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IRINA_888 [86]
2 years ago
15

RS= 6x-8

Mathematics
1 answer:
Airida [17]2 years ago
4 0

Answer:

The length of RS is 22

The length of UV is 22

Step-by-step explanation:

6x-8. ---------- equation (i)

4x+2 ------------ equation (ii)

Now,

equating equation i and ii, we get

6x-8 = 4x+2

or, 6x-4x = 2+8

or, 2x = 10

or, x = 10/2

.°. x = 5

Then,

length of RS = 6x-8

= 6×5-8

= 30-8

= 22

Again

length of UV = 4x+2

= 4×5+2

= 20+2

= 22

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Evaluate the expression you got in part f for d = 5.
Triss [41]

Answer:

Before you get started, take this readiness quiz.

Is n÷5 an expression or an equation? If you missed this problem, review Example 2.1.4.

Simplify 45. If you missed this problem, review Example 2.1.6.

Simplify 1+8•9. If you missed this problem, review Example 2.1.8.

Evaluate Algebraic Expressions

In the last section, we simplified expressions using the order of operations. In this section, we’ll evaluate expressions—again following the order of operations.

To evaluate an algebraic expression means to find the value of the expression when the variable is replaced by a given number. To evaluate an expression, we substitute the given number for the variable in the expression and then simplify the expression using the order of operations.

Example 2.3.1: evaluate

Evaluate x+7 when

x=3

x=12

Solution

To evaluate, substitute 3 for x in the expression, and then simplify.

x+7

Substitute.

3+7

Add.

10

When x=3, the expression x+7 has a value of 10.

To evaluate, substitute 12 for x in the expression, and then simplify.

x+7

Substitute.

12+7

Add.

19

When x=12, the expression x+7 has a value of 19. Notice that we got different results for parts (a) and (b) even though we started with the same expression. This is because the values used for x were different. When we evaluate an expression, the value varies depending on the value used for the variable.

exercise 2.3.1

Evaluate: y+4 when

y=6

y=15

Answer a

Answer b

exercise 2.3.2

Evaluate: a−5 when

a=9

a=17

Answer a

Answer b

Example 2.3.2

Evaluate 9x−2, when

x=5

x=1

Solution

Remember ab means a times b, so 9x means 9 times x.

To evaluate the expression when x=5, we substitute 5 for x, and then simplify.

9x−2

Substitute 5 for x.

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Multiply.

45−2

Subtract.

43

To evaluate the expression when x=1, we substitute 1 for x, and then simplify.

9x−2

Substitute 1 for x.

9⋅1−2

Multiply.

9−2

Subtract.

7

Notice that in part (a) that we wrote 9•5 and in part (b) we wrote 9(1). Both the dot and the parentheses tell us to multiply.

exercise 2.3.3

Evaluate: 8x−3, when

x=2

x=1

Answer a

Answer b

exercise 2.3.4

Evaluate: 4y−4, when

y=3

y=5

Answer a

Answer b

Example 2.3.3: evaluate

Evaluate x2 when x=10.

Solution

We substitute 10 for x, and then simplify the expression.

x2

Substitute 10 for x.

102

Use the definition of exponent.

Evaluate: 2x when x=6.

Answer

exercise 2.3.8

Evaluate: 3x when x=4.

Answer

Example 2.3.5: evaluate

Evaluate 3x+4y−6 when x=10 and y=2.

Solution

This expression contains two variables, so we must make two substitutions.

3x+4y−6

Substitute 10 for x and 2 for y.

3(10)+4(2)−6

Multiply.

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Add and subtract left to right.

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When x=10 and y=2, the expression 3x+4y−6 has a value of 32.

exercise 2.3.9

Evaluate: 2x+5y−4 when x=11 and y=3

Answer

exercise 2.3.10

Evaluate: 5x−2y−9 when x=7 and y=8

Answer

Example 2.3.6: evaluate

Evaluate 2x2+3x+8 when x=4.

Solution

We need to be careful when an expression has a variable with an exponent. In this expression, 2x2 means 2•x•x and is different from the expression (2x)2, which means 2x•2x.

2x2+3x+8

Substitute 4 for each x.

2(4)2+3(4)+8

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2(16)+3(4)+8

Multiply.

32+12+8

Add.

52

exercise 2.3.11

Evaluate: 3x2+4x+1 when x=3.

Answer

exercise 2.3.12

Evaluate: 6x2−4x−7 when x=2.

Answer

Identify Terms, Coefficients, and Like Terms

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8 0
3 years ago
283% of 305 estimate
saveliy_v [14]
2.83 x 305 ( of is always multiplication )
5 0
2 years ago
Read 2 more answers
Help please!!!
katovenus [111]

Answer:

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Step-by-step explanation:

<u>To solve quadratic systems,we always substitute one variable in terms if the other and then solve the equation.</u>

x + 2y = 6                                 ---------------(1)

y - 5 = (x-2)^{2}         ---------------(2)

y = (x-2)^{2} + 5         ---------------(3)

Substitute (3) in (1) ,

x + 2( (x-2)^{2} + 5 ) = 6

(a + b)^{2} =a^{2} + 2ab + b^{2}

x + 2( x^{2} - 4x + 4 + 5 ) = 6

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The roots of the quadratic equation ax^{2}  +bx+c is

x = \frac{(-b) + \sqrt{(-b)^{2}-4 \times ac }  }{2 \times a}  -----------(5)

According to equation (5),solution of (4) is

x =  \frac{7 + \sqrt{(-7)^{2}-4 \times 24 }  }{2 \times 2}

x =  \frac{7+\sqrt{49-96}}{4}

x = \frac{7+\sqrt{47}\times i }{4}

 

4 0
3 years ago
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1. c 2. d and 3,a hope i helped

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3 years ago
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5 0
2 years ago
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