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3 years ago

How do u solve question 30

Mathematics

1 answer:

Oksana_A [137]3 years ago

3 0

You just multiply the 2 numbers!

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1 2/5 as a improper friction

Oksanka [162]

Answer:

7/5

Step-by-step explanation:

8 0

3 years ago

Solve the equation: 3x + 34 when x = 4 Show work!

balandron [24]

3x + 34

x = 4

Substitute x:

3 times 4 = 12

12 + 34 = 46

The answer is 46.

Hope this helped

4 0

3 years ago

Read 2 more answers

Graph the system of equations. {x−y=64x+y=4

Fantom [35]

X-y=6 Equation 1
x+y=4 Equation 2

To graph the given system of equation, first find x and y-intercept of each equation.
x-y=6
When y=0
x=6 Point is (6,0)
When x=0
-y=6
y=-6 Point is (0,-6)

Now x-intercept and y-intercept for equation 2.
x+y=4
When x=0
y=4 Point is (0,4)
When y=0
x=4 Point is (4,0)

Now plot these points on the graph, the lines intersect each other at point (5,-1), which is the solution of the given system.

Answer: (5,-1)

5 0

3 years ago

Write the function in function notation. y=7b+8

sesenic [268]

F(b)=7b+8
you just need to change the x variable. make y into f(x) or in this case f(b)

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2 years ago

Factor the expression using the two different techniques listed for Parts 1(a) and 1(b).

Natalija [7]

Answer:

9a^4 b^10(2 +3a^6 b^5) (2 -3a^6 b^5)

see work below

Step-by-step explanation:

36a^4b^10 - 81a^16b^20

A) find the GCF

36a^4b^10 = 4*9 a^4b^10 = 2*2*3*3 a*a*a*a*b*b*b*b*b*b*b*b*b*b

81a^16b^20= 9*9a^16b^20= 3*3*3*3* a*a*a*a*a*a*a*a*a*a*a*a*a*a*a*a *b*b*b*b*b*b*b*b*b*b*b*b*b*b*b*b*b*b*b*b

The terms that appear in both terms is the GCF. The terms that remain are inside the parentheses.

The GCF is 3*3 a*a*a*a*b*b*b*b*b*b*b*b*b*b

36a^4b^10 - 81a^16b^20 = 3*3 a*a*a*a*b*b*b*b*b*b*b*b*b*b *

( 2*2 - 3*3a*a*a*a*a*a*a*a*a*a*a*a b*b*b*b*b*b*b*b*b*b)

Combining like terms

36a^4b^10 - 81a^16b^20 = 9a^4b^10(4-9a^12b^10)

The expression inside the parenthesis can be factored using the difference of squares

let x^2 =4 x =2

y^2 = 9a^12 b^10 y = 3a^6b^5

(x^2 -y^2) = (x+y)(x-y)

9a^4b^10(4-9a^12b^10) = 9a^4b^10 ( 2+3a^6b^5) ( 2-3a^6b^5)

b) difference of squares a^2 – b^2 = (a + b)(a – b)

let a^2 = 36a^4b^10

so a = 6a^2b^5

b^2 = 81a^16b^20

b = 9a^8 b^10

a^2 – b^2 = (a + b)(a – b)

36a^4b^10 - 81a^16b^20 = (6a^2b^5 +9a^8 b^10) (6a^2b^5 -9a^8 b^10)

We can factor a 3 a^2 b^5 out of the first term

3 a^2 b^5 (2 +3a^6 b^5) (6a^2b^5 -9a^8 b^10)

3 a^2 b^5 (2 +3a^6 b^5) 3 a^2 b^5 (2 -3a^6 b^5)

Multiply the terms outside the parentheses together

9a^4 b^10(2 +3a^6 b^5) (2 -3a^6 b^5)

4 0

3 years ago

Read 2 more answers