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2 years ago

2p – 10 – 4p – 7 can you help me?

Mathematics

2 answers:

Mrrafil [7]2 years ago

4 0

First combine like terms
2p-4p= -2p
-10-7= -17
final answer -2p-17

Mashutka [201]2 years ago

4 0

first combine like terms

2p-4p= -2p

-10-7= -17

final answer -2p-17

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Drag the tiles to the correct boxes to complete the pairs. Not all tiles will be used.

diamong [38]

Answer:

A-2, B-DNE*, C-3, D-DNE, E-4, F-1

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The first attachment shows the solutions to A and C.

The second attachment shows the solutions to E and F.

There are no real number solutions to systems B and D.

_____

In general, you can solve the linear equation for y, then substitute that into the quadratic. You can subtract the x-term on the left and complete the square to find the solutions.

(3-x) +12 = x^2 +x

15 = x^2 + 2x

16 = x^2 +2x +1 = (x +1)^2 . . . . add the square of half the x-coefficient to complete the square; next take the square root

±4 -1 = x = {-5, 3) . . . . . identifies the second solution set for system A

(x -1) -15 = x^2 +4x

-16 = x^2 +3x

-13.75 = x^2 +3x +2.25 = (x +1.5)^2

roots are complex: -1.5 ±i√13.75

(1-2x) +5 = x^2 -3x

6 = x^2 -x

6.25 = x^2 -x + .25 = (x -.5)^2

±2.5 +.5 = x = {-2, 3} . . . . . identifies the third solution set for system C

remaining problems are done in a similar way.

_____

* DNE = does not exist. There is no matching solution set for the complex numbers that are the solutions to this.

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<em>Hope this helps!</em>

<em />

<u>Brainliest would be great!</u>

<u />

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4 0

2 years ago

Solve each system of equations below by substitution

Anna007 [38]

Answer:

numero uno, ok

Step-by-step explanation:

6 0

2 years ago

So can anyone help me with #1 and 2 and show work I have no idea what I’m doing

Neporo4naja [7]

Answer:

x = 44*2

x = 62*2

x = 108/2

x = 160/2

12x = 60*2

12x = 120

x = 120/12

5x = 50/2

5x = 25

x = 25/5

3 0

2 years ago

The sum of the squares of 3 consecutive positive integers is 116. What are the numbers?

Elza [17]

You are supposed to use the third option: 3n2 + 6n + 5 = 116.
Take a look at the attachment to see how I got that solution.
Hope it helps! :)

5 0

2 years ago

Describe the transformation of the graph of f into the graph of g as either a horizontal or vertical stretch. f(x)=sqrt(x) and g

SpyIntel [72]

$\bf \qquad \qquad \qquad \qquad \textit{function transformations}
\\ \quad \\\\
% left side templates
\begin{array}{llll}
f(x)=&{{ A}}({{ B}}x+{{ C}})+{{ D}}
\\ \quad \\
y=&{{ A}}({{ B}}x+{{ C}})+{{ D}}
\\ \quad \\
f(x)=&{{ A}}\sqrt{{{ B}}x+{{ C}}}+{{ D}}
\\ \quad \\
f(x)=&{{ A}}(\mathbb{R})^{{{ B}}x+{{ C}}}+{{ D}}
\\ \quad \\
f(x)=&{{ A}} sin\left({{ B }}x+{{ C}} \right)+{{ D}}
\end{array}\\\\
--------------------\\\\$

$\bf \bullet \textit{ stretches or shrinks horizontally by } {{ A}}\cdot {{ B}}\\\\
\bullet \textit{ flips it upside-down if }{{ A}}\textit{ is negative}\\
\left. \qquad \right. \textit{reflection over the x-axis}
\\\\
\bullet \textit{ flips it sideways if }{{ B}}\textit{ is negative}\\
\left. \qquad \right. \textit{reflection over the y-axis}$

$\bf \bullet \textit{ horizontal shift by }\frac{{{ C}}}{{{ B}}}\\
\left. \qquad \right. if\ \frac{{{ C}}}{{{ B}}}\textit{ is negative, to the right}\\\\
\left. \qquad \right. if\ \frac{{{ C}}}{{{ B}}}\textit{ is positive, to the left}\\\\
\bullet \textit{ vertical shift by }{{ D}}\\
\left. \qquad \right. if\ {{ D}}\textit{ is negative, downwards}\\\\
\left. \qquad \right. if\ {{ D}}\textit{ is positive, upwards}\\\\
\bullet \textit{ period of }\frac{2\pi }{{{ B}}}$

with that template in mind, let's see $\bf f(x)=\sqrt{x}\qquad 
\begin{array}{llll}
g(x)=&\sqrt{0.5x}\\
&\quad \uparrow \\
&\quad B
\end{array}$

so B went form 1 on f(x), down to 0.5 or 1/2 on g(x)
B = 1/2, thus the graph is stretched by twice as much.

8 0

2 years ago

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