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2 years ago

14

he perimeter of a rectangle is twice the sum of its length and its width. the perimeter is 16 and the length is 2 meters more th

an twice the width. What is the length

1 answer:

Makovka662 [10]2 years ago

3 0

Answer:

Step-by-step explanation:

2(l+b)=16

Also it is given that l=2b+2

2(2b+2+b)=16

3b+2=8

b=2metre

l=2(2)+2

l=6metre

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ratelena [41]

Answer:

54 sq units

Step-by-step explanation:

The triangle at the top (FAB) has a base of 10 and a height of 3. Area of triangle is 1/2bh so 1/2(10 * 3) = 1/2 * 30 = 15 sq units

If you cut off the triangle you have a trapezoid, EBCD. A trapezoid is (b1 + b2)/2 * h, so (6 + 7)/2 * 6 = 13/2 * 6 = 6.5 * 6 = 39 sq units

Add them together: 15 + 39 = 54 sq units

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2 years ago

Which property would allow you to use mental computation to simplify the problem 27 + 15 + 3 + 5? additive identity additive inv

Fofino [41]

Answer:

Commutative Property

Step-by-step explanation:

Additive Identity: If you add a real number to zero or add zero to a real number, then you get the same real number back.

Additive Inverse: When you add the same number but reversed sign, you get 0.

Commutative Property: a + b = b + a

Addition distributive property: a(b+ c) = ab + ac.

In this case, we use the commutative property:

27 + 15 + 3 + 5 = 27 + 3 + 15 + 5.

We know that 27 + 3 = 30 and 15 + 5 = 20, so we get 50 quickly using mental math.

5 0

3 years ago

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A school administrator will assign each student in a group of n students to one of m classrooms. If 3 < m < 13 < n, is

tatiyna

Answer:

Hence to get same number of students in each classroom,the sufficient condition is that assign 13n students to each classroom.

Step-by-step explanation:

Given:

There are m classrooms and n be the students

3<m<13<n.

To Find:

Whether it is possible to assign each of n students to one of m classrooms with same no.of students.

Solution:

This problem is related to p/q form has to be integer in order to get same no of students assigned to the classroom.

As similar as ,n/m ratio

So 1st condition is that,

If it is possible to assign the n/m must be integer and n should be multiple of m,

when we assign 3n students to m classrooms ,we cannot say that 3n/m= integer so that n is greater than 13 i.e n=14 and m=6

hence they are not multiple of each other so they will not make same students in each classrooms.

Otherwise,n=14 and m=7 they will give same number but this condition is not sufficient condition to assign the student.

So 2nd condition is that ,

When we assign 13n students to m classrooms, as 13 is prime number and

3<m<13 which implies the 13n/m to be integer so n and m must be multiple of each other.

Suppose n=20 and m=5 classrooms

then 13*20=260 ,

260/5=52 students in each classroom,

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3 years ago

What is the answer to (5x3)-18

Fittoniya [83]

Answer:

-3

Step-by-step explanation:

5*3=15

15-18= -3

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2 years ago

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Solve for w, where w is a real number.

Ber [7]

Answer:

w= 9

Step-by-step explanation:

$\sqrt{ - 4w + 61} = w - 4$

Square both sides:

-4w +61= (w -4)²

$\boxed{(a - b)^{2} = a^2 -2ab + b^2 }$

Expand:

-4w +61= w² -2(w)(4) +4²

-4w +61= w² -8w +16

Simplify:

w² -8w +16 +4w -61= 0

w² -4w -45= 0

Factorize:

(w -9)(w +5)= 0

w -9= 0 or w +5= 0

w= 9 or w= -5 (reject)

Note:

-5 is rejected since we are only taking the positive value of the square root here. If the negative square root is taken into consideration, then w= -5 would give us -9 on both sides of the equation.

<u>Why </u><u>do </u><u>we </u><u>use </u><u>negative </u><u>square </u><u>root?</u>

When solving an equation such as x²= 4,

we have to consider than squaring any number removes the negative sign i.e., the result of a squared number is always positive.

In the case of x²= 4, x can be 2 or -2. Thus, whenever we introduce a square root, a '±' must be used.

However, back to our question, we did not introduce the square root so we should not consider the negative square root value.

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2 years ago

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