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jolli1 [7]
3 years ago
10

Please give me the answer to this problem

Mathematics
1 answer:
Darya [45]3 years ago
5 0

Total 27inch^

Calculation of 1 rectangle +1 triangle(11-7=4 is the base)

triangle    x^+4^=5^

                  x^+16=25

                  x^=9

                    x=3      (height=3)

Sides of triangle  Base=4   Hypotenuse=5     height =3 

area 1/2 4  x  3  =2x3=6

Area triangle =6

Area Rectanglel   7x3=21

21+6=27 inch^ total area



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6 0
3 years ago
Can you solve the inequality 2(x-3) is less than or equal to 10 without using the distributive property? Explain.
STatiana [176]

Answer:

Yes, without using the distributive property, we get the value for the given expression 2(x - 3)  ≤ 10  as   x ≤ 8

Step-by-step explanation:

Here, the given expression is:

2(x - 3)  ≤ 10

Yes, the given expression can be solved without using the DISTRIBUTIVE Property

Here, consider the given expression:

2(x - 3)  ≤ 10

Now, divide the  inequality by 2 on both sides, we get:

\frac{2(x-3)}{2}  \leq  \frac{10}{2}\\\implies (x - 3)  \leq 5

Now adding (+3) on both the sides, we get:

(x - 3)  ≤ 5    ⇒  (x - 3  + 3 )  ≤ 5 + 3  

or, x ≤ 8

Hence, without using the distributive property, we get the value of x ≤ 8 for the given expression 2(x - 3)  ≤ 10

6 0
3 years ago
For each system of equations, drag the true statement about its solution set to the box under the system?
natta225 [31]

Answer:

y = 4x + 2

y = 2(2x - 1)

Zero solutions.

4x + 2 can never be equal to 4x - 2

y = 3x - 4

y = 2x + 2

One solution

3x - 4 = 2x + 2 has one solution

Step-by-step explanation:

* Lets explain how to solve the problem

- The system of equation has zero number of solution if the coefficients

 of x and y are the same and the numerical terms are different

- The system of equation has infinity many solutions if the

   coefficients of x and y are the same and the numerical terms

   are the same

- The system of equation has one solution if at least one of the

  coefficient of x and y are different

* Lets solve the problem

∵ y = 4x + 2 ⇒ (1)

∵ y = 2(2x - 1) ⇒ (2)

- Lets simplify equation (2) by multiplying the bracket by 2

∴ y = 4x - 2

- The two equations have same coefficient of y and x and different

  numerical terms

∴ They have zero equation

y = 4x + 2

y = 2(2x - 1)

Zero solutions.

4x + 2 can never be equal to 4x - 2

∵ y = 3x - 4 ⇒ (1)

∵ y = 2x + 2 ⇒ (2)

- The coefficients of x and y are different, then there is one solution

- Equate equations (1) and (2)

∴ 3x - 4 = 2x + 2

- Subtract 2x from both sides

∴ x - 4 = 2

- Add 4 to both sides

∴ x = 6

- Substitute the value of x in equation (1) or (2) to find y

∴ y = 2(6) + 2

∴ y = 12 + 2 = 14

∴ y = 14

∴ The solution is (6 , 14)

y = 3x - 4

y = 2x + 2

One solution

3x - 4 = 2x + 2 has one solution

3 0
3 years ago
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