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allsm [11]
3 years ago
5

Evaluate the expression (18+7)+5³-9

Mathematics
2 answers:
wel3 years ago
5 0

Answer:

141

Step-by-step explanation:

lara31 [8.8K]3 years ago
3 0

Answer:

Step-by-step explanation:

18+7+5^3-9

25+125-9

150-9

141

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Isabella solved the following equation: 4x − 2x + 8 = 6(x + 4) Step Work Justification 1 4x − 2x + 8 = 6x + 24 Distributive Prop
AnnZ [28]

Answer:

step 3 has incorrect justification.

Step-by-step explanation:

because it was combine like terms but 8 is still in left hand side. So,combine like terms 3 has incorrect justification.

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3 years ago
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What is the volume of the shape below? Use 3.14 for 8in. 3 in​
musickatia [10]
The answer is 75.4

Explanation
7 0
3 years ago
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Archy [21]

(a) If the particle's position (measured with some unit) at time <em>t</em> is given by <em>s(t)</em>, where

s(t) = \dfrac{5t}{t^2+11}\,\mathrm{units}

then the velocity at time <em>t</em>, <em>v(t)</em>, is given by the derivative of <em>s(t)</em>,

v(t) = \dfrac{\mathrm ds}{\mathrm dt} = \dfrac{5(t^2+11)-5t(2t)}{(t^2+11)^2} = \boxed{\dfrac{-5t^2+55}{(t^2+11)^2}\,\dfrac{\rm units}{\rm s}}

(b) The velocity after 3 seconds is

v(3) = \dfrac{-5\cdot3^2+55}{(3^2+11)^2} = \dfrac{1}{40}\dfrac{\rm units}{\rm s} = \boxed{0.025\dfrac{\rm units}{\rm s}}

(c) The particle is at rest when its velocity is zero:

\dfrac{-5t^2+55}{(t^2+11)^2} = 0 \implies -5t^2+55 = 0 \implies t^2 = 11 \implies t=\pm\sqrt{11}\,\mathrm s \imples t \approx \boxed{3.317\,\mathrm s}

(d) The particle is moving in the positive direction when its position is increasing, or equivalently when its velocity is positive:

\dfrac{-5t^2+55}{(t^2+11)^2} > 0 \implies -5t^2+55>0 \implies -5t^2>-55 \implies t^2 < 11 \implies |t|

In interval notation, this happens for <em>t</em> in the interval (0, √11) or approximately (0, 3.317) s.

(e) The total distance traveled is given by the definite integral,

\displaystyle \int_0^8 |v(t)|\,\mathrm dt

By definition of absolute value, we have

|v(t)| = \begin{cases}v(t) & \text{if }v(t)\ge0 \\ -v(t) & \text{if }v(t)

In part (d), we've shown that <em>v(t)</em> > 0 when -√11 < <em>t</em> < √11, so we split up the integral at <em>t</em> = √11 as

\displaystyle \int_0^8 |v(t)|\,\mathrm dt = \int_0^{\sqrt{11}}v(t)\,\mathrm dt - \int_{\sqrt{11}}^8 v(t)\,\mathrm dt

and by the fundamental theorem of calculus, since we know <em>v(t)</em> is the derivative of <em>s(t)</em>, this reduces to

s(\sqrt{11})-s(0) - s(8) + s(\sqrt{11)) = 2s(\sqrt{11})-s(0)-s(8) = \dfrac5{\sqrt{11}}-0 - \dfrac8{15} \approx 0.974\,\mathrm{units}

7 0
2 years ago
Please help me out here ASAP. Thanks!
GuDViN [60]
   y - y₁ = m(x - x₁)
y - (-1) = 4[x - (-3)]
   y + 1 = 4(x + 3)
   y + 1 = 4(x) + 4(3)
   y + 1 = 4x + 12
       - 1           - 1
         y = 4x + 11

The answer is B.
4 0
3 years ago
A toy has various shaped objects that a child can push through matching holes. The area of the square hole is 14 square centimet
stiv31 [10]

Answer:

I am not sure the answer but I can say this from my experiences, to know the shapes and sizes, you may measure like I usually do.  

Step-by-step explanation:

I am not sure

7 0
3 years ago
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