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4 years ago

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The cost T, in hundreds of dollars, of tuition and fees at many community colleges can be approximated by Upper T equals one h

alf c plus 1
, where c is the number of credits for which a student registers. Graph the equation and use the graph to estimate the cost of tuition and fees when a student registers for 2 three-credit courses.

1 answer:

Rainbow [258]4 years ago

7 0

Course set of 8 and 2/3rds

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Please help me im really confused and i need an answer and how to do it!

Nutka1998 [239]

By solving the given equation of inequality " $4x+5 \geq 21$ ", we get " $x \geq 4$ ". A complete solution is below.

The given inequality equation is:

$4x+5 \geq 21$

By subtracting "5" from both sides of the above equation, we get

→ $4x+5 -5 \geq 21-5$

→ $4x\geq 16$

By applying cross-multiplication, we get

→ $x \geq \frac{16}{4}$

→ $x \geq 4$

Thus the answer above is the right one.

Learn more about inequality equation here:

brainly.com/question/12094806

8 0

3 years ago

Given: a||b, m∠ 6=128° Find: m∠1, m∠3

goldfiish [28.3K]

Answer:

m∠1 = 52°

m∠3 = 52°

Step-by-step explanation:

Step 1:

Since b is a straight line and is transversed by line m and we are given m∠6 = 128°, we can find m∠5, which is 180 - m∠6, giving us 52° for ∠5.

Step 2:

Because a and b are parallel, m∠5 and the m∠1 are the same, so we have m∠1 = 52° as one of our answers needed

Step 3: Because ∠3 is vertical to ∠1, we can use the vertical angles theory and say that m∠1 is equal to m∠3. Therefore, m∠3 = 52°

5 0

4 years ago

kalen's truck has a 60 gallon fuel tank The fuel gauge points to 1/4 how much fuel does he need to buy to fill the tank show You

goblinko [34]

He should need 45 more gallons. 60/4=15 15gallons for every 1/4 of the tank. 60-15=45

8 0

3 years ago

Find all values of x in the interval [0, 2π] that satisfy the equation. <br><br>6sin²(x) = 3

34kurt

Answer:

The solutions are π/4, 3π/4,5π/4,7π/4

Step-by-step explanation:

The given equation is

6sin²(x) = 3

Divide by 6 to get:

${ \sin}^{2} (x) = \frac{1}{2}$

This implies that;

$\sin(x) = \pm \frac{ \sqrt{2} }{2}$

If

$\sin(x) = \frac{ \sqrt{2} }{2}$

$x = \frac{\pi}{4}$

in the first quadrant

$x = \frac{3\pi}{4}$

in the second quadrant.

If

$\sin(x) = - \frac{ \sqrt{2} }{2}$

$x = \frac{5\pi}{4}$

in the third quadrant

$x = \frac{7\pi}{4}$

5 0

3 years ago

How to integrate this

Radda [10]

To integrate, let's first clean out the inside, and simplify:

$16\int {(sin x)^4} \ dx$

We can now safely substitute the value of u:

$u = sin x$

$du = cos (x) dx$

Now, we can write that:

$16\int{u^4} \ du$

This is much easier to integrate:

$16\int{u^4} \ du = 16[\frac{1}{5}u^5] + C$

Simplify:

$\frac{16}{5}u^5 + C$

Replace u (final answer):

$\frac{16}{5}sin^5x +C$

Hope I could help with your integration.

4 0

3 years ago