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adelina 88 [10]
2 years ago
15

What is the slope of this line A -1/2 B 1/2 C 2​

Mathematics
1 answer:
denpristay [2]2 years ago
4 0
-1/2 is your answer to this question
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A triangle has side lengths of 39in, 40 in, and 58 in. Classify it as acute, obtuse, or right.​
S_A_V [24]
The triangle is acute
7 0
3 years ago
Solve<br> y=2x2-3x-1<br> y=x-3
4vir4ik [10]

Answer:

The value of x is 1 and y = -2

Step-by-step explanation:

y = 2x²- 3x - 1;

y = x - 3

2x²- 3x - 1 = x - 3

2x² - 3x - 1 - x + 3 = 0

2x² - 4x + 2 = 0

take 2 as common

2(x² - 2x + 1) = 0

x² - 2x + 1 = 0

x² - x - x + 1 = 0

x(x - 1) - 1(x - 1) = 0

(x - 1) (x - 1) = 0

x - 1 = 0 or x - 1 = 0

x = 1 or x = 1

Now,

y = x - 3

y = 1 - 3

y = - 2

Thus, The value of x is 1 and y is -2

<u>-TheUnknown</u><u>Scientist</u>

4 0
3 years ago
Richard mows 1/3 of his yard in 1/2 hour. How much of his yard would Richard mow in 1 hour?
Vlada [557]
I hope this helps you




if he mows 1/3 yard in 1/2 hour


? yard in 1 hour



?.1/2=1.1/3


?=2/3 yard in 1 hour
3 0
3 years ago
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Solve for x <br> x+7/3=1<br> a) -10<br> b) -22/3<br> c) -3/7 <br> d) 4
Effectus [21]

Answer:

The answer is x = -4/3

Step-by-step explanation:

5 0
3 years ago
The point (1, −1) is on the terminal side of angle θ, in standard position. What are the values of sine, cosine, and tangent of
Lilit [14]

Check the picture below.

\bf (\stackrel{a}{1}~,~\stackrel{b}{-1})\qquad \impliedby \textit{let's find the hypotenuse} \\\\\\ \textit{using the pythagorean theorem} \\\\ c^2=a^2+b^2\implies c = \sqrt{a^2+b^2} \qquad \begin{cases} c=hypotenuse\\ a=adjacent\\ b=opposite\\ \end{cases} \\\\\\ c=\sqrt{1^2+(-1)^2}\implies c=\sqrt{2} \\\\[-0.35em] ~\dotfill

\bf sin(\theta ) \implies \cfrac{\stackrel{opposite}{-1}}{\stackrel{hypotenuse}{\sqrt{2}}}\implies \cfrac{-1}{\sqrt{2}}\cdot \cfrac{\sqrt{2}}{\sqrt{2}}\implies -\cfrac{\sqrt{2}}{(\sqrt{2})^2}\implies -\cfrac{\sqrt{2}}{2}

\bf cos(\theta ) \implies \cfrac{\stackrel{adjacent}{1}}{\stackrel{hypotenuse}{\sqrt{2}}}\implies \cfrac{1}{\sqrt{2}}\cdot \cfrac{\sqrt{2}}{\sqrt{2}}\implies \cfrac{\sqrt{2}}{(\sqrt{2})^2}\implies \cfrac{\sqrt{2}}{2} \\\\\\ tan(\theta ) = \cfrac{\stackrel{opposite}{-1}}{\stackrel{adjacent}{1}}\implies tan(\theta ) = -1

7 0
3 years ago
Read 2 more answers
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