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3 years ago

Find the value of x.

Mathematics

1 answer:

arsen [322]3 years ago

7 0

Answer:

x=30

Step-by-step explanation:

(x-1)+151=180

(x-1)=29

x=30

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A ladder is leaning against a vertical wall which is 5m high. The top of the ladder slides all the way down the wall so that the

Lynna [10]

Answer:

Do using Pythagoras theorem... Also before doing a question of height and distance always try to make the figure

Step-by-step explanation:

Use the two triangles ABC and ABD to first get the base and find the height of the ladder

7 0

3 years ago

Find 2x2 − 2z4 + y2 − x2 + z4 if x = −4, y = 3, and z = 2.<br><br> Is -3 the correct answer?

Ket [755]

We're given the algebraic expression 2x^2 - 2z^4 + y^2 - x^2 + z^4

By hypothesis, x = -4, y = 3 and z = 2

Let's replace each letter by its given value:

2x^2 - 2z^4 + y^2 - x^2 + z^4
2(-4)^2 - 2(2)^4 + (3)^2 - (-4)^2 + (2)^4
(2*16) - 2*16 + 9 - 16 + 16
32 - 32 + 9 - 16 + 16
9

So wrong. The correct answer is not -3, but 9.

Hope this helps! :D

6 0

3 years ago

Which of the following expressions is equivalent to the expression 2(7x+3)

Ira Lisetskai [31]

Answer:

14x+6 and (7x+7)+(3+3)

Step-by-step explanation:

4 0

3 years ago

1. Which of the following describes the end behavior of the function ƒ(x) = x^4 + 3x^3 – 2x + 7?

snow_tiger [21]

Hello, when x tends to $\infty$ the term with the highest degree will lead the behaviour.

In other words.

$\displaystyle \lim_{x\rightarrow+\infty} {x^4+3x^3-2x+7}\\\\=\lim_{x\rightarrow+\infty} {x^4}\\\\=+\infty\\\\\\\displaystyle \lim_{x\rightarrow-\infty} {x^4+3x^3-2x+7}\\\\=\lim_{x\rightarrow-\infty} {x^4}\\\\=+\infty$

So, the answer B is correct.

Thank you.

6 0

3 years ago

Questions attached as screenshot below:Please help me I need good explanations before final testI pay attention

Nikitich [7]

The acceleration of the particle is given by the formula mentioned below:

$a=\frac{d^2s}{dt^2}$

Differentiate the position vector with respect to t.

$\begin{gathered} \frac{ds(t)}{dt}=\frac{d}{dt}\sqrt[]{\mleft(t^3+1\mright)} \\ =-\frac{1}{2}(t^3+1)^{-\frac{1}{2}}\times3t^2 \\ =\frac{3}{2}\frac{t^2}{\sqrt{(t^3+1)}} \end{gathered}$

Differentiate both sides of the obtained equation with respect to t.

$\begin{gathered} \frac{d^2s(t)}{dx^2}=\frac{3}{2}(\frac{2t}{\sqrt[]{(t^3+1)}}+t^2(-\frac{3}{2})\times\frac{1}{(t^3+1)^{\frac{3}{2}}}) \\ =\frac{3t}{\sqrt[]{(t^3+1)}}-\frac{9}{4}\frac{t^2}{(t^3+1)^{\frac{3}{2}}} \end{gathered}$

Substitute t=2 in the above equation to obtain the acceleration of the particle at 2 seconds.

$\begin{gathered} a(t=1)=\frac{3}{\sqrt[]{2}}-\frac{9}{4\times2^{\frac{3}{2}}} \\ =1.32ft/sec^2 \end{gathered}$

The initial position is obtained at t=0. Substitute t=0 in the given position function.

$\begin{gathered} s(0)=-23\times0+65 \\ =65 \end{gathered}$

8 0

1 year ago