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KiRa [710]
3 years ago
10

Plz help I’m stuckkk

Mathematics
1 answer:
Shalnov [3]3 years ago
7 0

check the picture below.


notice, you simply have 6 rectangles, thus simply get the area of each, sum them up and that's the area of the net.


(5*4) + (4*3) + (5*3) + (5*4) + (5*3) + (4*3).

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How to find the real number solution of the equation x3+27x=9x2+27?
Alina [70]
Well think the answer is x = 14/9 but I don't think there is a full number for this equation.
3 0
3 years ago
Write an equation of the line that is perpendicular to the given line and that passes through the given point. y = −4x + 2; (−4,
torisob [31]
Y=1/4x-2 would be the equation for the line perpendicular to that line and crosses through that ordered pair!
8 0
3 years ago
In a seventh grade class, 14 out of 30 students are boys. In simplest terms, what is the ratio of the number of boys to the numb
Triss [41]

{Hello,\;There!}

They already give you the number of boys {14}, so all you need to do is subtract 30 - 14 which will give you the ratio of girls.

It does this because 14 takes up \frac{14}{30} of the seventh grade class.

\text{30 - 14 = 16}

  So your answer is going to be 14:16 un-simplified.

  However, since they ask you to put it in the simplest terms, we are going to simplify \frac{14}{16}.

  You can do this by dividing both numbers by 2.

=  \frac{14 \div 2}{16 \div 2}=\frac{7}{8}

\bold{So\;your\;answer\;is\;going\;to\;be}

= \frac{7}{8} or \text{7:8}

\rule{300}{1.0}

8 0
3 years ago
Fit a quadratic function to these three points: (-2,8)(0,-4),and (4,68)
aev [14]
Given three points
P1(-2,8)
P2(0,-4)
P3(4,68)
We need the quadratic equation that passes through all three points.

Solution:
We first assume the final equation to be
f(x)=ax^2+bx+c .............................(0)

Observations:
1. Points are not symmetric, so cannot find vertex visually.
2. Using the point (0,-4) we substitute x=0 into f(x) to get
f(0)=0+0+c=-4, hence c=-4.
3. We will use the two other points (P1 & P3) to set up a system of two equations to find a and b.
f(-2)=a(-2)^2+b(-2)-4=8 => 4a-2b-4=8.................(1)
f(4)=a(4^2)+b(4)-4=68 =>  16a+4b-4=68.............(2)
4. Solve system
2(1)+(2) => 24a+0b-12=84 => 24a=96 => a=96/24 => a=4 ......(3)
substitute (3) in (2) => 16(4)+4b-4=68 => b=8/4 => b=2  ..........(4)
5. Put values c=-4, a=4, b=2 into equation (0) to get

f(x)=4x^2+2x-4

Check:
f(-2)=4((-2)^2)+2(-2)-4=16-4-4=8
f(0)=0+0-4 = -4
f(4)=4(4^2)+2(4)-4=64+8-4=68
So all consistent, => solution ok.

6 0
4 years ago
For which pair of numbers is 4 a common factor.
dolphi86 [110]

Answer: 40 and 52

Step-by-step explanation:

40 =4 \times 10, 52=4\times 13, so both 40 and 52 have a factor of 4.

6 0
2 years ago
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