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3 years ago

9

Find the surface area of the figure below

1 answer:

Roman55 [17]3 years ago

8 0

Answer:

answer: 208

Hope that helps!

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Please help!! please please

Leona [35]

I honestly don'y know i'm really sorry.

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3 years ago

Quiz: Areas and Volumes of Similar Solids

Otrada [13]

Answer:

I think 320

Step-by-step explanation:

sorry if wrong but I think it 320

6 0

2 years ago

Hoochie [10]

Answer:

AT = 8 $\sqrt{2}$

Step-by-step explanation:

Using the sine ratio in the right triangle

and sin45° = $\frac{1}{\sqrt{2} }$

sin45° = $\frac{opposite}{hypotenuse}$ = $\frac{AB}{AT}$ = $\frac{8}{AT}$

Multiply both sides by AT

AT × sin45° = 8, that is

AT × $\frac{1}{\sqrt{2} }$ = 8

Multiply both sides by $\sqrt{2}$

AT = 8 $\sqrt{2}$

3 0

3 years ago

How does the circulatory system work at the tissue level?<br> Please help!

soldi70 [24.7K]

Answer:

Blood cells carry nutrients

Step-by-step explanation:

blood cells carry oxygen and other nutrients with them and that's how things like oxygen are supplied to tissues

8 0

3 years ago

please solve it please solve it please solve it please solve it please solve it please solve it please solve it please solve it

katovenus [111]

Since $\alpha$ and $\beta$ are roots of $ax^2+bx+c$ , we can factorize the quadratic in terms of the roots as

$ax^2+bx+c = a(x-\alpha)(x-\beta)$

Expanding the right side gives

$ax^2+bx+c = ax^2-a(\alpha+\beta)x+a\alpha\beta$

so that

$\alpha + \beta = -\dfrac ba \\\\ \alpha\beta = \dfrac ca$

A polynomial with roots $\alpha^2$ and $\beta^2$ would be

$(x-\alpha^2)(x-\beta^2)$

and expanding this gives

$x^2 - (\alpha^2+\beta^2)x+\alpha^2\beta^2$

Now,

$\alpha\beta = \dfrac ca \implies \alpha^2\beta^2 = (\alpha\beta)^2 = \dfrac{c^2}{a^2}$

and

$(\alpha+\beta)^2 = \alpha^2 + 2\alpha\beta + \beta^2 \\\\ \implies \alpha^2+\beta^2 = \left(-\dfrac ba\right)^2 -2\left(\dfrac ca\right) = \dfrac{b^2-2ac}{a^2}$

So we can write the second quadratic in terms of $a,b,c$ as

$(x-\alpha^2)(x-\beta^2) = x^2 - \dfrac{b^2-2ac}{a^2}x + \dfrac{c^2}{a^2}$

and to make things look cleaner, scale the whole expression by $a^2$ to get

$\boxed{a^2x^2 + (2ac-b^2)x + c^2}$

4 0

3 years ago