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Verizon [17]
3 years ago
15

Help with Geometry please!

Mathematics
1 answer:
Nesterboy [21]3 years ago
3 0
Its the fourth one hope this helps
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The top of the ladder makes a 13° angle with the building. The base of the ladder is 6 feet from the base of the building. How f
KonstantinChe [14]

tan x = opp/hyp  

x is the side of the building and 6 is the distance from the bottom of the ladder to the building

tan 13 = 6/x

multiply both sides by x

x tan 13 =6

divide both sides by tan 13

x = 6/ tan 13

x = 26

The ladder goes 26 ft up the building

5 0
3 years ago
A bill originally amounting Rs 1550, now amounts to Rs 1736 after adding VAT. Find the VAT rate.kaabil original amounting ​
Vlad [161]

Answer: 12%

Step-by-step explanation:

Original amount = Rs 1550

VAT = Rs 1736 - Rs 1550 = Rs 186

VAT rate = VAT/Original amount × 100

= 186/1550 × 100

= 18600/1550

= 12%

Therefore, the VAT rate is 12%.

Check:

= Rs 1550 + (12% × Rs 1550)

= Rs 1550 + $186

= Rs 1736

4 0
3 years ago
Please help!<br> Anwser photo below
Shkiper50 [21]

Answer:

C

Step-by-step explanation:

3 0
3 years ago
Read 2 more answers
Facts about pasical's education
Paha777 [63]
Blaise Pascal (June 19, 1623 - August 19, 1662) was a great contributor to math, science, and philosophy, especially Christian philosophy. Interesting Blaise Pascal Facts: Pascal's early education in France was conducted at home by his father due to the prodigious talent and understanding he showed as a child. Did u mean pascal instead of pasical becaus ethere is no such thing. Hope I helped :)
4 0
3 years ago
Find the absolute extrema of f(x) = e^{x^2+2x}f ( x ) = e x 2 + 2 x on the interval [-2,2][ − 2 , 2 ] first and then use the com
fredd [130]

f(x)=e^{x^2+2x}\implies f'(x)=2(x+1)e^{x^2+2x}

f has critical points where the derivative is 0:

2(x+1)e^{x^2+2x}=0\implies x+1=0\implies x=-1

The second derivative is

f''(x)=2e^{x^2+2x}+4(x+1)^2e^{x^2+2x}=2(2x^2+4x+3)e^{x^2+2x}

and f''(-1)=\frac2e>0, which indicates a local minimum at x=-1 with a value of f(-1)=\frac1e.

At the endpoints of [-2, 2], we have f(-2)=1 and f(2)=e^8, so that f has an absolute minimum of \frac1e and an absolute maximum of e^8 on [-2, 2].

So we have

\dfrac1e\le f(x)\le e^8

\implies\displaystyle\int_{-2}^2\frac{\mathrm dx}e\le\int_{-2}^2f(x)\,\mathrm dx\le\int_{-2}^2e^8\,\mathrm dx

\implies\boxed{\displaystyle\frac4e\le\int_{-2}^2f(x)\,\mathrm dx\le4e^8}

5 0
3 years ago
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