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3 years ago

Write an equation of a line perpendicular to the line y=-5x+6

Mathematics

1 answer:

FrozenT [24]3 years ago

3 0

Answer:

Step-by-step explanation:

From the condition of perpendicularity, for two lines to be perpendicular to each other the product of their gradients must equal to ( -1 ).

Same time, equation of a straight line is ; y = mx+ c where m us the gradient and c is the point of intersection of the line on y axis. Therefore, the gradient of this equation m1 ,= -5 and the gradient m2 of the second line will be 1/5. Therefore, the equation of the second line perpendicular to

y = -5x + 6 ; y = x/5 + 6 .

Therefore, y = x/5 + 6 is the new equation.

You might be interested in

A long distance telephone company offer two plans . The deluxe plan cost $30 a month and offers unlimited long distance calling.

Gnesinka [82]

Answer:

399 minutes a month

Step-by-step explanation:

As I understand the question the answer would, in other words, be how many minutes you can long-distance call with the economy plan for under $30.

The unit ratio is 5 cents per minute, which equates to 20 minutes worth of call time for one dollar (100/5).

The economy plan is $20 cheaper than the deluxe plan. $20 spent on long-distance calling gets you 400 mintues, but the question asks for an integer that would still leave remaining money.

Therefore the answer is 399 minutes of long-distance calling, which would leave five unspent cents.

8 0

2 years ago

What is the arc measure of an arc with length 4.189 cm and radius equal to 3 cm.'?

Mazyrski [523]

Given that the arc length is 4.189 cm and the radius is 3 cm, the size of the arc will found as follows;
C=theta/360 πd
suppose:
size of arc=theta=x
d=3*2=6 cm
hence;
4.189=x/360*π*6
4.189=0.0524x
x=4.189/0.0524
x=80.004°
The size of the arc length is 80.004°

6 0

3 years ago

Use two different methods to find an explain the formula for the area of a trapezoid that has parallel sides of length a and B a

evablogger [386]

Answer:

Formula of Trapezoid:

A = (a + b) × h / 2

The formula can be derived in different ways. for now, we have discussed two ways:

1. By using the formula of a triangle

2. By dividing into different sections

Step-by-step explanation:

1. By using the formula of a triangle

One of the ways to explain a formula for an area of a trapezoid using a formula for a triangle can be as follows.

Assume a trapezoid PQRS with lower base SR and upper base PQ (they are parallel) and sides PS and QR.

The image is attached below.

Connect vertices P and R with a diagonal.

Consider triangle ΔPQR as having a base PQ and an altitude from vertex R down to point M on base PQ (RM⊥PQ).

Its area is

S1= $\frac{1}{2} *PQ*RM$

Consider triangle ΔPRS as having a base SR and an altitude from vertex P up to point N on-base SR (PN⊥SR).

Its area is

S2= $\frac{1}{2} *SR*PN$

Altitudes RM and PN are equal and constitute the distance between two parallel bases PQ and SR.

They both are equal to the altitude of the trapezoid h.

Therefore, we can represent areas of our two triangles as

S1= $\frac{1}{2}*PQ*h$

S2= $\frac{1}{2}*SR*h$

Adding them together, we get the area of the whole trapezoid:

S=S1+S2= $\frac{1}{2} (PQ+SR)h,$

which is usually represented in words as "half-sum of the bases times the altitude".

2. By dividing into different sections

Trapezoid PQRS is shown below, with PQ parallel to RS.

Figure 1 - Trapezoid PQRS with PQ parallel to RS(image is attached below.)

We are going to derive the area of a trapezoid by dividing it into different sections.

If we drop another line from Q, then we will have two altitudes namely PT and QU.

Figure 2 - Trapezoid PQRS divided into two triangles and a rectangle. (image is attached below.)

From Figure 2, it is clear that Area of PQRS = Area of PST + Area of PQUT + Area of QRU. We have learned that the area of a triangle is the product of its base and altitude divided by 2, and the area of a rectangle is the product of its length and width. Hence, we can easily compute the area of PQRS. It is clear that

=> $A_{PQRS} = (\frac{ah}{2}) + b_{1}h + \frac{ch}{2}$