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MatroZZZ [7]
2 years ago
6

Solve each system by substitution.

Mathematics
1 answer:
VLD [36.1K]2 years ago
5 0

Answer:

b. no solution

Step-by-step explanation:

change to slope-int form

y=-4x-8

y=-4x+5/2

see how the slopes are the same but the y-int are different? they are parallel, so they never touch

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7h<br> t: h²k²<br> -7k<br> k² - h²
mezya [45]

Answer:

i'm pretty sure it's the second option

Step-by-step explanation:

4 0
1 year ago
The diameter of the circle is 9.7 m . Calculate the length of its circumference. Round to the nearest tenth and remember to use
RSB [31]

Answer:

30.4m

Step-by-step explanation:

Diameter(D) = 2r (radius)

D = 9.7m

r = 9.7m / 2

= 4.85m

C = 2πr

= 2π × 4.85m

= 30.47344873982099m

rounding - Leave it as is if the digit in the tenth is 4 or lower

- round it up if it's 5 or higher.

~~ 30.4m

5 0
2 years ago
Simplify the following <br> (-4r^5)(2r^8)
jolli1 [7]
Multiplying a - and a + equals a - , -8r18
8 0
3 years ago
Find the midpoint of the line segment whose endpoints are the given points. (-24.6, 38.1) and (-25.7, -17.7)
CaHeK987 [17]

Answer:

Midpoint is(-25.15,10.2)

Step-by-step explanation:

A line segment refers to line that has two endpoints.

Let (a,b),(c,d) be the endpoints of a line segment.

Midpoint of the line segment is given by (\frac{a+c}{2} ,\frac{b+d}{2})

Take the given points as follows:

(a,b)=(-24.6,38.1)\\(c,d)=(-25.7,-17.7)

Midpoint of a line segment =(\frac{-24.6-25.7}{2},\frac{28.1-17.7}{2})

=(\frac{-50.3}{2} ,\frac{20.4}{2} )\\=(-25.15,10.2)

7 0
3 years ago
3. Which of the following quadratic equations has no solution? A) 0 = −2(x − 5)2 + 3 B) 0 = −2(x − 5)(x + 3) C) 0 = 2(x − 5)2 +
klemol [59]

Answer:

D) 0 = 2(x + 5)(x + 3)

Step-by-step explanation:

Which of the following quadratic equations has no solution?

We have to solve the Quadratic equation for all the options in other to get a positive value as a solution for x.

A) 0 = −2(x − 5)2 + 3

0 = -2(x - 5) × 5

0 = (-2x + 10) × 5

0 = -10x + 50

10x = 50

x = 50/10

x = 5

Option A has a solution of 5

B) 0 = −2(x − 5)(x + 3)

Take each of the factors and equate them to zero

-2 = 0

= 0

x - 5 = 0

x = 5

x + 3 = 0

x = -3

Option B has a solution by one of its factors as a positive value of 5

C) 0 = 2(x − 5)2 + 3

0 = 2(x - 5) × 5

0 = (2x -10) × 5

0 = 10x -50

-10x = -50

x = -50/-10

x = 5

Option C has a solution of 5

D) 0 = 2(x + 5)(x + 3)

Take each of the factors and equate to zero

0 = 2

= 0

x + 5 = 0

x = -5

x + 3 = 0

x = -3

For option D, all the values of x are 0, or negative values of -5 and -3.

Therefore the Quadratic Equation for option D has no solution.

3 0
3 years ago
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