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3 years ago

Pls help n hurry i will give brainliest

Mathematics

1 answer:

AlekseyPX3 years ago

8 0

Answer: A) and B)

Step-by-step explanation:

So for the first one, a closed circle means that it includes the point and the graph proceeds in the positive direction. So it will be whatever number it starts on in the x axis is less than or equal to x.

The second one you find the distance it moves each second.

You do this by distance over time d/t. Then you divide the 408 by that number.

So 264cm / 11 seconds = 24 cm/sec

Then 408cm/ 24cm per sec = 17s

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What is 1/3 +1/4 equal in math

defon

The answer to your question would be 7/12 I hope I answered your question today!

7 0

3 years ago

Read 2 more answers

Find the midpoint of PQ P(4 1/3, 3 1/6), Q(-2 1/5, 3 2/3)

Paha777 [63]

Answer:

$(1\frac{1}{15},3\frac{5}{12})$

Step-by-step explanation:

To find the midpoint, you must average the x's then average the y's.

So the x's are $4 \frac{1}{3}$ and $-2\frac{1}{5}$ .

The y's are $3\frac{1}{6}$ and $3\frac{2}{3}$ .

To find an average of a data with 2 elements you must add the two elements then take that sum and divide it by 2.

So this is what we will do with the x's then the y's.

So first x:

$\frac{4\frac{1}{3}+-2\frac{1}{5}}{2}$

Write as improper fractions:

$\frac{\frac{13}{3}+\frac{-11}{5}}{2}$

Division by 2 is the same as multiplying by 1/2:

$\frac{13}{6}+\frac{-11}{10}$

The least common multiple of 6 and 10 is 30. We will multiply first fraction by 5/5 and 3/3 for the second so we can have the same denominator.

$\frac{65}{30}+\frac{-33}{30}$

Now that the bottoms are the same we can write as one fraction:

$\frac{65+-33}{30}$

Simplify:

$\frac{32}{30}$

If you want the answer as a mix fraction like your question began as we can do that by first figuring out how many 30's are in 32 and what is the remainder of that division.

There is one 30 and 32 and so 32-30=2 is the remainder.

$1\frac{2}{30}$

Reduce by dividing top and bottom by 2:

$1\frac{1}{15}$

Now for y:

$\frac{3\frac{1}{6}+3\frac{2}{3}}{2}$

Write the mix fractions as improper fractions:

$\frac{\frac{19}{6}+\frac{11}{3}}{2}$

Dividing by 2 is the same as multiplying by 1/2:

$\frac{19}{12}+\frac{11}{6}$

The least common multiple of 12 and 6 is 12 so I'm going to multiply the second fraction by 2/2 so the denominators will be the same:

$\frac{19}{12}+\frac{22}{12}$

Since the denominators are the same we can write as a single fraction:

$\frac{41}{12}$

How many 12's are in 41? 3 and so the remainder is 41-3(12)=41-36=5.

$3\frac{5}{12}$

So the midpoint is:

$(1\frac{1}{15},3\frac{5}{12})$ .

6 0

3 years ago

Read 2 more answers

What are the solutions to. x^2 - x - 20 = 0

nignag [31]

Step-by-step explanation:

x² - x - 20 = 0

x² - (5-4) x - 20 = 0 

x² - 5x + 4x -20 = 0

x ( x - 5) + 4 ( x - 5) = 0

(x - 5) (x+ 4) = 0

Either. 

x - 5 = 0

x = 5

Or, 

x + 4 = 0 

x = -4

3 0

3 years ago

A line includes the points (4, 8) and (3, 5). What is its equation in slope-intercept form? Write your answer using integers, pr

stich3 [128]

Answer:

A line includes the points (4, 8) and (3, 5) whose equation in slope-intercept form will be :

$y=3x-4$

Step-by-step explanation:

The point slope form: $(x_1,y_1) and(x_2,y_2)$

$(y-y_1)=m(x-x_1)$

where : m = Slope of the line

$m=\frac{y_2-y_1}{x_2-x_1}$

A line includes the points (4, 8) and (3, 5).The equation of the line will be:

$m=\frac{5-8}{3-4}=3$

$(y-8)=3\times (x-4)$

$y-8=3x-12$

$y=3x-4$

(y=mx+c), where c is intercept at y axis.

The above equation represents slope-intercept form of the line.

6 0

3 years ago

Which statement describes the inverse of m(x) = x^2 – 17x?

DochEvi [55]

Given:

The function is

$m(x)=x^2-17x$

To find:

The inverse of the given function.

Solution:

We have,

$m(x)=x^2-17x$

Substitute m(x)=y.

$y=x^2-17x$

Interchange x and y.

$x=y^2-17y$

Add square of half of coefficient of y , i.e., $\left(\dfrac{-17}{2}\right)^2$ on both sides,

$x+\left(\dfrac{-17}{2}\right)^2=y^2-17y+\left(\dfrac{-17}{2}\right)^2$

$x+\left(\dfrac{17}{2}\right)^2=y^2-17y+\left(\dfrac{17}{2}\right)^2$

$x+\left(\dfrac{17}{2}\right)^2=\left(y-\dfrac{17}{2}\right)^2$ $[\because (a-b)^2=a^2-2ab+b^2]$

Taking square root on both sides.

$\sqrt{x+\left(\dfrac{17}{2}\right)^2}=y-\dfrac{17}{2}$

Add $\dfrac{17}{2}$ on both sides.

$\sqrt{x+\left(\dfrac{17}{2}\right)^2}+\dfrac{17}{2}=y$

Substitute $y=m^{-1}(x)$ .

$m^{-1}(x)=\sqrt{x+(\dfrac{189}{4}})+\dfrac{17}{2}$

We know that, negative term inside the root is not real number. So,

$x+\left(\dfrac{17}{2}\right)^2\geq 0$

$x\geq -\left(\dfrac{17}{2}\right)^2$

Therefore, the restricted domain is $x\geq -\left(\dfrac{17}{2}\right)^2$ and the inverse function is $m^{-1}(x)=\sqrt{x+(\dfrac{189}{4}})+\dfrac{17}{2}$ .

Hence, option D is correct.

Note: In all the options square of $\dfrac{17}{2}$ is missing in restricted domain.

7 0

3 years ago