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3 years ago

A football team loses 15 yards on one play and gains 9 yards on another play. How many total yards did the team lose?

Mathematics

1 answer:

Licemer1 [7]3 years ago

4 0

Answer:

-6

Step-by-step explanation:

imagine you're on a line graph you start at zero and then go backward 15 spaces and move forward 9 spaces where does that leave you

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Can anyone solve?<br><br><img src="https://tex.z-dn.net/?f=%20%5Csf%7B5x%20%2B%20x%20%3D%20%3F%7D" id="TexFormula1" title=" \sf{

Lorico [155]

Answer:

$\huge \underline \color{cyan}{ \boxed{Question : - }}$ $\\ \\ \\ \\ \\ \\$ Wdym by similarities? $\\ \\ \\$ •Thanks for answering!~

Step-by-step explanation:

5 0

3 years ago

Read 2 more answers

two ferries start moving toward each other from opposite riverbanks, a and b. when they pass each other for the first time, the

Usimov [2.4K]

Let 'x' be the distance from THE far bank where 700 is the distance to the NEAR bank

boat one has travelled 700 (rate = 700/unit time) boat two has travelled x rate = x / unit time

boat one then travels x + 400 more and boat two travels 700 + (700+x -400) more when they meet

The time is the same rate x time = distance distance/rate = time equate the distances divided by the respective rates

(700 + x + 400)/700 = ( x + 700 + (700+x-400) )/x

1100x + x^2 = 1400x + 700000

x^2-300x -700000 = 0 quadratic formula yields x = 1000

One boat travels 700 the other 1000 whe they first meet.....width of river = 700+ 1000 = 1700 m

3 0

3 years ago

An object launched straight up at a speed of 29.4 meters per second has a height, h, in meters of h , t seconds after the object

Nostrana [21]

Answer:

h = 44.06 meters (maximum height)

the time the object takes to complete this whole path is 6 seconds, this is why the time at which the object reaches its maximum height will between 0 and 6 seconds

Step-by-step explanation:

To solve this question, we need to first recognize that this is a constant acceleration problem, specifically, it can be thought of as a projectile motion problem.

Recall, the equations of motion:

$1) v^2 - v_0^2 = 2a(s - s_0)\\2) s = v_0^2 + \frac{1}{2} at^2\\3) v = v_0 + at$

What do we already know?

$v_0 = 29.4 ms^-1$
The launch is straight up
$a = -9.81 ms^-2$ this is the gravitational acceleration $g$

$s_0 = 0 m$ , since our reference point is at s = 0, (the ground)

We can use use the Eq(1):

we know that when any object is launched up, at maximum height its velocity is going to be zero, $v = 0 ms^-2$

$v^2 - v_0^2 = 2a(s - s_0)\\0^2 - (29.4)^2 = 2(-9.81)(s- 0)\\s = 44.06 m$

this is the maximum height!

Why does t have to between zero and six?

We can answer this using a bit visualization, if you think about the second equation

$s = v_0 t - \frac{1}{2}at^2\\ s = 29.4t - 4.905t^2$

this is the equation of the whole trajectory that object makes.

and if you solve this by making $s = 0$ , you will get the times at which the object was at the ground. the times will be 0s and 5.99s.

so the amount of time the object takes to go through this whole path is 6 seconds and this why the object will only reach its maximum height in between this time interval.

hope this helps :)

5 0

3 years ago

Answer true or false for 1-10 (picture above)

Softa [21]

$\sqrt{a\cdot b}=\sqrt{a}\cdot\sqrt{b}\\\\\sqrt{a+b}\leq\sqrt{a}+\sqrt{b}\\\\\sqrt{a-b}\geq\sqrt{a}-\sqrt{b}\\----------------------\\1)\ \sqrt{72}=\sqrt{9\cdot8}=\sqrt{9}\cdot\sqrt{8}\qquad TRUE\\\\2)\ \sqrt{13-5}=\sqrt{13}-\sqrt5\qquad FALSE\\\\3)\ \sqrt{9\cdot10}=\sqrt9+\sqrt{10}\qquad FALSE\\\\4)\ \sqrt{18bx}=\sqrt{18}\cdot\sqrt{b}\cdot\sqrt{x}\qquad TRUE\ if\ b\geq0\ and\ x\geq0\\\\5)\ \sqrt{8\cdot12}=\sqrt8\cdot\sqrt{12}\qquad TRUE\\\\6)\ \sqrt{m+x}=\sqrt{m}+\sqrt{x}\qquad FALSE$

$7)\ \sqrt{100+64}=\sqrt{100}+\sqrt{64}\qquad FALSE\\\\8)\ \sqrt{30}=\sqrt{5\cdot6}=\sqrt5\cdot\sqrt6\qquad TRUE\\\\9)\ \sqrt{44}=\sqrt{22+22}=\sqrt{22}+\sqrt{22}\qquad FALSE\\\\10)\ \sqrt{10\cdot10}=\sqrt{10}\cdot\sqrt{10}\qquad TRUE$

3 0

3 years ago

Subtract.

Illusion [34]

Answer:−93 3−153−53

Step-by-step explanation:

wrong answer but the answer is on google

6 0

3 years ago