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2 years ago

8

Which expression represents the surface area of the regular square pyramid shown?

1 answer:

aliya0001 [1]2 years ago

3 0

Answer:

b^2 +2bl

Step-by-step explanation:

B = base length

L = slant height

The equation is for a regular square pyramid surface area

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Determine if (-2,6) and (4,12) are solutions to the system of equations:

kykrilka [37]

Answer:

Both $(-2,6)$ and $(4,12)$ are solutions to the system.

Step-by-step explanation:

In order to determine whether the two given points represent solutions to our system of equations, we must "plug" thos points into both equations and check that the equality remains valid.

Step 1: Plug $(-2,6)$ into $y=x+8$

$(6) = (-2)+8 = 6$

The solution verifies the equation.

Step 2: Plug $(-2,6)$ into $2y=x^2 + 8$

$2(6) = (-2)^2+8=4+8=12$

The solution verifies both equations. Therefore, $(-2,6)$ is a solution to this system.

Now we must check if the second point is also valid.

Step 3: Plug $(4,12)$ into $y=x+8$

$(12) = (4)+8 = 12$

Step 4: Plug $(4,12)$ into $2y=x^2 + 8$

$2(12) = (4)^2+8=16+8=24$

The solution verifies both equations. Therefore, $(4,12)$ is another solution to this system.

3 0

2 years ago

May someone please help me with this :)

Studentka2010 [4]

Answer:

i was taught this but i dont remember so I dont know if I can help sorry bud

8 0

3 years ago

I dont Understand this so can anyone help me please!! Thank you

storchak [24]

Youre correect answer is 70

3 0

3 years ago

Help me with this question

Llana [10]

The answer is B

(3^4*abc^-7)^6=(3^4*6)(a^6)(b^6)(c^6*-7)=(3^24*a^6*b^6)/(b^42)

6 0

2 years ago

Evaluate the line integral, where C is the given curve. C sin(x) dx + cos(y) dy, where C consists of the top half of the circle

kenny6666 [7]

Parameterize the circular part of $C$ (call it $C_1$ ) by

$x=4\cos t$

$y=4\sin t$

wih $0\le t\le\pi$ , and the linear part (call it $C_2$ ) by

$x=-4-t$

$y=4t$

with $0\le t\le1$ .

Then

$\displaystyle\int_C\sin x\,\mathrm dx+\cos y\,\mathrm dy=\left\{\int_{C_1}+\int_{C_2}\right\}\sin x\,\mathrm dx+\cos y\,\mathrm dy$

$=\displaystyle\int_0^\pi(-4\sin t\sin(4\cos t)+4\cos t\cos(4\sin t))\,\mathrm dt+\int_0^1(-\sin(-4-t)+\cos4t)\,\mathrm dt$

$=0+\displaystyle\int_0^1(\sin(t+4)+\cos4t)\,\mathrm dt$

$=\cos4-\cos5+\dfrac{\sin4}4$

7 0

2 years ago