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anzhelika [568]

3 years ago

11

What is the value of y in the equation 2 + y = −3? (1 point)

1 answer:

Ilya [14]3 years ago

8 0

Answer:

- 5

Step-by-step explanation:

2 + y = −3

Subtract 2 from both sides

2 + y - 2 = −3 - 2

Simplifying

y = -5

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Determine whether the three points are collinear.<br> (0, - 10). (-3,- 13), (2,-8)

LUCKY_DIMON [66]

Answer:

<h2>YES. These points are collinear.</h2>

Step-by-step explanation:

If three points are collinear, then the slopes are the same.

The formula of a slope:

$m=\dfrac{y_2-y_1}{x_2-x_1}$

For (0, -10) and (-3, -13):

$m=\dfrac{-13-(-10)}{-3-0}=\dfrac{-13+10}{-3}=\dfrac{-3}{-3}=1$

For (-3, -13) and (2, -8):

$m=\dfrac{-8-(-13)}{2-(-3)}=\dfrac{-8+13}{2+3}=\dfrac{5}{5}=1$

We can check the last pair (0, -10) and (2, -8):

$m=\dfrac{-8-(-10)}{2-0}=\dfrac{-8+10}{2}=\dfrac{2}{2}=1$

5 0

2 years ago

A plane takes off from an airport and flies at a speed of 400km/h on a course of 120° for 2 hours. the plane then changes its co

butalik [34]

Answer:

Distance from the airport = 894.43 km

Step-by-step explanation:

Displacement and Velocity

The velocity of an object assumed as constant in time can be computed as

$\displaystyle \vec{v}=\frac{\vec{x}}{t}$

Where $\vec x$ is the displacement. Both the velocity and displacement are vectors. The displacement can be computed from the above relation as

$\displaystyle \vec{x}=\vec{v}.t$

The plane goes at 400 Km/h on a course of 120° for 2 hours. We can compute the components of the velocity as

$\displaystyle \vec{v_1}=$

$\displaystyle \vec{v_1}=\ km/h$

The displacement of the plane in 2 hours is

$\displaystyle \vec{x_1}=\vec{v_1}.t_1=.(2)$

$\displaystyle \vec{x_1}=km$

Now the plane keeps the same speed but now its course is 210° for 1 hour. The components of the velocity are

$\displaystyle \vec{v_2}=$

$\displaystyle \vec{v_2}=km/h$

The displacement in 1 hour is

$\displaystyle \vec{x_2}=\vec{v_2}.t_2=.(1)$

$\displaystyle \vec{x_2}=km$

The total displacement is the vector sum of both

$\displaystyle \vec{x_t}=\vec{x_1}+\vec{x_2}=+$

$\displaystyle \vec{x_t}=km$

$\displaystyle \vec{x_t}=$

The distance from the airport is the module of the displacement:

$\displaystyle |\vec{x_t}|=\sqrt{(-746.41)^2+492.82^2}$

$\displaystyle |\vec{x_t}|=894.43\ km$

8 0

3 years ago

Find exact values for sin θ, cos θ and tan θ if csc θ = 3/2 and cos θ < 0.

kumpel [21]

Answer:

Part 1) $sin(\theta)=\frac{2}{3}$

Par 2) $cos(\theta)=-\frac{\sqrt{5}}{3}$

Part 3) $tan(\theta)=-\frac{2\sqrt{5}}{5}$

Step-by-step explanation:

step 1

Find the $sin(\theta)$

we have

$csc(\theta)=\frac{3}{2}$

Remember that

$csc(\theta)=\frac{1}{sin(\theta)}$

therefore

$sin(\theta)=\frac{2}{3}$

step 2

Find the $cos(\theta)$

we know that

$sin^{2}(\theta) +cos^{2}(\theta)=1$

we have

$sin(\theta)=\frac{2}{3}$

substitute

$(\frac{2}{3})^{2} +cos^{2}(\theta)=1$

$\frac{4}{9} +cos^{2}(\theta)=1$

$cos^{2}(\theta)=1-\frac{4}{9}$

$cos^{2}(\theta)=\frac{5}{9}$

square root both sides

$cos(\theta)=\pm\frac{\sqrt{5}}{3}$

we have that

$cos(\theta) < 0$ ---> given problem

so

$cos(\theta)=-\frac{\sqrt{5}}{3}$

step 3

Find the $tan(\theta)$

we know that

$tan(\theta)=\frac{sin(\theta)}{cos(\theta)}$

we have

$sin(\theta)=\frac{2}{3}$

$cos(\theta)=-\frac{\sqrt{5}}{3}$

substitute

$tan(\theta)=\frac{2}{3}:-\frac{\sqrt{5}}{3}=-\frac{2}{\sqrt{5}}$

Simplify

$tan(\theta)=-\frac{2\sqrt{5}}{5}$

6 0

2 years ago

Charles is doing renovations on his house. He has budgeted 25% of the cost for new

Rama09 [41]

Answer:

$14,400

<em>Brainliest, please!</em>

<em />

Step-by-step explanation:

25% is furniture, and furniture costs $3600.

So, 3600 x 4 is the total cost, which equals 14400

8 0

2 years ago

The number 84 can be expressed as the sum of 54 + 30. Which shows how to use the distributive

TEA [102]

Answer:I onesly dont no

Sorry I have no friends

Step-by-step explanation:

6 0

2 years ago