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4 years ago

I need help on my ixl

Mathematics

1 answer:

mojhsa [17]4 years ago

7 0

Area = W * L55 = 5 * L55/5 = L11 = L
Note "L" is the length.

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THERE IS A SS. Find the surface area of the cone. Use 3.14 for π.

kobusy [5.1K]

Area of a cone is v=1/3 pi r2^h
V=1/3 (3.14)(10)2^(30)
V=1/3 (3.14)(100)(30)
V=9420/3
V=3140

4 0

3 years ago

On a coordinate plane, the endpoints of line segment JK are J(−10,12) and K(8,−12). Point L lies on line segment JK and divides

lakkis [162]

Answer:

The coordinates of point L are (4.4 , -7.2) ⇒ answer A

Step-by-step explanation:

Assume that point L is (x , y)

Point L divides the line segment JK into two line segments such that

the ratio of JK to KL is 5 : 1

The coordinates of point J are (-10 , 12)

The coordinates of point K are (8 , -12)

∵ $x=\frac{x_{1}m_{2}+x_{2}m_{1}}{m_{1}+m_{2}}$

∵ $y=\frac{y_{1}m_{2}+y_{2}m_{1}}{m_{1}+m_{2}}$

Let $(x_{1},y_{1})$ = (-10 , 12) and $(x_{2},y_{2})$ = (8 , -12)

and $m_{1}:m_{2}$ = JL : KL

∵ JK : KL = 5 : 1

∵ JK = JL + KL

∴ 5 = JL + 1

Subtract 1 from both sides

∴ JL = 4

∴ JL : LK = 4 : 1

∴ $m_{1}:m_{2}$ = 4 : 1

∵ $x=\frac{(-10)(1)+(8)(4)}{4+1}$

∴ $x=\frac{-10+32}{5}$

∴ $x=\frac{22}{5}=4.4$

∴ The x-coordinate of point L is 4.4

∵ $y=\frac{(12)(1)+(-12)(4)}{4+1}$

∴ $y=\frac{12+(-48)}{5}$

∴ $y=\frac{-36}{5}=-7.2$

∴ The y-coordinate of point L is -7.2

* The coordinates of point L are (4.4 , -7.2)

6 0

3 years ago

What is the expanded form to write the prime factorization of 630

Monica [59]

I hope this helps you

3 0

3 years ago

What factor is missing? 24 + 12= ____(8+4)

Svetach [21]

3
Aoggianaiqos Ajaxzjka

3 0

3 years ago

Read 2 more answers

An engineer designed a valve that will regulate water pressure on an automobile engine. The engineer designed the valve such tha

amid [387]

Answer:

Given:

mean, u = 6.2

sample size, n = 180

Sample mean, X' = 6.3

s.d $\sigma$ = 0.9

Significance level = 0.05

The null and alternative hypothesis will be:

H0 : u = 6.2

H1 : u > 6.2

Degree of freedom = 180 - 1 = 179

Using t table, the t critical value,

t> t(0.05, 179) = 1.6534

The test statistic:

$t = \frac{X' - u}{\frac{\sigma}{\sqrt{n}}}$

$T = \frac{6.3 - 6.2}{\frac{0.9}{\sqrt{180}}} = 1.4907$

Since the test statistic(t calculated value) 1.4907 < t critical value (1.6534), we fail to reject the null hypothesis H0.

4 0

3 years ago