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Y_Kistochka [10]
3 years ago
7

I NEED HELP IM BEGGING YOU????:) Write the slope-intercept form of the equation for the line.

Mathematics
1 answer:
iogann1982 [59]3 years ago
8 0
It’s d because the slope is negative and it is 2/1 which can also be written as just 2. in the formula y=mx+b, m is the slope and b is the y-intercept. the y-intercept is where the line touches the y axis(in this case -1) so it would be written as y=2x+(-1)
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sammy [17]

Answer:

answer is A

Step-by-step explanation:

7 0
3 years ago
What is the solution to the equation
Dafna11 [192]
\sqrt{4t+5}=3-\sqrt{t+5}\\\\D:4t+5\geq0\ \wedge\ t+5\geq0\ \wedge\ \sqrt{t+5}\leq3\\\\t\geq-\dfrac{5}{4}\ \wedge\ t\geq-5\ \wedge\ t\leq6

therefore
D:x\in\left< -\dfrac{5}{4};\ 6\right>

\sqrt{4t+5}=3-\sqrt{t+5}\ \ \ |^2\\\\(\sqrt{4t+5})^2=(3-\sqrt{t+5})^2\ \ \ |use:(a-b)^2=a^2-2ab+b^2\\\\4t+5=3^2-2\cdot3\cdot\sqrt{t+5}+(\sqrt{t+5})^2\\\\4t+5=9-6\sqrt{t+5}+t+5\ \ \ \ |-t\\\\3t+5=14-6\sqrt{t+5}\ \ \ \ |-14\\\\3t-9=-6\sqrt{t+5}\ \ \ \ |change\ signs
9-3t=6\sqrt{t+5}\ \ \ \ |:3\\\\3-t=2\sqrt{t+5}\ \ \ \ |^2\\\\(3-t)^2=(2\sqrt{t+5})^2\\\\3^2-2\cdot3\cdot t+t^2=4(t+5)\\\\9-6t+t^2=4t+20\ \ \ |-4t-20\\\\t^2-10t-11=0\\\\t^2+t-11t-11=0\\\\t(t+1)-11(t+1)=0\\\\(t+1)(t-11)=0\iff t+1=0\ \vee\ t-11=0\\\\t=-1\in D\ \vee\ t=11\notin D
Answer: t = -1.


7 0
3 years ago
Read 2 more answers
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inysia [295]
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5 0
3 years ago
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Harrizon [31]
y=x^2-2x-15 (1) y=8x-40 (2)

8x-40=x^2-2x-15
X^2+10x-25=0
(x-5)^2
×=5 y=0
7 0
3 years ago
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ziro4ka [17]
Using the formula a² + b² = c² but reversing it to c² - b² = x² where x replaces the a for comprehension reasons

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√377 = 19.4~

Answer: x = 19.4
8 0
3 years ago
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