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Elena L [17]
3 years ago
13

What is 10y-3x+6=11 in standard form

Mathematics
1 answer:
disa [49]3 years ago
6 0
<span><span>3x</span><span><span>−10</span>y</span></span>=<span>−<span>5 is the answer I hope this helps! :)</span></span>
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What is the value of x in the equation below?<br> 12 – 2(x-1)=6
JulijaS [17]
X=25 in this equation. But if u divide by 12 it’s -25
5 0
3 years ago
Read 2 more answers
15. Simplify sin(90° - O) cos 0 - sin(180° +0) sin e.​
dalvyx [7]

Answer:  cos²(θ) + sin(θ)sin(e)

<u>Step-by-step explanation:</u>

sin (90° - θ)cos(Ф) - sin(180° + θ) sin(e)

Note the following identities:

sin (90° - θ) = cos(x)

sin (180° + θ) = -sin(x)

Substitute those identities into the expression:

   cos(x)cos(x)  -  -sin(x)sin(e)

=  cos²(x) + sin(x)sin(e)

4 0
3 years ago
The path of a ball thrown can be modeled by the equation f(x)=−16x^2+45x+6, where x is the horizontal distance. Use a transforma
andrey2020 [161]

Answer:

  g(x) = -4x^2 +(45/2)x +6

Step-by-step explanation:

A function is scaled horizontally by replacing x with x/k, where k is the expansion factor. Here, we want to expand the graph horizontally by a factor of 2, so the new function will be f(x/2).

  g(x) = f(x/2) = -16(x/2)^2 +45(x/2) +6

  g(x) = -4x^2 +(45/2)x +6

8 0
2 years ago
Measures of dispersion are used in finance as a proxy for risks. explain​
Bad White [126]

Answer:

Kindly check explanation

Step-by-step explanation:

Measures of dispersion are used to determine the variability of data points of samples ; it measures how much a given data deviates or behaves about the mean value or point. Measures of dispersion or variability include standard deviation and variance.

In finance, data with high degree of vatibalitu are considered as being unstable because this shows a high level of uncertainty about the average confidence of the output for such investment or business. Hence. Highly variable investments are cindisderdd volatile and risky

4 0
3 years ago
Given the points C(-1,-5) and D(-7,11) find the coordinates of point E on CD such that the ratio of CE to ED is 5:3.
Anon25 [30]

Answer:

The coordinates of point E on CD are \left(-\frac{19}{4},5 \right).

Step-by-step explanation:

Let be C = (-1,-5), D = (-7,11) and \frac{CE}{ED} = \frac{5}{3}. The given ratio can be translated vectorially into this:

\overrightarrow {CE} = \frac{5}{3} \cdot \overrightarrow{ED}

And let consider that each point is a vector with respect to origin:

\vec C = (-1, -5) and \vec {D} = (-7,11)

Then,

\vec E -\vec C = \frac{5}{3}\cdot (\vec D -\vec E)

\vec E +\frac{5}{3}\cdot \vec E = \frac{5}{3}\cdot \vec D + \vec C

\frac{8}{3}\cdot \vec E = \frac{5}{3}\cdot \vec D + \vec C

\vec E = \frac{5}{8}\cdot \vec D + \frac{3}{8}\cdot \vec C

\vec E = \frac{5}{8}\cdot (-7,11)+\frac{3}{8}\cdot (-1,-5)

\vec E = \left(-\frac{35}{8},\frac{55}{8}  \right)+\left(-\frac{3}{8},-\frac{15}{8}  \right)

\vec E = \left(-\frac{35}{8}-\frac{3}{8},\frac{55}{8}-\frac{15}{8}    \right)

\vec{E} = \left(-\frac{19}{4}, 5\right)

The coordinates of point E on CD are \left(-\frac{19}{4},5 \right).

6 0
3 years ago
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