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3 years ago

Find the perimeter of the figure to the nearest hundredth.

Mathematics

2 answers:

quester [9]3 years ago

8 0

The answer to the question is r2pi

Alenkinab [10]3 years ago

4 0

Answer: 48.84

Step-by-step explanation:

First, find the perimeter of the rectangle. 12+12+3+3=30ft. Next, find the perimeter of the semicircles. You can tell that if you put the semicircles together it would be a full circle. The formula for the perimeter of a circle is r2pi. The radius times 2 would be 6. Pi is approximately 3.14. 6 times 3.14 is 18.84. 18.84+30=48.84ft. Hope this helped :)

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What is the answer for this question 7(x+4)-7=48-2x

Alex777 [14]

Answer:

x = 3

Step-by-step explanation:

7(x + 4) - 7 = 48 - 2x ← distribute parenthesis and simplify left side

7x + 28 - 7 = 48 - 2x

7x + 21 = 48 - 2x ( add 2x to both sides )

9x + 21 = 48 ( subtract 21 from both sides )

9x = 27 ( divide both sides by 9 )

x = 3

8 0

2 years ago

Read 2 more answers

Let z=3+i, then find a. Z² b. |Z| c.<img src="https://tex.z-dn.net/?f=%5Csqrt%7BZ%7D" id="TexFormula1" title="\sq

zysi [14]

Given z = 3 + i, right away we can find

(a) square

z ² = (3 + i )² = 3² + 6i + i ² = 9 + 6i - 1 = 8 + 6i

(b) modulus

|z| = √(3² + 1²) = √(9 + 1) = √10

(d) polar form

First find the argument:

arg(z) = arctan(1/3)

Then

z = |z| exp(i arg(z))

z = √10 exp(i arctan(1/3))

z = √10 (cos(arctan(1/3)) + i sin(arctan(1/3))

Any complex number has 2 square roots. Using the polar form from part (d), we have

√z = √(√10) exp(i arctan(1/3) / 2)

and

√z = √(√10) exp(i (arctan(1/3) + 2π) / 2)

Then in standard rectangular form, we have

$\sqrt z = \sqrt[4]{10} \left(\cos\left(\dfrac12 \arctan\left(\dfrac13\right)\right) + i \sin\left(\dfrac12 \arctan\left(\dfrac13\right)\right)\right)$

and

$\sqrt z = \sqrt[4]{10} \left(\cos\left(\dfrac12 \arctan\left(\dfrac13\right) + \pi\right) + i \sin\left(\dfrac12 \arctan\left(\dfrac13\right) + \pi\right)\right)$

We can simplify this further. We know that z lies in the first quadrant, so

0 < arg(z) = arctan(1/3) < π/2

which means

0 < 1/2 arctan(1/3) < π/4

Then both cos(1/2 arctan(1/3)) and sin(1/2 arctan(1/3)) are positive. Using the half-angle identity, we then have

$\cos\left(\dfrac12 \arctan\left(\dfrac13\right)\right) = \sqrt{\dfrac{1+\cos\left(\arctan\left(\dfrac13\right)\right)}2}$

$\sin\left(\dfrac12 \arctan\left(\dfrac13\right)\right) = \sqrt{\dfrac{1-\cos\left(\arctan\left(\dfrac13\right)\right)}2}$

and since cos(x + π) = -cos(x) and sin(x + π) = -sin(x),

$\cos\left(\dfrac12 \arctan\left(\dfrac13\right)+\pi\right) = -\sqrt{\dfrac{1+\cos\left(\arctan\left(\dfrac13\right)\right)}2}$

$\sin\left(\dfrac12 \arctan\left(\dfrac13\right)+\pi\right) = -\sqrt{\dfrac{1-\cos\left(\arctan\left(\dfrac13\right)\right)}2}$

Now, arctan(1/3) is an angle y such that tan(y) = 1/3. In a right triangle satisfying this relation, we would see that cos(y) = 3/√10 and sin(y) = 1/√10. Then

$\cos\left(\dfrac12 \arctan\left(\dfrac13\right)\right) = \sqrt{\dfrac{1+\dfrac3{\sqrt{10}}}2} = \sqrt{\dfrac{10+3\sqrt{10}}{20}}$

$\sin\left(\dfrac12 \arctan\left(\dfrac13\right)\right) = \sqrt{\dfrac{1-\dfrac3{\sqrt{10}}}2} = \sqrt{\dfrac{10-3\sqrt{10}}{20}}$

$\cos\left(\dfrac12 \arctan\left(\dfrac13\right)+\pi\right) = -\sqrt{\dfrac{10-3\sqrt{10}}{20}}$

$\sin\left(\dfrac12 \arctan\left(\dfrac13\right)+\pi\right) = -\sqrt{\dfrac{10-3\sqrt{10}}{20}}$

So the two square roots of z are

$\boxed{\sqrt z = \sqrt[4]{10} \left(\sqrt{\dfrac{10+3\sqrt{10}}{20}} + i \sqrt{\dfrac{10-3\sqrt{10}}{20}}\right)}$

and

$\boxed{\sqrt z = -\sqrt[4]{10} \left(\sqrt{\dfrac{10+3\sqrt{10}}{20}} + i \sqrt{\dfrac{10-3\sqrt{10}}{20}}\right)}$

3 0

3 years ago

Read 2 more answers

Identify one characteristic of exponential growth.

vodka [1.7K]

If I were u I would chose B

7 0

2 years ago

The half-life of caffeine in a healthy adult is 4.8 hours. Jeremiah drinks 18 ounces of caffeinated

statuscvo [17]

We want to see how long will take a healthy adult to reduce the caffeine in his body to a 60%. We will find that the answer is 3.55 hours.

We know that the half-life of caffeine is 4.8 hours, this means that for a given initial quantity of coffee A, after 4.8 hours that quantity reduces to A/2.

So we can define the proportion of coffee that Jeremiah has in his body as:

P(t) = 1*e^{k*t}

Such that:

P(4.8 h) = 0.5 = 1*e^{k*4.8}

Then, if we apply the natural logarithm we get:

Ln(0.5) = Ln(e^{k*4.8})

Ln(0.5) = k*4.8

Ln(0.5)/4.8 = k = -0.144

Then the equation is:

P(t) = 1*e^{-0.144*t}

Now we want to find the time such that the caffeine in his body is the 60% of what he drank that morning, then we must solve:

P(t) = 0.6 = 1*e^{-0.144*t}

Again, we use the natural logarithm:

Ln(0.6) = Ln(e^{-0.144*t})

Ln(0.6) = -0.144*t

Ln(0.6)/-0.144 = t = 3.55

So after 3.55 hours only the 60% of the coffee that he drank that morning will still be in his body.

If you want to learn more, you can read:

brainly.com/question/19599469

7 0

3 years ago

The math club is selling T-shirts to raise money. Each T-shirt sold represents a profit of $2. The club has a total of 500 T-

vova2212 [387]

Answer:

2 < or equal to (t) < or equal to 1000

Step-by-step explanation:

2 is the profit of the (t) amount of t shirts so the amount should be greater than or equal too 1000 because if they have 500 shirts 500 x 2 is 1000

3 0

3 years ago