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3 years ago

I need help dont know how to do this.

Mathematics

1 answer:

UkoKoshka [18]3 years ago

8 0

To solve the question we need to calculate the equation of the straight line;
the equation is given by:
m=(y1-y)/(x2-x)
where;
m is the slope;
selecting 2 points from the graph say (0,2) and (2,0), the slope of the graoh will be:
m=(0-2)/(2-0)=-1
thus using point (-2,4) the equation will be given by:
-1=(y-4)/(x+2)
-1(x+2)=(y-4)
this can be written as:
-x-2=y-4
this can be written as:
y+2=-x+4
this can be written as:
y+2=-(x-2)
The answer is A.y+2=-(x-2)

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3 years ago

Help me asap i need it

s344n2d4d5 [400]

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3 years ago

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Answer:

Step-by-step explanation:

RQ < PR < PQ

TV < UV < UT

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$\dfrac{8}{x-3}=\dfrac{14}{x+6}\\\\8(x+6)=14(x-3)\\\\8x+48=14x-42\\\\48+42=14x-8x\\\\90=6x\\\\x=90:6=15$

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3 years ago

Quadrilateral ABCD has vertices A(1,0) B(5,0) C (7,2) D(3,2). Use slope to prove that ABCD is a parallelogram.

Galina-37 [17]

Answer:

$AD \parallel BC$

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Step-by-step explanation:

The vertices of quadrilateral ABCD are A(1,0) B(5,0) C (7,2) D(3,2).

The slope of side AB is

$M_{AB}=\frac{0-0}{5-1} =\frac{0}{4}=0$

The slope of side BC is

$M_{BC}=\frac{2-0}{7-5} =\frac{2}{2}=1$

The slope of side CD is

$M_{CD}=\frac{2-2}{3-7} =\frac{0}{-4}=0$

The slope of AD is

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$AD \parallel BC$

$AB \parallel CD$

We see that the opposite sides of the quadrilateral ABCD are equal.

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3 years ago

Item 14 A student solves the inequality −6v>42. Each step in the student's solution is shown below. Identify where the studen

Marizza181 [45]

Answer:

Step-by-step explanation:

Given the inequality solved by a student expressed as:

-6v>42

To get v, follow the simple steps

Step 1: multiply both sides by -1

-6v>42

-1(-6v)<-1(42)

6v < 42 (Note that when you multiply both sides of an inequality by a negative sign, the inequality sign will change)

Step 2: Divide through by 6

6v < 42

6v/6 < 42/6

v < 7

Hence the range of values of v are the values of v less than 7

Since we are not given the options, you can compare the solution given with that of the student to figure out the error. The major error that may happen is the different not changing the inequality sign after multiplying or dividing with a negative value as shown.

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2 years ago