We have to find x and x is the hypotenuse. We use this formula:
hyp²=side²+side²
So:
hyp²=side²+side²
hyp²= 12²+9²
hyp²= 225
hyp²= √225
hyp= 15
Therefore, the hypotenuse is 15 and x is 15.
HOPE THIS HELPS!
HAVE A GR8 EVENING ;-)
Answer:
d) one solution; (4, 1)
Step-by-step explanation:
It often works well to follow problem directions. A graph is attached, showing the one solution to be (4, 1).
_____
You know there will be one solution because the lines have different slopes. Each is in the form ...
y = mx + b
where m is the slope and b is the y-intercept.
The first line has slope -1 and y-intercept +5; the second line has slope 1 and y-intercept -3. The slope is the number of units of "rise" for each unit of "run", so it can be convenient to graph these by starting at the y-intercept and plotting points with those rise and run from the point you know.
Answer:


Step-by-step explanation:
<h3>Question-1:</h3>
so when <u>flash down</u><u> </u>occurs the rocket will be in the ground in other words the elevation(height) from ground level will be 0 therefore,
to figure out the time of flash down we can set h(t) to 0 by doing so we obtain:

to solve the equation can consider the quadratic formula given by

so let our a,b and c be -4.9,229 and 346 Thus substitute:

remove parentheses:

simplify square:

simplify multiplication:

simplify Substraction:

by simplifying we acquire:

since time can't be negative

hence,
at <u>4</u><u>8</u><u>.</u><u>2</u><u> </u>seconds splashdown occurs
<h3>Question-2:</h3>
to figure out the maximum height we have to figure out the maximum Time first in that case the following formula can be considered

let a and b be -4.9 and 229 respectively thus substitute:

simplify which yields:

now plug in the maximum t to the function:

simplify:

hence,
about <u>3</u><u>0</u><u>2</u><u>1</u><u>.</u><u>6</u><u> </u>meters high above sea-level the rocket gets at its peak?
Answer:
3^7 x 1/3^4; in other words the first choice
Step-by-step explanation: