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FinnZ [79.3K]
2 years ago
13

Please help marking brainlists 5-9c-4c=

Mathematics
1 answer:
Ivanshal [37]2 years ago
7 0

Answer:

the answer would be: 5-13c

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Expressions are equivalent???
Semenov [28]

Answer:

only the first option is equivalent

Step-by-step explanation:

4 0
3 years ago
Solve the following equation: lim x->0 sin3x/5x^3-4x.
goblinko [34]
Usual limit of sin is sinX/X--->1, when X--->0

sin3x/5x^3-4x=0/0?, sin3x/3x--->1 when x --->0, so sin3x/5x^3-4x=                    [3x. sin3x / 3x] /(5x^3-4x)=(sin3x / 3x) . (3x/5x^3-4x)
   =(sin3x / 3x) . (3/5x^2- 4)
finally lim  sin3x/5x^3-4x=lim (sin3x / 3x) .(3/5x^2- 4)=1x(3/-4)= - 3/4
                 x----->0            x---->0
3 0
3 years ago
-7πh = 98π <br> solve for h
ICE Princess25 [194]

Answer:

h = -14

Step-by-step explanation:

-7πh = 98π

⇔ -7h = 98   (Just Divide both sides by π)

⇔ h = 98 ÷ (-7) = -14

5 0
1 year ago
Read 2 more answers
Gabriel wants to paint the model. How
enyata [817]

Answer:

<h2>The area of the base is 144 square inches.</h2><h2>The area of each triangular face is 66 square inches.</h2><h2>Grabiel needs 408 square inches of paint.</h2>

Step-by-step explanation:

The complete problem is attached.

Notice that the figure is a square pyramid, where its base dimensions are 12 inches by 12 inches, which represents an area of

B=12 \times 12 = 144 \ in^{2}

The slant height of the pyramid is 11 inches, which allow us to find the area of each triangle face

A=\frac{1}{2}bh =\frac{1}{2}(12)(11)= 66 \ in^{2}

But there are four triangle faces, so A_{faces} =4(66)=264 \ in^{2}.

Therefore, the area of each triangular face is 66 square inches.

So, the total surface area would be the sum

S=144+264=408 \ in^{2}

Therefore, Gabriel needs 408 square inches to paint the whole model.

4 0
3 years ago
Since 2003 median home prices in Midvale, UT have been growing exponentially at roughly 4.7 % per year. If you had purchased a h
Kipish [7]

Answer:

The home would be worth $249000 during the year of 2012.

Step-by-step explanation:

The price of the home in t years after 2004 can be modeled by the following equation:

P(t) = P(0)(1+r)^{t}

In which P(0) is the price of the house in 2004 and r is the growth rate.

Since 2003 median home prices in Midvale, UT have been growing exponentially at roughly 4.7 % per year.

This means that r = 0.047

$172000 in 2004

This means that P(0) = 172000

What year would the home be worth $ 249000 ?

t years after 2004.

t is found when P(t) = 249000. So

P(t) = P(0)(1+r)^{t}

249000 = 172000(1.047)^{t}

(1.047)^{t} = \frac{249000}{172000}

\log{(1.047)^{t}} = \log{\frac{249000}{172000}}

t\log(1.047) = \log{\frac{249000}{172000}}

t = \frac{\log{\frac{249000}{172000}}}{\log(1.047)}

t = 8.05

2004 + 8.05 = 2012

The home would be worth $249000 during the year of 2012.

8 0
3 years ago
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