All categories

4 years ago

Find the distance between the points (1, 5) and (1, -4).

1 answer:

3 0

$\text {Distance = } \sqrt{(Y_2-Y1)^2 + (X_2-X_1)^2}$

$\text {Distance = } \sqrt{(5 - (-4))^2 + (1-1)^2}$

$\text {Distance = } \sqrt{81 + 0}$

$\text {Distance = } \sqrt{81}$

$\text {Distance = } 9 \text { units}$

You might be interested in

The fraction ½ can be written as a terminating decimal. true or false?

kvasek [131]

Yes. .50. hope this helps

4 0

3 years ago

Read 2 more answers

Need help please answer this question

Aleks04 [339]

Answer:

Step-by-step explanation:

Substitute the given values for x and y into the expression

5 × ( $\frac{60}{6(5)}$ ) - 1

= 5 × ( $\frac{60}{30}$ ) - 1

= (5 × 2) - 1

= 10 - 1

= 9 → C

4 0

3 years ago

What is the perimeter of DEFG, shown?

Murrr4er [49]

Answer:

C.34

Step-by-step explanation:

you find the length which is 10 squares and the width which is 7 squares. Since the shape is a quadrilateral you add all sides to find the perimeter= 2L+2W. 2(10)+2(7)=20+14=34

4 0

3 years ago

Factoring polynomials of 18x-60x^5 please explain the steps.

Afina-wow [57]

18x-60x^5
=18x-777600000x
=-777599982x

8 0

4 years ago

Read 2 more answers

A car that travels along the highway to Denver at a steady speed. When it begins, it is 300 miles from Denver. After 3 hours, it

umka21 [38]

Answer: $y=-40x+300$

Step-by-step explanation:

This car follows a line, hence its motion can be modeled by the Line equation.

In fact, we already have two points of the line, if we call $x$ the time in hours and $y$ the traveled distance in miles:

Point 1: $(x_{1},y_{1})=(0,300)$

Since we are told the car's initial position is 300 miles in the time 0 h

Point 2: $(x_{2},y_{2})=(3,180)$

Since we are told the car's final position is 180 miles after 3h

Let's find the Slope $m$ :

$m=\frac{y_{2}-y_{1}}{x_{2}-x_{1}}$

$m=-40$ This is the slope of the line

Now, the equation of the line is:

$y=mx+b$

We already know the slope, now we have to find the intersection point with the y-axis ( $b$ ) with any of the given points. Let's choose Point 1: $(x_{1},y_{1})=(0,300)$

$300=-40(0)+b$

Isolating $b$ :

$b=-300$

Then, the equation of the line is:

$y=-40 x+300$

7 0

4 years ago

Read 2 more answers