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Nadusha1986 [10]

2 years ago

5

Find the slope of the line which contains (-3, 7) and (1,5)

1 answer:

earnstyle [38]2 years ago

5 0

Answer:

-½

Step-by-step explanation:

Slope is (y2-y1)÷(x2-x1). Use these two coordinates. Substitute.

(5-7)÷(1-(-3))=(-2)÷4= -(1/2)

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Solve the simultaneous equations<br> y = 9 - X<br> y = 2x2 + 4x + 6

kenny6666 [7]

Answer:

$\mathrm{Therefore,\:the\:final\:solutions\:for\:}y=9-x,\:y=2x^2+4x+6\mathrm{\:are\:}$

$\begin{pmatrix}x=\frac{1}{2},\:&y=\frac{17}{2}\\ x=-3,\:&y=12\end{pmatrix}$

Step-by-step explanation:

Given the simultaneous equations

$y=9-x$

$y\:=\:2x^2\:+\:4x\:+\:6$

Subtract the equations

$y=9-x$

$-$

$\underline{y=2x^2+4x+6}$

$y-y=9-x-\left(2x^2+4x+6\right)$

$\mathrm{Refine}$

$x\left(2x+5\right)=3$

$\mathrm{Solve\:}\:x\left(2x+5\right)=3$

$2x^2+5x=3$ ∵ $\mathrm{Expand\:}x\left(2x+5\right):\quad 2x^2+5x$

$\mathrm{Subtract\:}3\mathrm{\:from\:both\:sides}$

$2x^2+5x-3=3-3$

$\mathrm{Solve\:with\:the\:quadratic\:formula}$

$\mathrm{Quadratic\:Equation\:Formula:}$

$\mathrm{For\:a\:quadratic\:equation\:of\:the\:form\:}ax^2+bx+c=0\mathrm{\:the\:solutions\:are\:}$

$x_{1,\:2}=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

$\mathrm{For\:}\quad a=2,\:b=5,\:c=-3:\quad x_{1,\:2}=\frac{-5\pm \sqrt{5^2-4\cdot \:2\left(-3\right)}}{2\cdot \:2}v\\$

$x=\frac{-5+\sqrt{5^2-4\cdot \:2\left(-3\right)}}{2\cdot \:2}$

$=\frac{-5+\sqrt{5^2+4\cdot \:2\cdot \:3}}{2\cdot \:2}$

$=\frac{-5+\sqrt{49}}{2\cdot \:2}$

$=\frac{-5+\sqrt{49}}{4}$

$=\frac{-5+7}{4}$

$=\frac{2}{4}$

$=\frac{1}{2}$

Similarly,

$x=\frac{-5-\sqrt{5^2-4\cdot \:2\left(-3\right)}}{2\cdot \:2}:\quad -3$

$\mathrm{The\:solutions\:to\:the\:quadratic\:equation\:are:}$

$x=\frac{1}{2},\:x=-3$

$\mathrm{Plug\:the\:solutions\:}x=\frac{1}{2},\:x=-3\mathrm{\:into\:}y=9-x$

$\mathrm{For\:}y=9-x\mathrm{,\:subsitute\:}x\mathrm{\:with\:}\frac{1}{2}:\quad y=\frac{17}{2}$

$\mathrm{For\:}y=9-x\mathrm{,\:subsitute\:}x\mathrm{\:with\:}-3:\quad y=12$

$\mathrm{Therefore,\:the\:final\:solutions\:for\:}y=9-x,\:y=2x^2+4x+6\mathrm{\:are\:}$

$\begin{pmatrix}x=\frac{1}{2},\:&y=\frac{17}{2}\\ x=-3,\:&y=12\end{pmatrix}$

3 0

3 years ago

5 points6) In a clothes bag there are 3 green socks (G), 2 blue socks (B), and 1 redsock (R). One sock is taken out of the bag a

ddd [48]

there are 6 combinations but if we change the order we can get more than 6, so the last option is out

the option 2 and 3 cant be because has two reds socks and just exist 1

s, the right option is the first

7 0

1 year ago

HURRYYY i neeeeeddd helppppppppp look at the photo ^^^

Oliga [24]

Can you show a better photo some of the question is cut off

8 0

2 years ago

Read 2 more answers

Please Help me To solve this Problem

mart [117]

You're given that φ is an angle that terminates in the third quadrant (III). This means that both cos(φ) and sin(φ), and thus sec(φ) and csc(φ), are negative.

Recall the Pythagorean identity,

cos²(φ) + sin²(φ) = 1

Multiply the equation uniformly by 1/cos²(φ),

cos²(φ)/cos²(φ) + sin²(φ)/cos²(φ) = 1/cos²(φ)

1 + tan²(φ) = sec²(φ)

Solve for sec(φ) :

sec(φ) = - √(1 + tan²(φ))

Given that cot(φ) = 1/4, we have tan(φ) = 1/cot(φ) = 1/(1/4) = 4. Then

sec(φ) = - √(1 + 4²) = -√17

3 0

2 years ago

I’m so confused please help I can’t afford another bad grade

iogann1982 [59]

Answer:

2

Step-by-step explanation:

x / (3y)

Let x = 18 and y = 3

18 / ( 3*3)

Determine the denominator first

18 / ( 9)

Divide

2

8 0

2 years ago