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3 years ago

5

Help please I need help

1 answer:

Llana [10]3 years ago

5 0

Answer:

#4 - y= 1/2x-2 #5- y=1/3x+3

x y x y

0 -2 0 3

2 -1 3 4

4 0 6 5

6 1 9 6

Step-by-step explanation:

Hope that helped :)

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Please help me this will mark brainiest thx I will appreciate it thank you

Savatey [412]

Answer:

z=40 degrees

x=90 degrees

y=40 degrees

Step-by-step explanation:

To solve z you know that 50+90+z= 180 degrees. So z equals 40. To solve x you subtract 90 from 180 to get 90 degrees. To solve you add 50+x+y. You know that x= 90. You would add the angles together to get 180 degrees. 50+90=140 and 180-140 equals 40. Sorry if i'm wrong.

6 0

3 years ago

write an enaquality that expresses the reason the lengths 5 feet , 10 feet and 20 feet could not make a triangle. explain how th

NikAS [45]

For any triangle, the 2 shortest sides added together have to be longer than the longest sides.

short side (10) + short side (5) needs to be greater than (>) longest side (20)

10 + 5 > 20

15 > 20

This inequality is incorrect, so the lengths cannot make a triangle.

Hope this helps :)

7 0

3 years ago

I will mark the brainiest! Please, and thanks! :)

Contact [7]

The best thing is to have a ruler with the appropriate units to construct these triangles;)

8 0

3 years ago

Help pls, math due tonight :(

Olenka [21]

9514 1404 393

Answer:

40 ft
t = 1 second

Step-by-step explanation:

A graphing calculator answers these questions easily.

The ball achieves a maximum height of 40 ft, 1 second after it is thrown.

__

The equation is usefully put into vertex form, as the vertex is the answer to the questions asked.

h(t) = -16(t^2 -2t) +24

h(t) = -16(t^2 -2t +1) +24 +16 . . . . . . complete the square

h(t) = -16(t -1)^2 +40 . . . . . . . . . vertex form

Compare this to the vertex form:

f(x) = a(x -h)^2 +k . . . . . . vertex (h, k); vertical stretch factor 'a'

We see the vertex of our height equation is ...

(h, k) = (1, 40)

The ball reaches a maximum height of 40 feet at t = 1 second after it is thrown.

8 0

3 years ago

Solve the initial value problem y'=2 cos 2x/(3+2y),y(0)=−1 and determine where the solution attains its maximum value.

zloy xaker [14]

Answer:

$y=\frac{-3\pm\sqrt{4sin2x+1}}{2}$

$x={\pi}{4}$

Step-by-step explanation:

We are given that

$y'=\frac{2cos2x}{3+2y}$

y(0)=-1

$\frac{dy}{dx}=\frac{2cos2x}{3+2y}$

$(3+2y)dy=2cos2x dx$

Taking integration on both sides then we get

$\int (3+2y)dy=2\int cos 2xdx$

$3y+y^2=sin2x+C$

Using formula

$\int x^n=\frac{x^{n+1}}{n+1}+C$

$\int cosx dx=sinx$

Substitute x=0 and y=-1

$-3+1=sin0+C$

$-2=C$

$sin0=0$

Substitute the value of C

$y^2+3y=sin2x-2$

$y^2+3y-sin 2x+2=0$

$y=\frac{-3\pm\sqrt{(3)^2-4(1)(-sin2x+2)}}{2}$

By using quadratic formula

$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$

$y=\frac{-3\pm\sqrt{9+4sin2x-8}}{2}=\frac{-3\pm\sqrt{4sin2x+1}}{2}$

Hence, the solution $y=\frac{-3\pm\sqrt{4sin2x+1}}{2}$

When the solution is maximum then y'=0

$\frac{2cos2x}{3+2y}=0$

$2cos2x=0$

$cos2x=0$

$cos2x=cos\frac{\pi}{2}$

$cos\frac{\pi}{2}=0$

$2x=\frac{\pi}{2}$

$x=\frac{\pi}{4}$

3 0

3 years ago