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3 years ago

Three times the greater of two consecutive odd integers is five less than four

Mathematics

1 answer:

Sergio [31]3 years ago

4 0

Answer:

Step-by-step explanation:

Let the smallest integer (odd) be 2x + 1

Let the larger of the integers be 2x + 3

3*(2x + 3) = 4(2x + 1) - 5 remove the brackets

6x + 9 = 8x + 4 - 5 Combine

6x + 9 = 8x - 1 Add 1 to both sides

6x + 9 + 1 = 8x Combine

6x + 10 = 8x Subtract 6x

10 = 2x Divide by 2

10/2 = x

x = 5

So the smaller odd number is 2x + 1 = 2*5 + 1 = 11

And the larger off number = 2x + 3 = 2*5 + 3 = 13

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manuel ate 1/3 of the crackers on a plate. his brother ate 1/4 of the crackers. there were 5 crackers left on the plate. how man

Neporo4naja [7]

Answer:

there would have been 12 crackers on the plate at the start

Step-by-step explanation:

So 1/3 +1/4 = 7/12

Therefore 5/12 left on the plate …ie 5.

So total to start with

5/5×12=12

Ans 12

5 0

2 years ago

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Inga [223]

Answer:

$y=-\sqrt{3}x+2$

Step-by-step explanation:

We want to find the equation of a straight line that cuts off an intercept of 2 from the y-axis, and whose perpendicular distance from the origin is 1.

We will let Point M be (x, y). As we know, Point R will be (0, 2) and Point O (the origin) will be (0, 0).

First, we can use the distance formula to determine values for M. The distance formula is given by:

$\displaystyle d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

Since we know that the distance between O and M is 1, d=1.

And we will let M(x, y) be (x₂, y₂) and O(0, 0) be (x₁, y₁). So:

$\displaystyle 1=\sqrt{(x-0)^2+(y-0)^2}$

Simplify:

$1=\sqrt{x^2+y^2}$

We can solve for y. Square both sides:

$1=x^2+y^2$

Rearranging gives:

$y^2=1-x^2$

Take the square root of both sides. Since M is in the first quadrant, we only need to worry about the positive case. Therefore:

$y=\sqrt{1-x^2}$

So, Point M is now given by (we substitute the above equation for y):

$M(x,\sqrt{1-x^2})$

We know that Segment OM is perpendicular to Line RM.

Therefore, their slopes will be negative reciprocals of each other.

So, let’s find the slope of each segment/line. We will use the slope formula given by:

$\displaystyle m=\frac{y_2-y_1}{x_2-x_1}$

Segment OM:

For OM, we have two points: O(0, 0) and M(x, √(1-x²)). So, the slope will be:

$\displaystyle m_{OM}=\frac{\sqrt{1-x^2}-0}{x-0}=\frac{\sqrt{1-x^2}}{x}$

Line RM:

For RM, we have the two points R(0, 2) and M(x, √(1-x²)). So, the slope will be:

$\displaystyle m_{RM}=\frac{\sqrt{1-x^2}-2}{x-0}=\frac{\sqrt{1-x^2}-2}{x}$

Since their slopes are negative reciprocals of each other, this means that:

$m_{OM}=-(m_{RM})^{-1}$

Substitute:

$\displaystyle \frac{\sqrt{1-x^2}}{x}=-\Big(\frac{\sqrt{1-x^2}-2}{x}\Big)^{-1}$

Now, we can solve for x. Simplify:

$\displaystyle \frac{\sqrt{1-x^2}}{x}=\frac{x}{2-\sqrt{1-x^2}}$

Cross-multiply:

$x(x)=\sqrt{1-x^2}(2-\sqrt{1-x^2})$

Distribute:

$x^2=2\sqrt{1-x^2}-(\sqrt{1-x^2})^2$

Simplify:

$x^2=2\sqrt{1-x^2}-(1-x^2)$

Distribute:

$x^2=2\sqrt{1-x^2}-1+x^2$

So:

$0=2\sqrt{1-x^2}-1$

Adding 1 and then dividing by 2 yields:

$\displaystyle \frac{1}{2}=\sqrt{1-x^2}$

Then:

$\displaystyle \frac{1}{4}=1-x^2$

Therefore, the value of x is:

$\displaystyle \begin{aligned}\frac{1}{4}-1&=-x^2\\-\frac{3}{4}&=-x^2\\ \frac{3}{4}&=x^2\\ \frac{\sqrt{3}}{2}&=x\end{aligned}$

Then, Point M will be:

$\begin{aligned} \displaystyle M(x,\sqrt{1-x^2})&=M(\frac{\sqrt{3}}{2}, \sqrt{1-\Big(\frac{\sqrt{3}}{2}\Big)^2)}\\M&=(\frac{\sqrt3}{2},\frac{1}{2})\end{aligned}$

Therefore, the slope of Line RM will be:

$\displaystyle \begin{aligned}m_{RM}&=\frac{\frac{1}{2}-2}{\frac{\sqrt{3}}{2}-0} \\ &=\frac{\frac{-3}{2}}{\frac{\sqrt{3}}{2}}\\&=-\frac{3}{\sqrt3}\\&=-\sqrt3\end{aligned}$