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mr_godi [17]
3 years ago
13

What is the value of the expression below when

Mathematics
1 answer:
DiKsa [7]3 years ago
4 0

Answer:

The value of the expression is 36.

Step-by-step explanation:

If substitute the x (in 4x) with 4, and y (in 10y) with 2, we get:

  • 4(4) + 10(2)
  • 16 + 20
  • = 36

Hope this helps!!

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Find the area and perimeter of the shape below. Help plzzz.
kozerog [31]
Find the area of each shape and add.Let's start with the first semi-circle(half-circle)
If the area of a circle is πr² , then the area of a semi-circle is 1/2πr² where is 22/7 or 3.14, and r is the radius which is 1.8 meters
A=1/2πr²
A=1/2*22/7*1.8*1.8
A=5.09 m²(rounded to nearest hundredth)
Since the semi-circles are two and have the same radius, multiply the area of the first one by two or go through the same process again.
So 5.09 *2=10.18meters square is the area of the two semi-circles
Now, let's find the area of a rectangle  which is length times width, where length is 6 meters and width is (1.8+1.8, because 1.8 is half the width)=3.6 meters
A=l*w
A=6*3.6
A=21.6meters square
Therefore the area of the shape is 10.18m²+21.6m²=219.888m²
4 0
4 years ago
Please help me, i will give brainliest to BEST answer
Travka [436]

three, twelve, five, and six

3 0
3 years ago
Read 2 more answers
Prove that if the set of vectors {u1, u2, u3} is linearly dependent and u4 is any vector, then the set {u1, u2, u3, u4} is linea
Lesechka [4]

Answer with Step-by-step explanation:

We are given that the set of vectors{u_1,u_2,u_3} is lineraly dependent set .

We have to prove that the set {u_1,u_2,u_3,u_4} is linearly dependent .

Linearly dependent vectors : If the vectors u_1,u_2.u_3,u_4

are linearly  dependent therefore the linear combination

a_1u_1+a_2u_2+a_3u_3+a_4u_4=0

Then ,there exit a scalar which is not equal to zero .

Let a_1\neq0 then the vector u_1 will be zero and remaining  other vectors are not zero.

Proof:

When u_1,u_2,u_3 are linearly dependent vectors therefore, linear combination of vectors of given set

a_1u_1+a_2u_2+a_3u_3=0

By definition of linearly dependent vector

There exist a scalar which is not equal to zero.

Suppose a_1\neq 0 then  u_1=0

The linear combination of the set {u_1,u_2,u_3,u_4}

a_1u_1+a_2u_2+a_3u_3+a_4u_4=0

When a_1\neq0\; and\; u_1=0

Therefore,the set {u_1,u_2,u_3,u_4} is linearly dependent because it contain a vector which is zero.

Hence, proved .

5 0
3 years ago
The measure of two sides are given. Between what two numbers must the third side fall?
AveGali [126]

Answer:

between 12 and 17

hope helps

3 0
3 years ago
Find the value of: ​
nasty-shy [4]

Answer:

the value is 6

Step-by-step explanation:

substitue

5 0
4 years ago
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