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3 years ago

What is the solution to l + (-3)=9

Mathematics

1 answer:

wariber [46]3 years ago

7 0

Answer:

Step-by-step explanation:

I + (-3) = 9

Lets isolate I:

9 - (-3) = I

12 = I

Check:

12 + (-3) = 9

9 = 9

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Draw the triangle with vertices a(1, 1), b(5, 3), c(1, 7). find the parametrization, including endpoints, and sketch to check. (

Firlakuza [10]

Part A:

The parametric equations $x=x_1+(x_2-x_1)t$ and $y=y_1+(y_2-y_1)t$ where $0\leq t\leq1$ describes the line segment that joins the points $(x_1,\ y_1)$ and $(x_2,\ y_2)$ .

The parameterization of the line joining points A(1, 1) and B(5, 3) are given by substituting $x_1=1,\ \ y_1=1,\ \ x_2=5,\ \ y_2=3$ .

Thus, we have

x = 1 + (5 - 1)t = 1 + 4t
y = 1 + (3 - 1)t = 1 + 2t

Therefore, the parameterization of line joining points A(1, 1) and B(5, 3) are x = 1 + 4t and y = 1 + 2t.

Part B:

The parametric equations $x=x_1+(x_2-x_1)t$ and $y=y_1+(y_2-y_1)t$ where $0\leq t\leq1$ describes the line segment that joins the points $(x_1,\ y_1)$ and $(x_2,\ y_2)$ .

The parameterization of the line joining points B(5, 3) and C(1, 7) are given by substituting $x_1=5,\ \ 
y_1=3,\ \ x_2=1,\ \ y_2=7$ .

Thus, we have

x = 5 + (1 - 5)t = 5 - 4t
y = 3 + (7 - 3)t = 3 + 4t

Therefore, the parameterization of line joining points B(5, 3) and C(1, 7) are x = 5 - 4t and y = 3 + 4t.

Part C:

The parametric equations $x=x_1+(x_2-x_1)t$ and $y=y_1+(y_2-y_1)t$ where $0\leq t\leq1$ describes the line segment that joins the points $(x_1,\ y_1)$ and $(x_2,\ y_2)$ .

The parameterization of the line joining points C(1, 7) and A(1, 1) are given by substituting $x_1=1,\ \ 
y_1=7,\ \ x_2=1,\ \ y_2=1$ .

Thus, we have

x = 1 + (1 - 1)t = 1
y = 7 + (1 - 7)t = 7 - 6t

Therefore, the parameterization of line joining points A(1, 1) and B(5, 3) are x = 1 and y = 7 - 6t.

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3 years ago

What is the probability that you will be able to draw all of the cards in the deck without being forced to stop because you have

Talja [164]

Step-by-step explanation:

Probability will be 0 According to me!!

<h2>Mark me BRAINLIEST </h2>

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3 years ago

Are s^2 ,2s^2 and 3s^2 equivalent

alexira [117]

No they are not quivalent

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3 years ago

Need help with trigonometry

vovangra [49]

Answer:

$\sf \tan(A) & =\dfrac{9\sqrt{22}}{44}$

Step-by-step explanation:

If angle C is the right angle, then side c is the hypotenuse.

Use Pythagoras' Theorem $a^2+b^2=c^2$ to find the length of side a:

Given:

b = 2√22
c = 13

$\implies a^2+(2 \sqrt{22})^2=13^2$

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$\implies a^2=81$

$\implies a=\sqrt{81}$

$\implies a=9$

<u>Tan Trig Ratio</u>

$\sf \tan(\theta)=\dfrac{O}{A}$

where:

$\theta$ is the angle
O is the side opposite the angle
A is the side adjacent the angle

Given:

$\theta$ = A
O = side opposite angle A = a = 9
A = side adjacent angle A = b = 2√22

$\begin{aligned}\implies \sf \tan(A) & =\dfrac{9}{2\sqrt{22}}\\\\ & =\dfrac{9}{2\sqrt{22}} \times \dfrac{\sqrt{22}}{\sqrt{22}}\\\\ & = \dfrac{9\sqrt{22}}{44} \end{aligned}$

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andrew11 [14]

Just do 10.5 × 50 and whatever answer you get label as grams

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