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3 years ago

Use the drawing tool(s) to form the correct answer on the provided graph.

Mathematics

1 answer:

RoseWind [281]3 years ago

4 0

Answer: Mark points and draw a lines threw them (-1, -10) and (3, 10)

Step-by-step explanation:

Drawing the inverse of a line means taking two points and reversing them, two points on the graph and reversing them would be like this,

(-10, -1) and making it (-1, -10)

and a different point

(10, 3) and making it (3, 10)

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find all possible value of the given variable

mamaluj [8]

$1.\\ \\ h^2+5h=0 \\ \\h(x+5)=0\\ \\x=0 \ \ \ or \ \ \ x+5 =0\ \ |-5\\ \\x+5-5=0-5\\ \\x=0 \ \ \ or \ \ \ x=-5$

$2.\\ \\ z^2-z=0\\ \\z(x-1)=0\\ \\z=0 \ \ \ or \ \ \ z-1 =0 \ \ | +1\\ \\z-1+1 =0 +1 \\ \\x=0 \ \ \ or \ \ \ z=1$

$3.\\ \\m^2+13m+40=0 \\ \\a=1 ,\ b=13, \ c=40 \\ \\\Delta =b^2-4ac =13^2-4\cdot 1\cdot 40=169 - 1600=-1431 \\ \\and \ we \ know \ when \ \Delta \ is \ negative, \ theres \ no \solution$

$4.\\ \\z^2-3z=0 \\ \\ (z-3)=0\\ \\z=0 \ \ \ or \ \ \ z-3 =0\ \ |+3\\ \\ z-3+3=0+3\\ \\z=0 \ \ \ or \ \ \ z=3$

$5.\\ \\q^2+7q=0 \\ \\q(q+7)=0\\ \\q=0 \ \ \ or \ \ \ q+7 =0\ \ |-7\\ \\q+7-7=0-7\\ \\q=0 \ \ \ or \ \ \ q=-7$

$6.\\ \\k^2+2k=0\\ \\k(k+2)=0\\ \\k=0 \ \ \ or \ \ \ k+2 =0\ \ |-2\\ \\k+2-2=0-2\\ \\k=0 \ \ \ or \ \ \ k=-2$

$7. \\ \\ x^2-3x-70=0 \\ \\a=1,\ b=-3, \ c=-70 \\ \\\Delta =b^2-4ac = (-3)^2-4\cdot 1\cdot (-70)= 9+280=289\\ \\ x_{1}=\frac{-b-\sqrt{\Delta} }{2a}=\frac{3-\sqrt{289}}{2 }=\frac{ 3-17}{2}=\frac{-14}{2}=-7$

$x_{2}=\frac{-b+\sqrt{\Delta} }{2a}=\frac{3+\sqrt{289}}{2 }=\frac{ 3+17}{2}=\frac{20}{2}=10\\ \\(x+7)(x-10)=0$

$8.\\ \\q^2+7q-60=0 \\ \\a=1,\ b=7, \ q=-60 \\ \\\Delta =b^2-4ac = 7^2-4\cdot 1\cdot (-60)=49+240=289 \\ \\ x_{1}=\frac{-b-\sqrt{\Delta} }{2a}=\frac{-7-\sqrt{289}}{2 }=\frac{ -7-17}{2}=\frac{-24}{2}=-12$

$x_{2}=\frac{-b+\sqrt{\Delta} }{2a}=\frac{-7+\sqrt{289}}{2 }=\frac{ -7+17}{2}=\frac{ 10}{2}= 5\\ \\(x+12)(x-5)=0$

$9.\\ \\z^2+9z-36=0 \\ \\a=1,\ b=9, \ q=-36 \\ \\\Delta =b^2-4ac = 9^2-4\cdot 1\cdot (-36)= 81+144=225\\ \\ x_{1}=\frac{-b-\sqrt{\Delta} }{2a}=\frac{-9-\sqrt{225}}{2 }=\frac{ -9-15}{2}=\frac{-24}{2}=-12$

$x_{2}=\frac{-b+\sqrt{\Delta} }{2a}=\frac{-9+\sqrt{225}}{2 }=\frac{ -9+15}{2}=\frac{6}{2}=3\\ \\(x+11)(x-3)=0$

$10.\\ \\d^2-13d+22=0 \\ \\a=1,\ b=-13, \ q=22 \\ \\\Delta =b^2-4ac = (-13)^2-4\cdot 1\cdot 22= 169-88=81\\ \\ d_{1}=\frac{-b-\sqrt{\Delta} }{2a}=\frac{13-\sqrt{81}}{2 }=\frac{ 13-9}{2}=\frac{4}{2}=2$

$d_{2}=\frac{-b+\sqrt{\Delta} }{2a}=\frac{13+\sqrt{81}}{2 }=\frac{ 13+9}{2}=\frac{22}{2}=11\\ \\(d-2)(d-11)=0$

7 0

3 years ago

Factorise each of the following algebraic expressions completely,

LekaFEV [45]

Answer:

see explanation

Step-by-step explanation:

(a)

Given

2k - 6k² + 4k³ ← factor out 2k from each term

= 2k(1 - 3k + 2k²)

To factor the quadratic

Consider the factors of the product of the constant term ( 1) and the coefficient of the k² term (+ 2) which sum to give the coefficient of the k- term (- 3)

The factors are - 1 and - 2

Use these factors to split the k- term

1 - k - 2k + 2k² ( factor the first/second and third/fourth terms )

1(1 - k) - 2k(1 - k) ← factor out (1 - k) from each term

= (1 - k)(1 - 2k)

1 - 3k + 2k² = (1 - k)(1 - 2k) and

2k - 6k² + 4k³ = 2k(1 - k)(1 - 2k)

(b)

Given

2ax - 4ay + 3bx - 6by ( factor the first/second and third/fourth terms )

= 2a(x - 2y) + 3b(x - 2y) ← factor out (x - 2y) from each term

= (x - 2y)(2a + 3b)

8 0

3 years ago

Find the marginal profit of C(x)=5x^2, R(x)=x^3+7x+10 and find all values of x for which the marginal profit is zero.

jonny [76]

Marginal profit, P(x) = R(x) - C(x)

P(x) = x³ - 5x² + 7x + 10

x³ - 5x² + 7x + 10 = 0
If you solve it, you get x = -0.839 where P(x) is zero

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3 years ago

What is the degree of the polynomial given below?

Ivenika [448]

Answer:

it is 3 i am pretty sure

Step-by-step explanation:

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3 years ago

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15 - 5 x = 10 x + 45

Tomtit [17]

Answer:

x=-2

Step-by-step explanation:

add 5x , then subtract 45, then divide by 15

6 0

3 years ago

Read 2 more answers