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3 years ago

Find the equation of the line that is parallel to y = 3x - 2 and contains the points ( 2,11 )

Mathematics

1 answer:

antiseptic1488 [7]3 years ago

7 0

Answer:

Below

Step-by-step explanation:

Let y' = mx + b be the equation of the second line

● m is the slope

● b is the y-intercept

The lines are parallel so they have the

same slope.

● y' = 3x + b

The second line crosses the point with the coordinates (2,11)

Replace y' by 11 and x by 2 to find b

● 11 = 3×2 + b

● 11 = 6 + b

Substract 6 from both sides

● 11 - 6 = 6 + b -6

● b = 5

● y' = 3x + 5

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If p:q=2/3:3 and p:r=3/4:2/3 calculate the ratio p:q:r. Giving the answer in its simplest form

adelina 88 [10]

Answer:

Given

if p:q=2/3:3 and p:r=3/4:1/2, calculate the ratio p:q:r giving your answer in its simplest form

We need to find the ratio p:q:r

Given p:q = 2/3 : 3 = 2/3 / 3 = 2/9

and p : r = 3/4 : 1/2 = 3/4 / 1/2 = 3/2

Now p/q = 2/9 and p/r = 3/2

We need to make p equal numerators so we get

p/q = 2/9 x 3/3 = 6/27 and

p/r = 2/3 x 3/2 = 6/4

Therefore p : q : r = 6 : 27 : 4

6 0

4 years ago

For the figures below, assume they are made of semicircles, quarter circles and squares. For each shape, find the area and perim

ICE Princess25 [194]

Answer:

Part a) The area of the figure is $\frac{9}{2}(4+\pi )\ cm^{2}$

Part b) The perimeter of the figure is $3(2+2\sqrt{2}+ \pi)\ cm$

Step-by-step explanation:

Step 1

Find the area of the figure

In this problem we have that

The figure ABC is the half of a square and the other figure is a semicircle

Find the area of the figure ABC

we have

$AB=6\ cm, BC=6\ cm$

The area of the half square ABC is equal to find the area of triangle ABC

$A1=\frac{1}{2}*6*6=18\ cm^{2}$

Find the area of the semicircle

The area of the semicircle is equal to

$A2=\pi r^{2}/2$

we have that

$r=6/2=3\ cm$

substitute

$A2=\pi (3)^{2}/2$

$A2=(9/2) \pi\ cm^{2}$

The area of the figure is equal to

$18\ cm^{2}+(9/2) \pi\ cm^{2}= \frac{9}{2}(4+\pi )\ cm^{2}$

Step 2

Find the perimeter of the figure

The perimeter of the figure is equal to

$P=AB+AC+length\ CB$

we have

$AB=6\ cm$

Applying Pythagoras theorem

$AC=\sqrt{6^{2}+6^{2}}\\AC=6\sqrt{2}\ cm$

Remember that

the circumference of a semicircle is equal to

$C=\frac{1}{2}2\pi r=\pi r$

$r=6/2=3\ cm$

$C=\pi(3)$

$C=3 \pi\ cm$

The perimeter of the figure is equal to

$P=6\ cm+6\sqrt{2}\ cm+3 \pi\ cm$

Simplify

$P=3(2+2\sqrt{2}+ \pi)\ cm$

5 0

3 years ago

What is the probability of rolling a product of 30 when rolling two dice?

Nimfa-mama [501]

It has a probability of 2 out of 36 chances as there are only two ways it can happen out of 36 different combinations that can be rolled. Product means multipication also in Maths

The first probability on the first roll is that Dice 1 must be a 5 and Dice 2 must be a 6 so that 5x6=30 

The second probability on the second roll is that Dice 1 must be 6 while Dice 2 must be 5 so that 6x5=30 

That is the only two ways out of 36 different combinations that can be made by rolling to Die. Therefore the probability is 2 out of 36

8 0

3 years ago

Learning Task 4. Solve the problem by applying the sum and product of roots

Delvig [45]

Answer:

Thus, the dimensions of the metal plate are 10 dm and 8 dm.

Step-by-step explanation:

For a quadratic equation:

$x^2+bx+c=0$

The sum of the roots is -b and the product is c. Note the leading coefficient is 1.

We know the perimeter of the rectangular metal plate is 36 dm and its area is 80 dm^2. Being L and W its dimensions, then:

P=2(L+W)=36

A=L.W=80

Note both formulas are closely related to the roots of the quadratic equation, we only need to adjust the data for the perimeter to be exactly the sum of L+W and not double of it.

Thus we use the semi perimeter instead as P/2=L+W=18

The quadratic equation is, then:

$x^2-18x+80=0$