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3 years ago

Please help me with this problem!!

Mathematics

1 answer:

aleksandrvk [35]3 years ago

7 0

Answer:

=51 units^2

Step-by-step explanation:

When using the area formula, he needs to be careful to label correctly.

The sides opposite the angles have the same letters

Angle A = unknown a = 17

Angle B = 45 side b = 12

Angle C = 30 side c=8

Area = 1/2 ab sin C

= 1/2 ( 12*17) sin 30

=51 units^2

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Need help with 73, 74, 75, & 76. Find the value of X.

bija089 [108]

Answer:

see explanation

Step-by-step explanation:

73.

18 = 3x $\sqrt{3}$ (divide both sides by 3 )

6 = x $\sqrt{3}$ ( multiply both sides by $\sqrt{3}$ )

6 $\sqrt{3}$ = 3x [ $\sqrt{3}$ × $\sqrt{3}$ = 3 ]

divide both sides by 3

x = 2 $\sqrt{3}$

74.

24 = 2x $\sqrt{2}$ ( divide both sides by 2 )

12 = x $\sqrt{2}$ ( multiply both sides by $\sqrt{2}$ )

12 $\sqrt{2}$ = 2x ( divide both sides by 2 )

x = 6 $\sqrt{2}$

75.

9 $\sqrt{2}$ × x = 18 $\sqrt{2}$ ( divide both sides by 9 $\sqrt{2}$ )

x = 2

76.

2 = x × $\frac{4}{\sqrt{3} }$ ( multiply both sides by $\sqrt{3}$ )

2 $\sqrt{3}$ = 4x ( divide both sides by 4 )

x = $\frac{\sqrt{3} }{2}$

8 0

3 years ago

Read 2 more answers

A person's metabolic rate is the rate at which the body consumes energy. Metabolic rate is important in studies of weight gain,

cupoosta [38]

Let m be mean

Mean= sum/ n

Mean= (1720+1687+1367+1614+1460+1867+1436) / 7

m= 11151 / 7

M= 1593

Mean= 1593

Standard deviation

|x-m|^2

For 1st: |1720-1593|^2=8836

For 2nd: |1687-1593|^2=10201

For 3rd: |1367-1593|^2=51076

For 4th: |1614-1593|^2=441

For 5th: |1460-1593|^2=1689

For 6th: |1867-1593|^2=75076

For 7th: |1436-1593|^2=24649

Summation of |x-m|^2 = 171968

Standard deviation sample formula is:

S.D = sqrt((summation of |x-m|^2) / n-1)

S.D=sqrt(171968/6)

S.D=sqrt(28661.33)

S.D=169.30

Standard deviation is 169.30

3 0

3 years ago

3. Solve the inequality –2(z + 5) + 20 > 6.

12345 [234]

$\boxed{-2\left(z+5\right)+20>6\quad :\quad \begin{bmatrix}\mathrm{Solution:}\:&\:z$

Solution:

Given inequality is:

$-2(z+5)+20>6$

We have to solve the given inequality

$-2\left(z+5\right)+20>6\\\\\mathrm{Subtract\:}20\mathrm{\:from\:both\:sides}\\\\-2\left(z+5\right)+20-20>6-20\\\\\mathrm{Simplify}\\\\-2\left(z+5\right)>-14$

$Multiply\ both\ sides\ by\ -1\ \left(reverse\:the\:inequality\right)$

Whenever we multiply or divide an inequality by a negative number, we must flip the inequality sign

$\left(-2\left(z+5\right)\right)\left(-1\right)$

Thus the solution to inequality is:

$-2\left(z+5\right)+20>6\quad :\quad \begin{bmatrix}\mathrm{Solution:}\:&\:z$

4 0

3 years ago

A. If she buys a pair of jeans for $102.25 and two shirts that each cost $44.27

slava [35]

Answer:

$481.54

Step-by-step explanation:

672.33-(102.25+(2)44.27)

672.33-(102.25+88.54)

672.33-190.79

481.54

6 0

3 years ago

The coordinate plane below represents a city. Points A through F are schools in the city. graph of coordinate plane. Point A is

Firlakuza [10]

Part A;
There are many system of inequalities that can be created such that only contain points C and F in the overlapping shaded regions.

Any system of inequalities which is satisfied by (2, 2) and (3, 4) but is not stisfied by (-3, -4), (-4, 3), (1, -2) and (5, -4) can serve.

An example of such system of equation is
x > 0
y > 0
The system of equation above represent all the points in the first quadrant of the coordinate system.
The area above the x-axis and to the right of the y-axis is shaded.

Part 2:
It can be verified that points C and F are solutions to the system of inequalities above by substituting the coordinates of points C and F into the system of equations and see whether they are true.

Substituting C(2, 2) into the system we have:
2 > 0
2 > 0
as can be seen the two inequalities above are true, hence point C is a solution to the set of inequalities.

Part C:
Given that Natalie can only attend a school in her designated zone and that Natalie's zone is defined by y < −2x + 2.

To identify the schools that Natalie is allowed to attend, we substitute the coordinates of the points A to F into the inequality defining Natalie's zone.

For point A(-3, -4): -4 < -2(-3) + 2; -4 < 6 + 2; -4 < 8 which is true

For point B(-4, 3): 3 < -2(-4) + 2; 3 < 8 + 2; 3 < 10 which is true

For point C(2, 2): 2 < -2(2) + 2; 2 < -4 + 2; 2 < -2 which is false

For point D(1, -2): -2 < -2(1) + 2; -2 < -2 + 2; -2 < 0 which is true

For point E(5, -4): -4 < -2(5) + 2; -4 < -10 + 2; -4 < -8 which is false

For point F(3, 4): 4 < -2(3) + 2; 4 < -6 + 2; 4 < -4 which is false

Therefore, the schools that Natalie is allowed to attend are the schools at point A, B and D.

7 0

3 years ago