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stiv31 [10]
3 years ago
12

5. How can you get the variable alone in the equation x + 37 = 94?

Mathematics
1 answer:
dlinn [17]3 years ago
4 0
The answer is d.because you have to subtract from both side
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Which of the functions graphed below is continuous?
vaieri [72.5K]

Answer:

The answer is the second one

5 0
3 years ago
Find the value g(-2), where g is defined below.<br> g(x) = 2x^2- 4x – 4
DanielleElmas [232]

Answer:

12

Step-by-step explanation:

For this question, simply plug in -2 for x:

g(-2)=2(-2)^2-4(-2)-4\\g(-2)=2(4)+8-4\\g(-2)=8+8-4\\g(-2)=12

Hope this helps!!

8 0
3 years ago
A ball is thrown in the air from a height of 3 m with an initial velocity of 12 m/s. To the nearest tenth of a second, how long
stepladder [879]

Answer:  1.4 seconds

<u>Step-by-step explanation:</u>

The equation is: h(t) = at² + v₀t + h₀          where

  • a is the acceleration (in this case it is gravity)
  • v₀ is the initial velocity
  • h₀ is the initial height

Given:  

  • a = -9.81 (if it wasn't given in your textbook, you can look it up)
  • v₀ = 12
  • h₀ = 3

Since we are trying to find out when it lands on the ground, h(t) = 0

EQUATION:   0 = 9.81t² + 12t + 3

Use the quadratic equation to find the x-intercepts

                        a=-9.81, b=12, c=3

x=\dfrac{-b \pm \sqrt{b^2-4ac}}{2a}\\\\\\x=\dfrac{-(12)\pm \sqrt{(12)^2-4(-9.81)(3)}}{2(-9.81)}\\\\\\x=\dfrac{-12\pm 16.2}{-19.62}\\\\\\x=\dfrac{-12+ 16.2}{-19.62}=-0.2\qquad x=\dfrac{-12- 16.2}{-19.62}=\large\boxed{1.4}\\

Note: Negative time (-0.2) is not valid

4 0
3 years ago
I mark as brainliest i promise​
Nataly_w [17]

Answer:

32 units

Step-by-step explanation:

2 x 4 x 4 = 32

you mulyiply the height the width and the length

5 0
2 years ago
Read 2 more answers
Work out the angle of elevation from the
insens350 [35]

\quad \huge \quad \quad \boxed{ \tt \:Answer }

\qquad \tt \rightarrow \:Angle \:\: of \:\: elevation= 30°

____________________________________

\large \tt Solution  \: :

Let the angle of elevation be " x "

\qquad \tt \rightarrow \:  \cos(x)  =  \dfrac{9 \sqrt{3} }{18}

\qquad \tt \rightarrow \:  \cos(x)  =  \dfrac{ \sqrt{3} }{2}

\qquad \tt \rightarrow \: x =  \cos {}^{ - 1}  \bigg( \cfrac{ \sqrt{3} }{2} \bigg )

\qquad \tt \rightarrow \: x = 30 \degree \:  \:  or \:  \:  \cfrac{ \pi}{6}   \:  \: rad

Answered by : ❝ AǫᴜᴀWɪᴢ ❞

8 0
2 years ago
Read 2 more answers
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