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2 years ago

8

Santos flipped a coin 500 times. The coin landed heads up 225 times. Find the ratio of heads to a total number of coin flips. Ex

press as a simplified ratio.

1 answer:

m_a_m_a [10]2 years ago

8 0

Answer:

225:500=9:20

Step-by-step explanation:

225/5=45

45/5=9

500/5=100

100/5=20

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David can jog 4 miles in. 75 minutes.what is David’s jogging rate per hour?

weeeeeb [17]

Answer: 3.2 miles

Step-by-step explanation: Trust me bro. 1000% right

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2 years ago

Adrian has some dimes and some quarters. He has at least 13 coins worth no more than $2.35 combined. If Adrian has 11 dimes, det

tresset_1 [31]

Answer:

11 dimes equals 110

Step-by-step explanation:

so you have 3.45 cents.

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3 years ago

3. Which of the following is the surface area of this right triangular prism? (

Ostrovityanka [42]

160in 5x8=40x4=160 is the answer to ur question

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2 years ago

If the original square had a side length of

irina [24]

Answer:

Part a) The new rectangle labeled in the attached figure N 2

Part b) The diagram of the new rectangle with their areas in the attached figure N 3, and the trinomial is $x^{2} +11x+28$

Part c) The area of the second rectangle is 54 in^2

Part d) see the explanation

Step-by-step explanation:

The complete question in the attached figure N 1

Part a) If the original square is shown below with side lengths marked with x, label the second diagram to represent the new rectangle constructed by increasing the sides as described above

we know that

The dimensions of the new rectangle will be

$Length=(x+4)\ in$

$width=(x+7)\ in$

The diagram of the new rectangle in the attached figure N 2

Part b) Label each portion of the second diagram with their areas in terms of x (when applicable) State the product of (x+4) and (x+7) as a trinomial

The diagram of the new rectangle with their areas in the attached figure N 3

we have that

To find out the area of each portion, multiply its length by its width

$A1=(x)(x)=x^{2}\ in^2$

$A2=(4)(x)=4x\ in^2$

$A3=(x)(7)=7x\ in^2$

$A4=(4)(7)=28\ in^2$

The total area of the second rectangle is the sum of the four areas

$A=A1+A2+A3+A4$

State the product of (x+4) and (x+7) as a trinomial

$(x+4)(x+7)=x^{2}+7x+4x+28=x^{2} +11x+28$

Part c) If the original square had a side length of x = 2 inches, then what is the area of the second rectangle?

we know that

The area of the second rectangle is equal to

$A=A1+A2+A3+A4$

For x=2 in

substitute the value of x in the area of each portion

$A1=(2)(2)=4\ in^2$

$A2=(4)(2)=8\ in^2$

$A3=(2)(7)=14\ in^2$

$A4=(4)(7)=28\ in^2$

$A=4+8+14+28$

$A=54\ in^2$

Part d) Verify that the trinomial you found in Part b) has the same value as Part c) for x=2 in

We have that

The trinomial is

$A(x)=x^{2} +11x+28$

For x=2 in

substitute and solve for A(x)

$A(2)=2^{2} +11(2)+28$

$A(2)=4 +22+28$

$A(2)=54\ in^2$ ----> verified

therefore

The trinomial represent the total area of the second rectangle

7 0

3 years ago

3. If ‘A’ can finish a work in ‘n’ days then part of work finished in 1 day is: (a) 1 – n

zaharov [31]

Answer:

If ‘A’ can finish a work in ‘n’ days then part of work finished in 1 day

will be $\frac{1}{n}$ .

Step-by-step explanation:

From the question, it is clear that

If ‘A’ can finish a work in ‘n’ days, then
we have to determine the part of work finished in 1 day.

So

let ' $n$ ' be the number of days $A$ takes to complete the whole work

Let the whole job be denoted as ' $1$ '

Thus, the part of work finished in 1 day will be:

$\:\frac{Total\:work}{Time\:taken\:by\:A\:to\:complete\:the\:work}\:=\frac{1}{n}$

Therefore, If ‘A’ can finish a work in ‘n’ days then part of work finished in 1 day will be $\frac{1}{n}$ .

Keywords: work, word problem

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3 0

2 years ago