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andrew-mc [135]
2 years ago
7

What is the lateral area of a right rectangular prism that has rectangular bases measuring 3" by 5" and a height of 6"?

Mathematics
2 answers:
lyudmila [28]2 years ago
7 0

The lateral area of a right rectangular prism that has rectangular bases measuring 3''{\text{ by 5''}} and a height of 6'' is \boxed{96{\text{ i}}{{\text{n}}^2}}. Choice (A) is correct.

Further explanation:

Given:

The options are as follows,

(a).{\text{LA}} = 96{\text{ i}}{{\text{n}}^2}

(b).{\text{LA}} = 95{\text{ i}}{{\text{n}}^2}

(c).{\text{LA}} = 93{\text{ i}}{{\text{n}}^2}

(d).{\text{LA}} = 98{\text{ i}}{{\text{n}}^2}

Explanation:

The perimeter of the base can be calculated as follows,

\begin{aligned}P&= 2 \times \left( {3 + 5} \right)\\&= 2 \times 8\\&= 16{\text{ in}}\\\end{aligned}

The lateral area of a right rectangular prism that has rectangular bases can be obtained as follows,

\begin{aligned}LA &= P \times h\\&= 16 \times 6\\ &= 96{\text{ i}}{{\text{n}}^2}\\\end{aligned}

The lateral area of a right rectangular prism that has rectangular bases measuring 3''{\text{ by 5''}} and a height of 6'' is \boxed{96{\text{ i}}{{\text{n}}^2}}. Choice (A) is correct.

Learn more:

  1. If the clothing maker bought 500 m2 of this fabric, how much money did he lose? use 1tepiz=0.625dollar and 0.9144m=1yard brainly.com/question/2479097.
  2. Suppose that you find the volume of all the oceans to be 1.4×109km3 in a reference book. to find the mass, you can use the density of water, also found in this reference book, but first you must convert the volume to cubic meters. what is this volume in cubic meters? brainly.com/question/1446243.

Answer details:

Grade: High school

Subject: Mathematics

Chapter: Number system

Keywords: lateral area, right rectangular, prism, rectangle, area, rectangular bases, measuring, height, base area, rectangular prism, height of prism.

Mademuasel [1]2 years ago
6 0

Answer:

The answer is the option A

96\ in^{2}

Step-by-step explanation:

we know that

The lateral area of a right rectangular prism is equal to

LA=P*h

where

P is the perimeter of the base

h is the height of the prism

Find the perimeter

P=2*(3+5)=16\ in

h=6\ in

substitute in the formula

LA=16*6=96\ in^{2}

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Answer:

0.

Step-by-step explanation:

Fair dice are dice that have 6 sides, and the probability of rolling a side is the same as rolling another.

Since each die has 6 sides, the most you can get from the two dice are 6 + 6 = 12. Therefore, getting a 13 is impossible. So, there is a probability of 0.

Hope this helps!

4 0
1 year ago
Which system of equations can be used to find the roots of the equation 12 x 3-5x=2 x 2+x+6
kirill [66]

Answer:

Option (a) is correct.

The system of equation becomes

y=12x^3-5x\\\\ y=2x^2+x+6

Step-by-step explanation:

Given : Equation  12x^3-5x=2x^2+x+6

We have to construct a  system of equations that  can be used to find the roots of the equation 12x^3-5x=2x^2+x+6

Consider the given equation 12x^3-5x=2x^2+x+6

To construct a system of equation put both sides of the given equation equal to a same variable.

Let the variable be "y", Then the equation 12x^3-5x=2x^2+x+6

becomes,

12x^3-5x=y=2x^2+x+6

Thus, The system of equation becomes

y=12x^3-5x\\\\ y=2x^2+x+6

Option (a) is correct.

   

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2 years ago
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Calculate the unpaid balance, finance charge, and new balance using the unpaid balance method. Note: interest rate is given as a
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Unpaid balance 283.63-60=223.63
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5 0
1 year ago
If the point (4,-1) is a point on the graph of f then f
AlekseyPX

If the point (4,-1) is a point on the graph of y = f(x). The corresponding point on the graph of y = g(x) is: (4,-1/2), (2,-1), (-1,-1), (1,4), (4,-4), (-12,1)

  1. g(x) = \frac{1}{2} f(x) => (4,-1/2)
  2. g(x) = f(x-2) => (2,-1)
  3. g(x) = f(-x) => (-1,-1)
  4. g(x) = f(4x) => (1,4)
  5. g(x) = 4f(x) => (4,-4)
  6. g(x) = -f(x) => (-12,1)
<h3>Further explanation </h3>
  1. Dividing the function by 2 divides all the y-values by 2 as well. So to get the new point, we will take the y-value (-1) and divide it by 2 to get 2.  Therefore, the new point is  (4,-1/2)
  2. Subtracting 2 from the input of the function makes all of the x-values increase by 2 (in order to compensate for the subtraction). We will need to add 2 to the x-value (4) to get 2.  Therefore, the new point is (2,-1)
  3. Making the input of the function negative will multiply every x-value by  1.  To get the new point, we will take the x-value (4) and multiply it by -1  to get. Therefore, the new point is (-1,-1)
  4. Multiplying the input of the function by 4 makes all of the x-values be divided by 4 (in order to compensate for the multiplication). We will need to divide the x-value (4) by 4 to get 1.  Therefore, the new point is  (1,4)
  5. Multiplying the whole function by  4 increases all y-values by a factor of 4 , so the new y-value will be 4  times the original value (4) or -4.  Therefore, the new point is  (4,-4)
  6. Multiplying the whole function by -1 also multiplies every y-value by -1,  so the new y-value will be  -1 times the original value (-1). or 1. Therefore, the new point is (-12,1)

<h3>Learn more</h3>
  1. Learn more about corresponding point  brainly.com/question/10218370
  2. Learn more about point on the graph brainly.com/question/11297347
  3. Learn more about the graph brainly.com/question/11534295

<h3>Answer details</h3>

Grade:  9

Subject:  mathematics

Chapter:  corresponding point

Keywords:   corresponding point, the graph, point on the graph

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In the derivation of the quadratic formula by completing the square, the equation (x+b over 2a)^2=-4ac+b^2 over 4a^2 is created
vredina [299]

Answer:

The result of applying the square root property of equality to this equation is x=\dfrac{-b\pm \sqrt{b^2-4ac}}{2a}.

Step-by-step explanation:

Consider the provided equation.

\left(x+\dfrac{b}{2a}\right)^2=\dfrac{-4ac+b^2}{4a^2}

As the above equation is formed by perfect square trinomial so simply applying the square root property as shown:

\sqrt{(x+\dfrac{b}{2a})^2}=\pm \dfrac{\sqrt{-4ac+b^2}}{\sqrt{4a^2}}\\x+\dfrac{b}{2a}=\pm \dfrac{\sqrt{b^2-4ac}}{2a}

Isolate the variable x.

x=-\dfrac{b}{2a}\pm \dfrac{\sqrt{b^2-4ac}}{2a}\\x=\dfrac{-b\pm \sqrt{b^2-4ac}}{2a}

Hence, the result of applying the square root property of equality to this equation is x=\dfrac{-b\pm \sqrt{b^2-4ac}}{2a}.

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