0.5 0.45 0.405 4.5 least to greatest

If a 45 degree 45 degree 90 degree triangle has a hypotenuse length of 12 find the leg length

AURORKA [14]

Answer:

<h2>Leg length = 8.5</h2><h2 />

Step-by-step explanation:

let x represent the leg length

if we use the Pythagorean theorem we get:

x² + x² = 12²

then

2x² = 12²

Then

x² = 144/2 = 72

then

x = √(72)

=8,485281374239

7 0

2 years ago

An object is thrown upward from the top of a 192 ft building with an initial velocity of 64 feet

Airida [17]

Answer:

t=6

Step-by-step explanation:

ground height = 0

(are you sure your formula is correct? isn't it - 16t²?)

if h=16t² +64t+192 is true then

16t² +64t+192 = 0

t² + 4t + 12 = 0

t = (-4 ± √(4² - 4*12)) / 2*1 = (-4 ± √-32) / 2 = -2 ± 2√-2

There is no solution of t

if it is h= - 16t² +64t+192

0 =- 16t² + 64t + 192

t² - 4t - 12 = 0

(t + 2) (t -6) = 0

t should be positive

t = 6 sec

7 0

3 years ago

Alana has 2 1/4 bags of chocolate chips that she wants to use in 3 batches of chocolate chip cookies. How much of the chocolate

drek231 [11]

Answer:

1 1/3 bags of chocolate chips

Step-by-step explanation:

If Alana has 2 1/4 bags of chocolate chips that she wants to use in 3 batches of chocolate chip cookies, then we can say;

2 1/4 bags of cookies = 3 batches of chips

To determine how much of the chocolate chips will she use in each batch of cookies, we can say;

1 bag of cookies = x batches of chips

Equating both expressions

2 1/4 bags of cookies = 3 batches of chips

1 bag of cookies = x batches of chips

Cross multiply

2 1/4 x = 3

9/4 x = 3

9x/4 = 3

9x = 12

x = 12/9

x = 4/3

x = 1 1/3

Hence Alana will need 1 1/3 bags of chocolate chips

8 0

2 years ago

Press on the picture

scoray [572]

B is the correct answer

7 0

2 years ago

Read 2 more answers

Please help me on this problem

Anna71 [15]

Answer:

The pairs of integer having two real solution for $ax^{2} -6x+c = 0$ are

$a = -4, c = 5$
$a = 1, c = 6$
$a = 2, c = 3$
$a = 3, c = 3$

Step-by-step explanation:

Given

$ax^{2} -6x+c = 0$

Now we will solve the equation by putting all the 6 pairs so we get the following

$-3x^{2} -6x-5 = 0$ for $a = -3 , c=-5$

$-4x^{2} -6x+5 = 0$ for $a = -4 , c=5$

$1x^{2} -6x+6 = 0$ for $a = 1 , c=6$

$2x^{2} -6x+3 = 0$ for $a = 2 , c=3$

$3x^{2} -6x+3 = 0$ for $a = 3 , c=3$

$5x^{2} -6x+4 = 0$ for $a = 5 , c=4$

The above all are Quadratic equations inn general form $ax^{2} +bx+c=0$

where we have a,b and c constant values

So for a real Solution we must have

$Disciminant , b^{2} -4\timesa\timesc \geq 0$

for $a = -3 , c=-5$ we have

$Discriminant =-24$ which is less than 0 ∴ not a real solution.

for $a = -4 , c=5$ we have

$Discriminant = 116$ which is greater than 0 ∴ a real solution.

for $a = 1 , c=6$ we have

$Discriminant =12$ which is greater than 0 ∴ a real solution.

for $a = 2 , c=3$ we have

$Discriminant =12$ which is greater than 0 ∴ a real solution.

for $a = 3 , c=3$ we have

$Discriminant =0$ which is equal to 0 ∴ a real solution.

for $a = 5 , c=4$ we have

$Discriminant =-44$ which is less than 0 ∴ not a real solution.

7 0

3 years ago