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3 years ago

14

Maria solved the equation -5 n = -12. Her answer was n = 2. She is fairly confident about her answer, but wants to do a quick ch

eck of her solution using estimation. Which of the following shows a good check of her answer?

1 answer:

777dan777 [17]3 years ago

3 0

Plug in 2 to see if she is correct. Hope this helps!

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Which system of linear inequalities has the point (3, -2) in its solution set?

Licemer1 [7]

Answer:

None of the given system of linear inequalities

Step-by-step explanation:

Given

$(x,y) =(3,-2)$

Required

The line inequalities with the above solution

The first set of linear inequalities, we have:

$y < -3$

$y \ge \frac{2}{3}x - 4$

$y < -3$ implies that the values of y is -4,-5.....

While $(x,y) =(3,-2)$ implies that y = -2

Hence, the first set is wrong

The second set of linear inequalities, we have:

$y > - 2$

$y \ge \frac{2}{3}x - 4$

$y > - 2$ implies that the values of y is -1,0.....

While $(x,y) =(3,-2)$ implies that y = -2

Hence, the second set is wrong

5 0

3 years ago

On a coordinate plane, a shape is plotted with vertices of (3, 1), (0, 4), (3, 7), and (6, 4). what is the area of the shape if

QveST [7]

Let
<span>A (3, 1)
B (0, 4)
C(3, 7)
D (6, 4)

step 1
find the distance AB
d=</span>√[(y2-y1)²+(x2-x1)²]------> dAB=√[(4-1)²+(0-3)²]-----> dAB=√18 cm

step 2
find the distance CD
d=√[(y2-y1)²+(x2-x1)²]------> dCD=√[(4-7)²+(6-3)²]-----> dCD=√18 cm

step 3
find the distance AD
d=√[(y2-y1)²+(x2-x1)²]------> dAD=√[(4-1)²+(6-3)²]-----> dAD=√18 cm

step 4
find the distance BC
d=√[(y2-y1)²+(x2-x1)²]------> dBC=√[(7-4)²+(3-0)²]-----> dBC=√18 cm

step 5
find slope AB and CD
m=(y2-y1)/(x2-x1)
mAB=-1
mCD=-1
AB and CD are parallel and AB=CD

step 6
find slope AD and BC
m=(y2-y1)/(x2-x1)
mAD=1
mBC=1
AD and BC are parallel and AD=BC
and
AB and AD are perpendicular
BC and CD are perpendicular

therefore
the shape is a square wit length side √18 cm

area of a square=b²
b is the length side of a square
area of a square=(√18)²------> 18 cm²

the answer is
18 cm²

see the attached figure

6 0

3 years ago

Value of 8 power of -3

lozanna [386]

Answer:

Step-by-step explanation: $8^{-3} =\frac{1}{8^{3} } =\frac{1}{8X8X8} =\frac{1}{512}$

7 0

4 years ago

A circular swimming pool has a diameter of 40 meters. The depth of the pool is constant along west-east lines and increases line

Nataliya [291]

Answer:

Volume is $2000\pi\ m^{3}$

Solution:

As per the question:

Diameter, d = 40 m

Radius, r = 20 m

Now,

From north to south, we consider this vertical distance as 'y' and height, h varies linearly as a function of y:

iff

h(y) = cy + d

Then

when y = 1 m

h(- 20) = 1 m

1 = c.(- 20) + d = - 20c + d (1)

when y = 9 m

h(20) = 9 m

9 = c.20 + d = 20c + d (2)

Adding eqn (1) and (2)

d = 5 m

Using d = 5 in eqn (2), we get:

$c = \frac{1}{5}$

Therefore,

$h(y) = \frac{1}{5}y + 5$

Now, the Volume of the pool is given by:

$V = \int h(y)dA$

where

A = $r\theta$

$A = rdr\ d\theta$

Thus

$V = \int (\frac{1}{5}y + 5)dA$

$V = \int_{0}^{2\pi}\int_{0}^{20} (\frac{1}{5}rsin\theta + 5) rdr\ d\theta$

$V = \int_{0}^{2\pi}\int_{0}^{20} (\frac{1}{5}r^{2}sin\theta + 5r}) dr\ d\theta$

$V = \int_{0}^{2\pi} (\frac{1}{15}20^{3}sin\theta + 1000) d\theta$

$V = [- 533.33cos\theta + 1000\theta]_{0}^{2\pi}$

$V = 0 + 2\pi \times 1000 = 2000\pi\ m^{3}$

7 0

3 years ago

How you do you solve for it??

AysviL [449]

Sam is dead, im sorry to have to tell you over line, i love you mom
With love,
Mike

6 0

3 years ago