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3 years ago

find the permimter of the quadrilateral if x=2 the perimiter is in inches the perimiters are (4x2+8x) (31 ) (7x+30) (3x2-5x+20)

Mathematics

2 answers:

d1i1m1o1n [39]3 years ago

8 0

Answer:

115

Step-by-step explanation:

4*2+8(2) = 24

7(2)+30 = 44

3*2-5(2)+20 = 6-10+20 = -4+20 = 16

24+31+44+16

115

kirill115 [55]3 years ago

6 0

7(2) + 30 = 14+30 = 44
3(2^2) -5(2)+20 = 12 - 10 + 20 = 22
4(2^2) +8(2) = 16 + 16 = 32

P = 31 + 44 + 22 + 32 = 129

The perimeter is 129 inches

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What is the sum or whatever of this equation. This equation is V = πx3^2x8

navik [9.2K]

Answer:

$V=226.08$

Step-by-step explanation:

To evaluate $V$ from the following expression:

$V=\pi\times3^2\times 8$

Taking $\pi=3.14$

$V=3.14\times 3^2\times 8$

$V=3.14\times 9\times 8$

∴ $V=226.08$

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3 years ago

Emily and her children went into a grocery store and she bought $20.80 worth of bananas and peaches. Each banana costs $0.80 and

MakcuM [25]

Answer:

The number of bananas that Emily bought was 6 and the number of peaches that Emily bought was 8

Step-by-step explanation:

<u><em>The complete question is</em></u>

Emily and her children went into a grocery store and she bought $20.80 worth of bananas and peaches. Each banana costs $0.80 and each peach costs $2. She bought a total of 14 peaches and bananas altogether. Determine the number of peaches and the number of bananas that Emily bought

Let

x ----> the number of bananas that Emily bought

y ----> the number of peaches that Emily bought

we know that

She bought a total of 14 bananas and peaches altogether

$x+y=14$ -----> equation A

She bought $20.80 worth of bananas and peaches

$0.80x+2y=20.80$ -----> equation B

Solve the system by graphing

Remember that the solution is the intersection point both graphs

using a graphing tool

The solution is the point (6,8)

see the attached figure

therefore

The number of bananas that Emily bought was 6 and the number of peaches that Emily bought was 8

3 0

3 years ago

There are 7800 grams of lodine that have a half-life of 8 days. How much lodine will remain after 40 days?

anygoal [31]

Given that we know the initial mass of Iodine and its half-life, we want to see how much will remain after 40 days.

After 40 days, 244.17 grams of Iodine will remain.

<h3 /><h3>The half-life of materials and how to use it:</h3>

The half-life of a material is the time it takes for that amount of material to reduce to its half.

We can model the amount of Iodine as:

A(t) = A*e^{k*t}

Where A is the initial amount, in this case, 7800g.
k is a constant that depends on the half-life.
t is the time in days.

Replacing what we know, we get:

A(t) = 7800g*e^{k*t}

Now we use the fact that the half-life is 8 days, this means that:

e^{k*8} = 1/2

ln(e^{k*8}) = ln(1/2)

k*8 = ln(1/2)

k = ln(1/2)/8 = -0.0866

Then the function is:

A(t) = 7800g*e^{-0.0866*t}

So now we just need to evaluate this in t = 40.

A(40) = 7800g*e^{-0.0866*40} = 244.17g

So, after 40 days, 244.17 grams of Iodine will remain.

If you want to learn more about half-life and decays, you can read:

brainly.com/question/11152793

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2 years ago

Classwork: Multiply, Add, and Subtract Polyno

V125BC [204]

I hope this was helpful

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1 year ago

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Darina [25.2K]

Answer:

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Answer is 12

Step-by-step explanation:

Add them together and divide by the numbers of (things)

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2 years ago

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