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Anna71 [15]
3 years ago
14

Help me with 2, 8, and 10 plz

Mathematics
1 answer:
OverLord2011 [107]3 years ago
3 0

For #2:

<span> 3x^2 + 4 – 2x^2 + 6</span>

<span><span>You need to combine your like terms:
</span> 3x^2 – 2x^2= x^2</span>

4 + 6 = 10

This will now give you:

<span>X^2 +10</span>
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I need help ASAP<br> (Please show work)
schepotkina [342]
A)24 boards/1 * 47.5 inch/boards * 1 ft/12in = 95 ft.
according to dimensional analysis the answer is 95 ft.
b) 95/8 is approximately 12 so 12 boards are needed. 

8 0
3 years ago
What is the answer too 2 plus 2<br> A.4<br> B.8<br> C.6
lyudmila [28]

The answer will be A.4

4 0
3 years ago
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After doing some work in the house, Bob and Carol want to put a concrete patio on the side of the house to keep people from trac
12345 [234]

Answer:

V=3.098 yd³

Step-by-step explanation:

To answer this question we need to work with formula of the Volume of rectangular prism, the patio. Once it is required 1 inch deep. Some relations will be useful.

As 1 yard = 1 feet/3

1 yard = 1 inch/36

1 feet=12 inches

a) 23 feet 9 inches

23 ft 9''=23 +9/12=23.75 ft

b)10 feet  1 inch

10 feet  1" = 10ft + 1/12=10.083 ft

c)4 inches: 4/12 =0.33

V= w.l.h

V=23.75*10.083*0.33=79.025 ft³

Converting from cubic yard to cubic feet:

V≈79.025 : 27 yd³=2.926 yd³

Considering the possible spillage due to uneven base. ( adding 5%of V)

V≈ 2.92+0.05*(2.92)=3.098 yd³

4 0
3 years ago
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(a) Find the equation of the normal to the hyperbola x = 4t, y = 4/t at the point (8,2).
gtnhenbr [62]

The slope of the tangent line to the curve at (8, 2) is given by the derivative \frac{dy}{dx} at that point. By the chain rule,

\dfrac{dy}{dx} = \dfrac{dy}{dt} \times \dfrac{dt}{dx} = \dfrac{\frac{dy}{dt}}{\frac{dx}{dt}}

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x = 4t \implies \dfrac{dx}{dt} = 4

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Then

\dfrac{dy}{dx} = \dfrac{-\frac4{t^2}}4 = -\dfrac1{t^2}

We have x=8 and y=2 when t=2, so the slope at the given point is \frac{dy}{dx} = -\frac14.

The normal line to the same point is perpendicular to the tangent line, so its slope is +4. Then using the point-slope formula for a line, the normal line has equation

y - 2 = 4 (x - 8) \implies \boxed{y = 4x - 30}

Alternatively, we can eliminate the parameter and express y explicitly in terms of x :

x = 4t \implies t = \dfrac x4 \implies y = \dfrac4t = \dfrac4{\frac x4} = \dfrac{16}x

Then the slope of the tangent line is

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5 0
1 year ago
Combining like terms<br><br> simplify the expression<br> -r - 10r
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-r-10r

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