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2 years ago

URGENT HELPP!!!NO LINKS

Mathematics

1 answer:

Scrat [10]2 years ago

7 0

Answer:

25 I believe

Step-by-step explanation:

think has 5 letters

only using it once ofc

5×5= 25

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Prove 2/sqrt3cosx+sinx=sec(pi/6-x)

dlinn [17]

Thus L.H.S = R.H.S that is 2/√3cosx + sinx = sec(Π/6-x) is proved

We have to prove that

2/√3cosx + sinx = sec(Π/6-x)

To prove this we will solve the right-hand side of the equation which is

R.H.S = sec(Π/6-x)

= 1/cos(Π/6-x)

[As secƟ = 1/cosƟ)

= 1/[cos Π/6cosx + sin Π/6sinx]

[As cos (X-Y) = cosXcosY + sinXsinY , which is a trigonometry identity where X = Π/6 and Y = x]

= 1/[√3/2cosx + 1/2sinx]

= 1/(√3cosx + sinx]/2

= 2/√3cosx + sinx

R.H.S = L.H.S

Hence 2/√3cosx + sinx = sec(Π/6-x) is proved

Learn more about trigonometry here : brainly.com/question/7331447

#SPJ9

5 0

1 year ago

NO LINKS!!! Part 2: Find the Lateral Area, Total Surface Area, and Volume. Round your answer to two decimal places.

slava [35]

Answer:

<h3>Question 7</h3>

Lateral Surface Area

The bases of a triangular prism are the triangles.

Therefore, the Lateral Surface Area (L.A.) is the total surface area excluding the areas of the triangles (bases).

$\implies \sf L.A.=2(10 \times 6)+(3 \times 6)=138\:\:m^2$

Total Surface Area

Area of the isosceles triangle:

$\implies \sf A=\dfrac{1}{2}\times base \times height=\dfrac{1}{2}\cdot3 \cdot \sqrt{10^2-1.5^2}=\dfrac{3\sqrt{391}}{4}\:m^2$

Total surface area:

$\implies \sf T.A.=2\:bases+L.A.=2\left(\dfrac{3\sqrt{391}}{4}\right)+138=167.66\:\:m^2\:(2\:d.p.)$

Volume

$\sf \implies Vol.=area\:of\:base \times height=\left(\dfrac{3\sqrt{391}}{4}\right) \times 6=88.98\:\:m^3\:(2\:d.p.)$

<h3>Question 8</h3>

Lateral Surface Area

The bases of a hexagonal prism are the pentagons.

Therefore, the Lateral Surface Area (L.A.) is the total surface area excluding the areas of the pentagons (bases).

$\implies \sf L.A.=5(5 \times 6)=150\:\:cm^2$

Total Surface Area

Area of a pentagon:

$\sf A=\dfrac{1}{4}\sqrt{5(5+2\sqrt{5})}a^2$

where a is the side length.

Therefore:

$\implies \sf A=\dfrac{1}{4}\sqrt{5(5+2\sqrt{5})}\cdot 5^2=43.01\:\:cm^2\:(2\:d.p.)$

Total surface area:

$\sf \implies T.A.=2\:bases+L.A.=2(43.01)+150=236.02\:\:cm^2\:(2\:d.p.)$

Volume

$\sf \implies Vol.=area\:of\:base \times height=43.011193... \times 6=258.07\:\:cm^3\:(2\:d.p.)$

8 0

1 year ago

What is the answer to thiss

lana [24]

Hi Student!

This question is fairly simple because it gives us an equation and they also give us a value for the variable that is within the equation and they tell us evaluate the expression. So let's plug in the values and solve.

Plug in the values

$m^2 + 5$

$(9)^2 + 5$

Factor out the exponent

$(9*9) + 5$

$81 + 5$

Combine

$86$

Therefore, the final answer that we would get when substituting m with 9 in the given equation is that we get 86.

5 0

1 year ago

Evaluate the definite integral using the graph of f(x) (Image included)

Tanya [424]

a) The first integral corresponds to the area under y = f(x) on the interval [0, 3], which is a right triangle with base 3 and height 5, hence the integral is

$\displaystyle \int_0^3 f(x) \, dx = \frac12 \times 3 \times 5 = \boxed{\frac{15}2}$

b) The integral is zero since the areas under the curve over [3, 4] and [4, 5] are equal but opposite in sign. In other words, on the interval [3, 5], f(x) is symmetric and odd about x = 4, so

$\displaystyle \int_3^5 f(x) \, dx = \int_3^4 f(x) \, dx + \int_4^5 f(x) \, dx = \int_3^4 f(x) \, dx - \int_3^4 f(x) \, dx = \boxed{0}$