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2 years ago

6

Need help in all asap plz

1 answer:

makkiz [27]2 years ago

6 0

1. 1 by 0.305
2. 1 by 0.946
3. 1 by 0.454
4. 1 by 3.79

Basically, you multiply the measurement you got to what you are trying to convert it to, like I listed above ^.

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Write an equivalent fraction for 10/2

Scorpion4ik [409]

10/2 is an improper fraction. So, change it to a mixed number. 10÷2=5

5 is the mixed number and is the equivalent fraction. You can also do 5/1 which is also equivalent.

7 0

3 years ago

horrorfan [7]

Answer:

6 square root of 2

Step-by-step explanation:

sin theta = opposite / hypotenuse

sin 45 = 6/x

Multiply by x

x sin 45 = 6

Divide by sin 45

x sin 45/ sin 45 = 6 / sin 45

x = 6/ sin 45

x = 6/ (1/sqrt(2))

x = 6 sqrt(2)

3 0

2 years ago

Read 2 more answers

Please help me with the following questions. Thanks!

Natali5045456 [20]

Answer:

The answer to your questions are in bold

Step-by-step explanation:

a)

C = $\left[\begin{array}{ccc}-6&6\\-2&4\\\end{array}\right]$

= -24 + 12

= -12

b) -1 7 -4 -1 -1 - 4 7 - 1 - 5 6

-2 -6 -8 8 = -2 - 8 -6 + 8 -10 2

2 -3 2 -7 2 + 2 -3 - 7 4 -10

-1 10 -6 5 -1 - 6 10 + 5 -7 15

5 0

3 years ago

Let f(x) be an exponential functionWhich transformations to the graph of f(x) would describe e - f(x - 3) ?

gayaneshka [121]

Look at the graph below carefully

Observe the results of shifting ={2}^{x}f(x)=2x

vertically:

The domain, (−∞,∞) remains unchanged.

When the function is shifted up 3 units to ={2}^{x}+3g(x)=2x +3:

The y-intercept shifts up 3 units to (0,4).

The asymptote shifts up 3 units to y=3y=3.

The range becomes (3,∞).

When the function is shifted down 3 units to ={2}^{x}-3h(x)=2 x −3:

The y-intercept shifts down 3 units to (0,−2).

The asymptote also shifts down 3 units to y=-3y=−3.

The range becomes (−3,∞).

3 0

2 years ago

Read 2 more answers

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