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3 years ago

26/3 + 6 = 24 please answer!!!

Mathematics

1 answer:

telo118 [61]3 years ago

3 0

Answer:

The answer is false.

26/3 is 8.66. 8.66 + 6 is 14.66 which doesn't equal 24

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PLEASE HELP ME with this question

vovangra [49]

Answer:

abc are correct

Step-by-step explanation:

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2 years ago

What is 238 rouned to the nearst houndred

Alexeev081 [22]

200 because:
250 or higher is 300
249 or lower is 200

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2 years ago

Read 2 more answers

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

Can somebody please help me and tell me what the error is!! PLEASE I NEED HELP ELSE I WILL FAIL MY CLASS AND GO TO SUMMER SCHOOL

VikaD [51]

Answer:

I think that is enlargement

Step-by-step explanation:

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3 years ago

I need help wit one & two please help

seropon [69]

23 is less than 25 which means it's closer to 20, so you would round it down to 20. 1.75 is over 0.50, 1/2, so you would round it up to 2. If you multiply 2 and 20 you would get 40.

Number two gets the same explanation just with different numbers.

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3 years ago