynomial with 0 to find the zeroes.

$x^2-x-12=0$

Splitting the middle term, we get

$x^2-4x+3x-12=0$

$x(x-4)+3(x-4)=0$

$(x+3)(x-4)=0$

$x=-3,4$

The zeroes of the given polynomial are -3 and 4.

The zeroes of a quadratic polynomial are negatives of the zeroes of the given polynomial. So, the zeroes of the required polynomial are 3 and -4.

A quadratic polynomial is defined as:

$x^2-(\text{Sum of zeroes})x+\text{Product of zeroes}$

$x^2-(3+(-4))x+(3)(-4)$

$x^2-(-1)x+(-12)$

$x^2+x-12$

Therefore, the required polynomial is $x^2+x-12$ .

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3 years ago

28 students are in Ms.Whitfield's class. 22 students passed their recent test. What percentage passed the test? Answer to the ne

nirvana33 [79]

22/28 converted to a percentage is 78.6%. (rounded to the nearest tenth)

78.6% of the class passed the test.

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A Parallelogram must be a rectangle when it’s •A.)Diagnols are perpendicular •B.Diagnols are congruent •C. opposite sides are pa

Verizon [17]

Answer:A

Step-by-step explanation: the diagonals face each other making them perpendicular making both sides parallel to each other, being straight.

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3 years ago

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5,622 lbs = _____ tons _____ lbs?

artcher [175]

Answer:

5,622 = abt 2.811 U.S. tons

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3 years ago

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How would I find the integral of <img src="https://tex.z-dn.net/?f=%5Cint%5Cfrac%7Btdt%7D%7Bt%5E4%2B2%7D" id="TexFormula1" title

kotegsom [21]

Let $t=\sqrt y$ , so that $t^2=y$ , $t^4=y^2$ , and $\mathrm dt=\dfrac{\mathrm dy}{2\sqrt y}$ . Then

$\displaystyle\int\frac t{t^4+2}\,\mathrm dt=\int\frac{\sqrt y}{2\sqrt y(y^2+2)}\,\mathrm dy=\frac12\int\frac{\mathrm dy}{y^2+2}$

Now let $y=\sqrt2\tan z$ , so that $\mathrm dy=\sqrt2\sec^2z\,\mathrm dz$ . Then

$\displaystyle\frac12\int\frac{\mathrm dy}{y^2+2}=\frac12\int\frac{\sqrt2\sec^2z}{(\sqrt2\tan z)+2}\,\mathrm dz=\frac{\sqrt2}4\int\frac{\sec^2z}{\tan^2z+1}\,\mathrm dz=\frac1{2\sqrt2}\int\mathrm dz=\dfrac1{2\sqrt2}z+C$

Transform back to $y$ to get

$\dfrac1{2\sqrt2}\arctan\left(\dfrac y{\sqrt2}\right)+C$

and again to get back a result in terms of $t$ .

$\dfrac1{2\sqrt2}\arctan\left(\dfrac{t^2}{\sqrt2}\right)+C$

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3 years ago