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ser-zykov [4K]
3 years ago
11

Write the equation of the line that passes through the point (-2, 7) and is PARALLEL to the line y = -4x + 1.​

Mathematics
1 answer:
qwelly [4]3 years ago
3 0

Answer:

The equation is y=x-9

Step-by-step explanation:

If we want a line parallel to the line in the problem it will have the same slope. The equation given in the problem is already in slope intercept form so we can take the slope for the new equation directly from this equation.

The slope-intercept form of a linear equation is:  y=mx+b

Where  m  is the slope and  b  is the y-intercept value.  

y=1x-2

Therefore the slope is  

m=1

We can now use the point-slope formula to write the equation of the parallel line:

The point-slope formula states:  

(y-y1)=m(x-x1)

Where  m  is the slope and  (x1,y1) is a point the line passes through.  

Substituting the information from the problem gives:

(y- (-7)=1(x-2)

(y+7)=1(x-2)

Or, we can solve for  y  to put the equation in slope-intercept form:

y+7=x-2

y+7-7=x-2-7

y+0=x-9

y=x-9

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Answer:

y = 5x +4

Step-by-step explanation:

x has to equal 1

y = 5(1) +4 = 9

8 0
3 years ago
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Step-by-step explanation: YOU´RE STUPID

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If 2.5 mol of dust particles were laid end to end along the equator, how many times would they encircle the planet? The circumfe
Natalka [10]

Answer:

They encircle the planet 3.76\times 10^{11} times.

Step-by-step explanation:

Consider the provided information.

We have 2.5 mole of dust particles and the Avogadro's number is 6.022\times 10^{23}

Thus, the number of dust particles is:

2.5\times 6.022\times 10^{23}=15.055\times 10^{23}

Diameter of a dust particles is 10μm and the circumference of earth is 40,076 km.

Convert the measurement in meters.

Diameter: 10\mu m\times \frac{10^{-6}m}{\mu m} =10^{-5}m

If we line up the particles the distance they could cover is:

15.055\times 10^{23}\times 10^{-5}=15.055\times 10^{18}=1.5055\times 10^{19}

Circumference in meters:

40,076km\times \frac{1000m}{1km}=40,076,000 m

Therefore,

\frac{1.5055\times 10^{19}}{40,076,000} = 3.76\times 10^{11}

Hence, they encircle the planet 3.76\times 10^{11} times.

8 0
3 years ago
(no links pls) find the value of m
Nina [5.8K]

Answer:

D.

<P= 5

I hope this helps:)

6 0
3 years ago
A figure is composed of a rectangle and a right triangle. What is the area of the figure?
strojnjashka [21]

QUESTION 1

The figure is a tra-pezoid.

The area of a trapezoid is given by the formula;

A=\frac{1}{2}(Sum\:of\:parallel\:sides)\times h

We substitute the values to obtain;

Area=\frac{1}{2}(13.2cm+8.4cm)\times 6cm

Area=\frac{1}{2}(21.6cm)\times 6cm

Area=64.8cm^2

QUESTION 2

The approximate length of the ribbon is equal to the circumference of the circular table cloth.

We can find this by using the formula for calculating the circumference of a circle.

C=2\pi r

where r=3.5ft is the radius of the circular table cloth.

We substitute this value and \pi=3.14 into the formula to get;

C=2(3.14)(3.5)ft

C=21.98ft

The approximate length of the ribbon is 22.0ft to the nearest tenth.

QUESTION 3

Type of quadrilateral:Rectangle

Explanation: A rectangle has 4 angles that are right angles.

The two pairs of opposite sides of a rectangle are also congruent.

QUESTION 4.

The circumference of a semi circle is calculated using the formula;

C=\pi r

where r=12.5cm is the radius of the semicircle.

C=3.14\times 12.5cm

C=39.25cm

QUESTION 5

The area of the entire figure is the area of the semicircle plus the area of the isosceles triangle.

Area=\frac{1}{2}\pi r^2+\frac{1}{2}bh

Area=\frac{1}{2}\times3.14\times12.5^2+\frac{1}{2}\times25\times 24

Area=345.3125+300

Area=645.3125cm^2

Area\approx645cm^2

QUESTION 6

The area of the parallelogram is given by the formula;

A=bh

From the diagram, b=y\:in.,h=3in..

Given, A=23.7 inches squared.

We substitute the values to obtain;

23.7=3y

y=\frac{23.7}{3}

y=7.9in.

QUESTION 7

The area of a rectangle is given by the formula;

Area=l\times b

The area of the bigger rectangle =9\times15=135in^2

The area of the smaller rectangle =8\times 13=104in^2

The area of the  shaded region =135-104=31in^2

5 0
4 years ago
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