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3 years ago

Can someone tell me everything they know about Introducing Proportional Relationships?

Mathematics

1 answer:

Nookie1986 [14]3 years ago

4 0

Answer:

1.Radius if half of diameter

2.If you want to get to circumference it diameter x 3.14

3.To find area of circle r x r x 3.14

4.From circumferenc to diameter it is circumfrence

divided by 3.14

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In parallelogram DEFG if m_DEF=110° find m_FGD. E 110° D G

Sergio039 [100]

Answer:

E=x

or 110= x ( opposite angles of parallelogram are congurent)

E+D = 180 ( angles on st line)

110+D =180

D=70

D=F ( opposite angles of parallelogram are congurent)

so F=70

3 0

3 years ago

What is the next term in the sequence below? -324,108,-36,12

STALIN [3.7K]

Answer:

-4

Step-by-step explanation:

bc it is

.......................

4 0

3 years ago

The area of a quadrilateral is 64 in^2. The length of the quadrilateral is 32 in. How wide is the quadrilateral?

Mars2501 [29]

Answer: 2 inches

Explanation: The area formula is length times width (L*w=a). We know that the area is 64in^2 (a=64) and the length is 32in (L=32), so we put those values into the equation and get 32*w=64. We then get w by itself by dividing both sides by 32, and get w=2. so the width is 2 inches.

7 0

3 years ago

Solve P = 2(1 + w) for I

Paraphin [41]

Answer:

Step-by-step explanation:

2(l + w) = P

l + w = P/2

l = P/2 - w

Option 2

5 0

3 years ago

Two forces with magnitudes of 150 and 75 pounds act on an object at angles of 30° and 150°, respectively. Find the direction and

Anastaziya [24]

The problem is modelled in the first picture shown below

To work out the resultant vector, we modelled the vectors 150N and 75N as triangle AOB is shown in the second picture with AB as the resultant vector.

We use the cosine rule to work out the length AB
$AB^{2}= 75^{2}+ 150^{2}-(2*75*150*cos(60))$
$AB^{2} =28125-11250$
$AB^{2}=16875$
$AB= \sqrt{16875} =130$ (nearest whole number)

The third picture shows the full diagram of the vectors

To work out the direction of the resultant vector, we use the sin rule to find the size of angle A and angle B

Angle A
$\frac{130}{sin(60)}= \frac{75}{sin(A)}$
$130sin(A)=75sin(60)$
$sin(A)= \frac{75sin(60)}{130}$
$sin(A)=0.4996300406$
$A= sin^{-1}(0.4996300406)$
$A=30$ (rounded to nearest whole number)

Angle B
$B=180-60-30=90$

Direction is 60° toward negative x-axis

Answer: Magnitude 130N and direction 60° toward negative x-axis

3 0

3 years ago