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3 years ago

I need help finding the slope of a line

Mathematics

1 answer:

solniwko [45]3 years ago

6 0

Answer:

Your answer is C=1/5

Step-by-step explanation:

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The Ferrels save $150 each month for their next summer vacation. Write an equation that they can use to find s, their savings, a

jeka57 [31]

One equation you can use is:
$150*M=S

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3 years ago

Find the area of this triangle

VARVARA [1.3K]

The answer is 308 by multiple

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3 years ago

Read 2 more answers

pls pls pls, I beg you to answer this question only if you know the correct answer please please please I beg you

Likurg_2 [28]

Answer:

Median is the middle line of the box, Lower quartile is the number of the lower end of the box, Upper quartile is the line at the larger end of the box, Maximum is the largest number in the box plot (end of right whisker), Minimum is the lowest number (end of left whisker).

Step-by-step explanation:

32.

Median: 88

Maximum (largest number): 102

Minimum (lowest number): 38

Interquartile range:

96 - 72 = 24

25 % higher

75 % lower?

You can see the range is more towards the right of the box meaning they have higher scores, and that the median or middle is a higher number so it means that most of the students understood and did well on the test.

???.

First quartile: 16

Second: 19

Third: 20

Range: 4

34. Since they are already in least to greatest order, all we have to do is graph it!

85 - 99 is 2

100 - 114 is 1

115 - 129 is 5

130 - 144 is 6

3 0

2 years ago

How do you solve X/2-5=10

Afina-wow [57]

Answer: x2-5=10

Two solutions were found :

x = ± √15 = ± 3.8730

Reformatting the input :

Changes made to your input should not affect the solution:

(1): "x2" was replaced by "x^2".

Rearrange:

Rearrange the equation by subtracting what is to the right of the equal sign from both sides of the equation :

x^2-5-(10)=0

Step by step solution :

Step 1 :

Trying to factor as a Difference of Squares :

1.1 Factoring: x2-15

Theory : A difference of two perfect squares, A2 - B2 can be factored into (A+B) • (A-B)

Proof : (A+B) • (A-B) =

A2 - AB + BA - B2 =

A2 - AB + AB - B2 =

A2 - B2

Note : AB = BA is the commutative property of multiplication.

Note : - AB + AB equals zero and is therefore eliminated from the expression.

Check : 15 is not a square !!

Ruling : Binomial can not be factored as the difference of two perfect squares.

Equation at the end of step 1 :

x2 - 15 = 0

Step 2 :

Solving a Single Variable Equation :

2.1 Solve : x2-15 = 0

Add 15 to both sides of the equation :

x2 = 15

When two things are equal, their square roots are equal. Taking the square root of the two sides of the equation we get:

x = ± √ 15

The equation has two real solutions

These solutions are x = ± √15 = ± 3.8730

Two solutions were found :

x = ± √15 = ± 3.8730

Step-by-step explanation:

5 0

3 years ago

(A)using geometry vocabulary, describe a sequence of transformations that maps figure P (-1,2)(-1,4) (-4,2) (-4,4) onto figure Q

andrey2020 [161]

Before we proceed on determining the transformation happening on this problem, it's better to see first the location of the figure by drawing it in a cartesian coordinate plane. We have

If we observe the figures and the coordinates of the plot, we can see that there is a difference of 1 on the x coordinates of P and y coordinates of Q. Therefore, the first transformation that we consider here is the movement of figure P by 1 unit to the left. We have

$\begin{gathered} P_1=(-1-1,2_{})=(-2,2) \\ P_2=(-1-1,4)=(-2,4) \\ P_3=(-4-1,2)=(-5,2) \\ P_4=(-4-1,4)=(-5,4) \end{gathered}$

This transformation changes the location of figure P into

The next transformation will be the rotation of the red dotted figure on the figure above by 90 degrees counterclockwise. With this transformation, the coordinates will transform as

$P_{ccw,90}=(-y,x)$

Hence, for the rotation, we have the new coordinates.

$\begin{gathered} P_1^{\prime}=(-2,-2) \\ P_2^{\prime}=(-4,-2) \\ P_3^{\prime}=(-2,-5) \\ P_4^{\prime}=(-4,-5) \end{gathered}$

The transformed image, which is represented as NMPO, will now be at

For the last transformation, we will be reflecting the figure NMPO over the <em>y</em> axis. This changes the coordinates as

$P_{\text{rotation,y}-\text{axis}}=(-x,y)$

We now have the new coordinates:

$\begin{gathered} P^{\doubleprime}_1=(2,-2)=Q_1_{}_{} \\ P_2^{\doubleprime}=(4,-2)=Q_3 \\ P_3^{\doubleprime}=(2,-5)=Q_2 \\ P_4^{\doubleprime}_{}=(4,-5)=Q_4_{} \end{gathered}$

As you can see, they have the same coordinates as figure Q.

The mapping rules for the sequence described above are as follows:

First transformation (moving one unit to the left (x-1,y))

$\begin{gathered} P_1(-1,2)\rightarrow P_1(-1-1,2)\rightarrow P_1(-2,2) \\ P_2(-1,4)\rightarrow P_1(-1-1,4)\rightarrow P_2(-2,4) \\ P_3(-4,2)\rightarrow P_1(-4-1,2)\rightarrow P_3(-5,2) \\ P_4(-4,4)\rightarrow P_1(-4-1,4)\rightarrow P_4(-5,4) \end{gathered}$

Second transformation (rotation counter clockwise (-y,x))

$\begin{gathered} P_1(-2,2)\rightarrow P^{\prime}_1(-2,-2)_{} \\ P_2(-2,4)\rightarrow P^{\prime}_2(-4,-2) \\ P_3(-5,2)\rightarrow P^{\prime}_3(-2,-5)_{} \\ P_4(-5,4)\rightarrow P^{\prime}_4(-4,-5)_{} \end{gathered}$

Third Transformation (reflection over y-axis (-x,y))

$\begin{gathered} P^{\prime}_1(-2,-2)\rightarrow P^{\doubleprime}_1(-(-2),-2)\rightarrow P^{\doubleprime}_1=(2,-2)=Q_1 \\ P^{\prime}_2(-4,-2)\rightarrow P^{\doubleprime}_1(-(-4),-2)\rightarrow P^{\doubleprime}_1=(4,-2)=Q_3 \\ P^{\prime}_3(-2,-5)\rightarrow P^{\doubleprime}_1(-(-2),-5)\rightarrow P^{\doubleprime}_1=(2,-5)=Q_2 \\ P^{\prime}_4(-4,-5)\rightarrow P^{\doubleprime}_1(-(-4),-5)\rightarrow P^{\doubleprime}_1=(4,-5)=Q_4 \end{gathered}$

7 0

1 year ago