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OLEGan [10]
3 years ago
12

PLZ i need anwse ASAP

Mathematics
1 answer:
KatRina [158]3 years ago
3 0

Surface area is the sum of area of all the faces .

In the net diagram, we have 6 faces. So we have to find the area of each face and add them to find the surface area .

And since each face is rectangle in shape, so we will use the formula of area of rectangle, which is

Area= length*width

So the required surface area is

A = (11*5)+(11*4)+(11*5) +(11*4)+(5*4)+(5*4) \\A = 55+44+55+44+20+20 =238 mm^2

And that's the required surface area of the rectangular prism .

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Please help with all!!
EastWind [94]

Answer:

1. x=7

2. x= -12

3.x=12

4.x=9

5. x= -5

6. x=7

Step-by-step explanation:


4 0
3 years ago
Which number is less than -3 or greater than 5?
Anna71 [15]

Answer:

4

Step-by-step explanation:

7 0
4 years ago
A function f(x) is transferred from the parent function p(x) = |x| by a reflection over the x-axis, vertical stretched by a fact
Dmitry [639]
Refrection over the x-axis changes |x| to -|x|.
Vertical stretch by a factor of 2.5 changes the function to -2.5|x|
Shifting the function down by 3 units changes the function to -2.5|x| - 3

Therefore, f(x) = -2.5|x| - 3
7 0
4 years ago
In a class of 10, there are 2 students who forgot their lunch. If the teacher chooses 2 students, what is the probability that b
Anuta_ua [19.1K]

At the beginning, the chances that the teacher will choose a student who has forgotten his lunch are 2/10, or 1/5.

If the teacher now chooses another student from the remaining 9 students, the chances of his choosing a student who has forgotten his lunch are 1/9, since one of the forgetful students has already been removed from the original group of 10 students, and there are only 9 in the group remaining.

The probability that both students forgot lunch is (1/5)(1/9), or 1/45.

7 0
3 years ago
A sheet of paper 90 cm-by-66 cm is made into an open box (i.e. there's no top), by cutting x-cm squares out of each corner and f
NNADVOKAT [17]

Answer:

26 - \sqrt{181} cm

Step-by-step explanation:

The volume of the box is:

V = height * length * width

V = x*(66 - 2*x)*(90 - 2*x)

V = (66*x - 2*x^2)*(90 - 2*x)

V = 5940*x - 132*x^2 - 180*x^2 + 4*x^3

V = 4*x^3 - 312*x^2 + 5940*x

where x is the length of the sides of the squares,  in cm.

The mathematical problem is :

Maximize: V = 4*x^3 - 312*x^2 + 5940*x

subject to:

x > 0

2*x < 66 <=> x < 33

In the maximum, the first derivative of V, dV/dx, is equal to zero

dV/dx = 12*x^2 - 624*x + 5940

From quadratic formula

x = \frac{-b \pm \sqrt{b^2 - 4(a)(c)}}{2(a)}

x = \frac{624 \pm \sqrt{(-624)^2 - 4(12)(5940)}}{2(12)}

x = \frac{624 \pm \sqrt{104256}}{24}

x = \frac{624 \pm \sqrt{2^6*3^2*181}}{24}

x = \frac{624 \pm 8*3*\sqrt{181}}{24}

x_1 = \frac{624 + 24*\sqrt{181}}{24}

x_1 = 26 + \sqrt{181}

x_2 = \frac{624 - 24*\sqrt{181}}{24}

x_2 = 26 - \sqrt{181}

But x_1 > 33, then is not the correct answer.

5 0
3 years ago
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