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3 years ago

Solve the following equation algebraically x^2=192 a. 13.86 b. -13.86, 13.86 c. -14.36, 14.36 d. -96, 96

Mathematics

1 answer:

nirvana33 [79]3 years ago

4 0

Answer:

idrk but it think u multiply them easily

Step-by-step explanation:

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Heeelp

alukav5142 [94]

Answer: see below

Step-by-step explanation

Let 2 + a = 11 x

Let 35 - b = 11 y

Where x and y are any unknown integer

subtract the two equations

- 33 + a + b = 11 (x+y)

a+ b = 11 (x+ y) +33

a+ b = 11 (x+y) + 3 (11)

a+ b = 11(x+ y+3)

Which proves that a+b is a factor of 11

6 0

2 years ago

Find the absolute minimum and absolute maximum values of f on the given interval. f(t) = 3 (*sqaure root sign*) t (20 − t), [0,

mafiozo [28]

9514 1404 393

Answer:

maximum: 15∛5 ≈ 25.6496392002
minimum: 0

Step-by-step explanation:

The minimum will be found at the ends of the interval, where f(t) = 0.

The maximum is found in the middle of the interval, where f'(t) = 0.

$f(t)=\sqrt[3]{t}(20-t)\\\\f'(t)=\dfrac{20-t}{3\sqrt[3]{t^2}}-\sqrt[3]{t}=\sqrt[3]{t}\left(\dfrac{4(5-t)}{3t}\right)$

This derivative is zero when the numerator is zero, at t=5. The function is a maximum at that point. The value there is ...

f(5) = (∛5)(20-5) = 15∛5

The absolute maximum on the interval is 15∛5 at t=5.

5 0

3 years ago

1)Sheyna drive to the lake and back. It took two hours less time to get there than it did to get back. The average speed on the

ipn [44]

Answer:

8 hours

Step-by-step explanation:

Given:

Sheyna drives to the lake with average speed of 60 mph and

$v_1 = 60\ mph$

Sheyna drives back from the lake with average speed of 36 mph

$v_2 = 36\ mph$

It took 2 hours less time to get there than it did to get back.

Let $t_1$ be the time taken to drive to lake.

Let $t_2$ be the time taken to drive back from lake.

$t_2-t_1 = 2$ hrs ..... (1)

To find:

Total time taken = ?

$t_1+t_2 = ?$

Solution:

Let D be the distance to lake.

Formula for time is given as:

$Time =\dfrac{Distance}{Speed }$

$t_1 = \dfrac{D}{60}\ hrs$

$t_2 = \dfrac{D}{36}\ hrs$

Putting in equation (1):

$\dfrac{D}{36}-\dfrac{D}{60} = 2\\\Rightarrow \dfrac{5D-3D}{180} = 2\\\Rightarrow \dfrac{2D}{180} = 2\\\Rightarrow D = 180\ miles$

So,

$t_1 = \dfrac{180}{60}\ hrs = 3 \ hrs$

$t_2 = \dfrac{180}{36}\ hrs = 5\ hrs$

So, the answer is:

$t_1+t_2 = \bold{8\ hrs}$

8 0

3 years ago

For the<br> expression y + 6-3 determine the coefficient for the variable term.<br> (2 points)

aksik [14]

Answer:

The answer is 1.

Step-by-step explanation:

The number is usually in front of the variable, if there is no number, it is a 1.

6 0

3 years ago

Helpppp pleaseeeee and could you put an explenation??? pleaseeee

Illusion [34]

Answer:

Step-by-step explanation:

The angles would be supplementary

45 + 5x + 35 = 180

5x + 80 = 180

5x = 100

x = 20

To find out why they are supplementary, refer to the transversal below

∠3 = ∠7 (corresponding angles)

∠8 = 180 - ∠7 (supplementary angles)

And since ∠7 = ∠3...

∠8 = 180 - ∠3 which is why (in the problem) those two angles are supplementary (adding to 180 degrees)

4 0

3 years ago