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3 years ago

May someone help me out on this question?

Mathematics

1 answer:

bulgar [2K]3 years ago

6 0

Answer:

y = 2 for x < 0

x for 0 ≤ x < 3

3 for x ≥ 3

Step-by-step explanation:

From the graph attached,

Equation of line (1),

y = 2

Where x < 0

Let the equation of line (2) is,

y = mx + b

y-intercept 'b' = 0 [line (2) passes through origin]

Slope of the line 'm' = $\frac{\text{Rise}}{\text{Run}}$

= $\frac{3}{3}$

= 1

Therefore, equation of line (2) will be,

y = x

where 0 ≤ x < 3

Equation of line (3) will be,

y = 3

Where x ≥ 3

Therefore, piecewise linear function will be,

y = 2 for x < 0

x for 0 ≤ x < 3

3 for x ≥ 3

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Becky put $100 into in the account to use for school expenses. The account earns 11% interest, compounded annually. How much wil

Mnenie [13.5K]

Answer: $187 will be in the account after 6 years.

Step-by-step explanation:

We would apply the formula for determining compound interest which is expressed as

A = P(1+r/n)^nt

Where

A = total amount in the account at the end of t years

r represents the interest rate.

n represents the periodic interval at which it was compounded.

P represents the principal or initial amount deposited

From the information given,

P = $100

r = 11% = 11/100 = 0.11

n = 1 because it was compounded once in a year.

t = 6 years

Therefore,.

A = 100(1 + 0.11/1)^1 × 6

A = 100(1 + 0.11)^6

A = 100(1.11)^6

A = $187

5 0

3 years ago

The volume of water remaining in a hot tub when it is being drained satisfies the differential equation dV/dt = −3(V)^1/2 , wher

dimaraw [331]

The given function is a variable separable differential equation. Combine like terms, integrate, apply the appropriate limits, and express V in terms of t. This is done as follows:

dV/dt = -3(V)^1/2
dV/-3V^1/2 = dt
$\int\limits^V_m { \frac{1}{03 \sqrt{V} } } \, dV = \int\limits^t_0 {} \, dt$

m here is the initial V which is 225. Then after integrating,

-2/3 (√V - √225) = t
-2/3 (√V - 15) = t

$V= \sqrt{ \frac{-3}{2}t+15 }$

That is the expression for V at time t. I hope I was able to help. Have a good day.

5 0

3 years ago

When Tomas simplified the expression -2.6 + (-5.4), he got 2.8. what mistake did Tomas likely make?

vladimir2022 [97]

Tomas used incorrectly the rule of signs, the expression should be simplified as follows:

-2.6 + (-5.4)

-2.6 -5.4 = 8

<h3></h3><h3>What mistake did Tomas likely make?</h3>

Here Tomas wants to perform an addition between two numbers:

-2.6 + (-5.4)

And the outcome that Tomas gets is:

-2.6 + (-5.4) = 2.8

Here his mistake seems to bee that he thought the second number was a positive number, and he solved the operation:

-2.6 + 5.4

So, he used wrong the rule of signs.

Remember that the rule of signs says that:

(-)*(+) = (+)*(-) = (-)

Using that, we can rewrite the original expression:

-2.6 + (-5.4)

to:

-2.6 - 5.4

Solving that, we get:

-2.6 - 5.4 = -8

Which is the outcome that Tomas would have gotten if he had used correctly the rule of signs.

If you want to learn more about the rule of signs:

brainly.com/question/13333620

#SPJ1

5 0

2 years ago

Find the length of the third side. If necessary, round to the nearest tenth. 10 14

abruzzese [7]

Answer:

9.8 units

Explanation:

According to Pythagoras theorem, the square of length of the hypotenuse is equal to the sum of squares of lengths of other two sides.

Let the third side be x units.

Then,

x² + 10² = 14²

=> x² + 100 = 196

=> x² = 196 - 100

=> x² = 96

=> x = √96

=> x = 4√6

=> x = 9.79....

=> x = 9.8 (Rounding to the nearest tenths)

So, the length of the third side is 9.8 units.

5 0

3 years ago

(x+16)°<br> (4x-5)°<br> Find the measure of the angle.<br> I

o-na [289]

$4x - 5 = x + 16$

Subtract the sides of the equation plus 5

$4x = x + 16 + 5$

$4x = x + 21$

Subtract the sides of the equation minus x

$3x = 21$

Divided the sides of the equation by 3

$\frac{3}{3} x = \frac{21}{3} \\$

$x = 7$

So the measure of the angle equals :

$4(7) - 5 = 28 - 5 = 23$

_________________________________

And we're done.

Thanks for watching buddy good luck.

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6 0

3 years ago