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2 years ago

A. 4x^5+3x+2 B. 4x^2+3x+2x^5 C. 4x^5+3+2/x^5 D. 18x^5+12x+6

Mathematics

1 answer:

aalyn [17]2 years ago

3 0

The answer is C. 4x^5 + 3 + 2/x^5

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De lenguaje común a algebraico:La altura (h) de un triángulo equilátero es igual a la raíz de la diferencia

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3 years ago

the diameter of a cylinder is 14cm.the height of the cylinder is 9cm. calculate the Volume and curved surface area

kicyunya [14]

Answer:

The volume is 1386 cm³ and curved surface area is 396 cm².

Step-by-step explanation:

Given that,

The diameter of a cylinder is 14 cm

Height of the cylinder is 9 cm

We need to find the volume and curved surface area .

If r is the radius, r = 14/2 = 7 cm

Volume of a cylinder is given by :

$V=\pi r^2 h\\\\V=\dfrac{22}{7}\times (7)^2 \times 9\\\\V=1386\ \text{cm}^3$

The curved surface area of the cylinder is given by :

$A=2\pi r h\\\\A=2\times \dfrac{22}{7}\times 7\times 9\\\\A=396\ \text{cm}^2$

So, the volume is 1386 cm³ and curved surface area is 396 cm².

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2 years ago

What are the endpoint coordinates for the midsegment of △BCD that is parallel to BC?

natulia [17]

Answer:

<em>The endpoint coordinates for the mid-segment are: $(-2,-1)$ and $(1,0)$ </em>

Step-by-step explanation:

According to the given diagram, the coordinates of the vertices of $\triangle BCD$ are: $B(-3,1), C(3,3)$ and $D(-1,-3)$

Now, the endpoints of the mid-segment of $\triangle BCD$ which is parallel to $BC$ will be the mid-points of sides $BD$ and $CD$ .

<u>Formula for the coordinate of mid-point</u> : $(\frac{x_{1}+x_{2}}{2}, \frac{y_{1}+y_{2}}{2})$ , where $(x_{1}, y_{1})$ and $(x_{2}, y_{2})$ are two endpoints.

So, the mid-point of $BD$ will be: $(\frac{-3-1}{2},\frac{1-3}{2})=(\frac{-4}{2},\frac{-2}{2})=(-2,-1)$

and the mid-point of $CD$ will be: $(\frac{3-1}{2},\frac{3-3}{2})=(\frac{2}{2},\frac{0}{2})=(1,0)$

Thus, the endpoint coordinates for the mid-segment of $\triangle BCD$ that is parallel to $BC$ are $(-2,-1)$ and $(1,0)$

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