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2 years ago

Help please i need help if i get this one wrong i need to start over

Mathematics

1 answer:

earnstyle [38]2 years ago

6 0

Answer:

Step-by-step explanation:

Right/left by 3 units because in the equation, it added 3. Right because it is positive and Right/Left because this represents the change in the x-axis.

Up/Down by 5 units because in the equation, it added 5. Up because it is positive and Up/Down because this represnts the change in the y-axis.

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Solve by elimination: 3y-4x=29 2y+5=4

Vaselesa [24]

The solution is x = -61/8 and y = -1/2.

Step-by-step explanation:

The given equations are -4x + 3y = 29 and 2y + 5 = 4

Step 1 : There is no x term in the second equation.

Step 2 : Solve for y, 2y = 4 -5

y = -1/2

Step 3 : Substitute y = -1/2 in first equation and solve for x.

Step 4 : -4x + 3(-1/2) = 29

-4x -(3/2) = 29

-8x -3 = 29 $\times$ 2

-8x = 58 + 3

x = -61/8

5 0

2 years ago

Write the equation of the line that passes through the given points. (0,-1) and (-11,-4)

solong [7]

Answer:

Line equation

y=0.2727272727272727x−1 Slope 0.27 Intercept -1

Step-by-step explanation:

5 0

2 years ago

-1/2 + 4/9

stira [4]

Answer:

-1/18

Step-by-step explanation:

First, you have to change both fractions to have like denominators. The LCM of 2 and 9 is 18.

Second, you have to multiply -1/2 by 9/9 to get a denominator of 18. -1/2 x 9/9 is - 9/18.

Third, you have to multiply 4/9 by 2/2 to get a denominator of 18. 4/9 x 2/2 = 8/18.

Fourth, -9/18 + 8/18 = -1/18

Hope this helps!

4 0

2 years ago

What are the zeros of the quadratic function f(x) = 2x2 – 10x – 3?

Natasha2012 [34]

Answer:

$x=\frac{50+\sqrt{31}}{2},\frac{50-\sqrt{31}}{2}$ are zeroes of given quadratic equation.

Step-by-step explanation:

We have been a quadratic equation:

$2x^2-10x-3$

We need to find the zeroes of quadratic equation

We have a formula to find zeroes of a quadratic equation:

$x=\frac{b^2\pm\sqrt{D}}{2a}\text{where}D=\sqrt{b^2-4ac}$

General form of quadratic equation is $ax^2+bx+c$

On comparing general equation with b given equation we get

a=2,b=-10,c=-3

On substituting the values in formula we get

$D=\sqrt{(-10)^2-4(2)(-3)}$

$\Rightarrow D=\sqrt{100+24}=\sqrt{124}$

Now substituting D in $x=\frac{b^2\pm\sqrt{D}}{2a}$ we get

$x=\frac{(-10)^2\pm\sqrt{124}}{2\cdot 2}$

$x=\frac{100\pm\sqrt{124}}{4}$

$x=\frac{100\pm2\sqrt{31}}{4}$

$x=\frac{50\pm\sqrt{31}}{2}$

Therefore, $x=\frac{50+\sqrt{31}}{2},\frac{50-\sqrt{31}}{2}$

5 0

2 years ago

Read 2 more answers

Help please and please hurry this assignment is due 11:59

IceJOKER [234]

You find the area of trapezium: 45
Then lower shape, triangle: 13.5
Then you add both:
B) 58.5 square feet:))

4 0

1 year ago