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3 years ago

What is the surface area of the square pyramid represented by the net below?

Mathematics

2 answers:

dmitriy555 [2]3 years ago

5 0

Answer:

Step-by-step explanation:

Morgarella [4.7K]3 years ago

4 0

Answer: 96

Step-by-step explanation:

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A family of two adults and four children is going to an amusement park.Admission is $21.75 for adults and $15.25 for children.Wh

Nimfa-mama [501]

The total cost would be $104.50

3 0

3 years ago

Please help me solve this :)

REY [17]

Answer:

105 cm

Step-by-step explanation:

An easy way to do this is by stacking the table on top of the other table

120 + 90 = 210

then you must how tallone table is, since the tables are the same, we divide by 2

210 ÷ 2 = 105

look at the picture

7 0

2 years ago

Read 2 more answers

Line l contains the points (1,5) and (4, -4). point P has the coordinates (-1, 1)

Colt1911 [192]

Answer:

The distance from p to l is $\sqrt{10}\ units$

Step-by-step explanation:

we know that

The distance between point p from line l is equal to the perpendicular segment from line l to point p

step 1

<em>Find the slope of line l</em>

we have the points

(1,5) and (4, -4)

The formula to calculate the slope between two points is equal to

$m=\frac{y2-y1}{x2-x1}$

substitute the values

$m=\frac{-4-5}{4-1}$

$m=\frac{-9}{3}$

$m=-3$

step 2

Find the equation of the line l

The equation in point slope form is equal to

$y-y1=m(x-x1)$

we have

$m=-3$

$point\ (1,5)$

substitute

$y-5=-3(x-1)$ -----> equation A

step 3

Find the slope of the line perpendicular to the line l

Remember that

If two lines are perpendicular, then their slopes are opposite reciprocal (The product of their slopes is equal to -1)

$m_1*m_2=-1$

we have

$m_1=-3$ ---> slope of line l

therefore

$m_2=\frac{1}{3}$ ----> slope of the line perpendicular to line l

step 4

Find the equation of the line perpendicular to line l that passes through the point p

The equation in point slope form is equal to

$y-y1=m(x-x1)$

we have

$m=\frac{1}{3}$

$point\ p(-1,1)$

substitute

$y-1=\frac{1}{3}(x+1)$ -----> equation B

step 5

Solve the system of equations

$y-5=-3(x-1)$ -----> equation A

$y-1=\frac{1}{3}(x+1)$ -----> equation B

Solve the system by graphing

The solution of the system is the intersection point both graphs

The solution is the point q(2,2)

see the attached figure

step 6

we know that

The distance between the point p and the line l is equal to the distance between the point p and the point q

the formula to calculate the distance between two points is equal to

$d=\sqrt{(y2-y1)^{2}+(x2-x1)^{2}}$

we have the points

p(-1,1) and q(2,2)

substitute the values

$d_p_q=\sqrt{(2-1)^{2}+(2+1)^{2}}$

$d_p_q=\sqrt{(1)^{2}+(3)^{2}}$

$d_p_q=\sqrt{10}\ units$

3 0

3 years ago

Write an equation for the line that is parallel to the given line and that passes through the given point.

Tom [10]

Answer:

y - 11 = 7(x - 3)

y - 11 = 7x - 21

y = 7x -10

3 0

2 years ago

ALGEBRA 1 PLEASE HELP

solong [7]

Subtract the equations straight down. So we'll subtract the x terms together
5x - 5x = 0x = 0
meaning the x terms effectively go away after subtraction

Then we do the y terms together in the same exact order
y-3y = -2y

Then finally the right hand side values
6 - (-5) = 6+5 = 11

All of that work done indicates we have 0x-2y = 11 which is the same as -2y = 11. Divide both sides by -2 and we end up with y = -11/2

This answer isn't listed so I'm thinking there's a typo your teacher made somewhere.

8 0

3 years ago