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3 years ago

Express each ratio in simplest form 18:30=

Mathematics

2 answers:

netineya [11]3 years ago

6 0

The answer is 3:5 bro.

MatroZZZ [7]3 years ago

4 0

Express each ratio in simplest form 18:30
 $\frac{18}{30} = \frac{3}{5}$

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Number of Packets

Alex

Answer:

13. $D = 18$

14. $S = (x, y + 6)$

Step-by-step explanation:

13.

Given

$M = (-8,2)$

$S = (10,2)$

Required

Determine the distance between M and S

Distance (D) is calculated using:

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

Where

$(x_1,y_1) = (-8,2)$

$(x_2,y_2) = (10,2)$

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

$D = \sqrt{(10 - (-8))^2 + (2 - 2)^2}$

$D = \sqrt{(10 +8)^2 + 0^2}$

$D = \sqrt{18^2 + 0}$

$D = \sqrt{18^2}$

$D = 18$

Hence, the distance between M and S is 18 units

14.

The coordinate of S and P are not given,

So, I'll just use (x,y) for P

i.e.

$P=(x,y)$

Required

Determine the coordinates of S

If S is 6 units above P, then the coordinates of S is

$S = (x, y + 6)$

i.e. we add the units to the y coordinate of P.

8 0

3 years ago

Write equation parallel to x=5 through the point (_6,6)

uranmaximum [27]

Answer:

$\boldsymbol x\boldsymbol=\boldsymbol - \bold6$

Step-by-step explanation:

Graph of x=5 is vertical straight line parallel to y-axis and perpendicular to x-axis. The points on graph have the x-coordinate as 5 and y-coordinate varies.

Similarly, line parallel to x=5 or y-axis passing through point (-6,6), in that all the points have same x-coordinate as -6. So the equation of line will be x=-6

For simplicity of visualization graph of both the equation x=5 and x=-6 are attached.

7 0

3 years ago

The paint can holds 2.5 litres and the length is 19cm what is the radius

Alekssandra [29.7K]

10.75 is the radius for cylinder

8 0

3 years ago

Just number 23 super confused

vodka [1.7K]

Answer:

x = -7

Step-by-step explanation:

This is an exponential equation.

$\frac{1}{2}^x=128$

We need to take log(or ln) of both sides and use logarithm property to solve this.

So, let's take ln of both sides, so we have:

$ln(\frac{1}{2}^x)=ln(128)$

Now, we will use this property shown below to further simplify:

$ln(a^b)=bln(a)$

So, we now have:

$ln(\frac{1}{2}^x)=ln(128)\\xln(\frac{1}{2})=ln(128)\\x=\frac{ln(128)}{ln(\frac{1}{2})}\\x=-7$

The value of x is -7

6 0

3 years ago

Identify the zeros of the function f(x) = 2x^2 − 4x + 5 using the Quadratic Formula

Nostrana [21]

For this case we have that by definition, the roots, or also called zeros, of the quadratic function are those values of x for which the expression is 0.

Then, we must find the roots of:

$2x ^ 2-4x + 5 = 0$

Where:

$a = 2\\b = -4\\c = 5$

We have to: $x = \frac {-b \pm \sqrt {b ^ 2-4 (a) (c)}} {2 (a)}$

Substituting we have:

$x = \frac {- (- 4) \pm \sqrt {(- 4) ^ 2-4 (2) (5)}} {2 (2)}\\x = \frac {4 \pm \sqrt {16-40}} {4}\\x = \frac {4 \pm \sqrt {-24}} {4}$

By definition we have to:

$i ^ 2 = -1$

So:

$x = \frac {4 \pm \sqrt {24i ^ 2}} {4}\\x = \frac {4 \pm i \sqrt {24}} {4}\\x = \frac {4 \pm i \sqrt {2 ^ 2 * 6}} {4}\\x = \frac {4 \pm 2i \sqrt {6}} {4}\\x = \frac {2 \pm i \sqrt {6}} {2}$

Thus, we have two roots:

$x_ {1} = \frac {2 + i \sqrt {6}} {2}\\x_ {2} = \frac {2-i \sqrt {6}} {2}$

Answer:

$x_ {1} = \frac {2 + i \sqrt {6}} {2}\\x_ {2} = \frac {2-i \sqrt {6}} {2}$

3 0

3 years ago