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2 years ago

Please help me find the surface are of these solids

Mathematics

2 answers:

vovikov84 [41]2 years ago

6 0

Answer:

184 in^3

Step-by-step explanation:

The formula to find the surface area of this solid is :

2 × (lh + lw + wh) (l: length, w: width, h: height)

2 × (2×6 + 2×10 + 6×10) = 184

Alja [10]2 years ago

5 0

the answer is 184.

Step-by-step explanation:

we learned to answer these tipes of questions by doing the heart method.

You might be interested in

If the spinner is spun twice, what is the probability that the numbers that are spun don't match?

WITCHER [35]

Answer:

25%

Step-by-step explanation:

There are 4 spots and to get the same you would land on the same 1

Which would be 1/4=25%

Hope this helps!

6 0

3 years ago

Read 2 more answers

Order from greatest to least:show your work❗️<br><br> 12,-4,8,-3,0

Aneli [31]

The answer would be 12, 8, 0, -3, -4.

The answer is that because 12 is a positive number so that's higher than negative, 8 goes second because its positive and higher than the others, 0 is third because negatives numbers are LOWER than 0, not higher, -3 is next because higher negative numbers are always lower, not greater than smaller numbers because -8 is LOWER than -4 since its on the left of a number line.

6 0

3 years ago

Which of the following sets are subspaces of R3 ?

Ratling [72]

Answer:

The following are the solution to the given points:

Step-by-step explanation:

for point A:

$\to A={(x,y,z)|3x+8y-5z=2} \\\\\to for(x_1, y_1, z_1),(x_2, y_2, z_2) \varepsilon A\\\\ a(x_1, y_1, z_1)+b(x_2, y_2, z_2) = (ax_1+bx_2,ay_1+by_2,az_1+bz_2)$

$=3(aX_l +bX_2) + 8(ay_1 + by_2) — 5(az_1+bz_2)\\\\=a(3X_l+8y_1- 5z_1)+b (3X_2+8y_2—5z_2)\\\\=2(a+b)$

The set A is not part of the subspace $R^3$

for point B:

$\to B={(x,y,z)|-4x-9y+7z=0}\\\\\to for(x_1,y_1,z_1),(x_2, y_2, z_2) \varepsilon B \\\\\to a(x_1, y_1, z_1)+b(x_2, y_2, z_2) = (ax_1+bx_2,ay_1+by_2,az_1+bz_2)$

$=-4(aX_l +bX_2) -9(ay_1 + by_2) +7(az_1+bz_2)\\\\=a(-4X_l-9y_1+7z_1)+b (-4X_2-9y_2+7z_2)\\\\=0$

$\to a(x_1,y_1,z_1)+b(x_2, y_2, z_2) \varepsilon B$

The set B is part of the subspace $R^3$

for point C: $\to C={(x,y,z)|x$

In this, the scalar multiplication can't behold

$\to for (-2,-1,2) \varepsilon C$

$\to -1(-2,-1,2)= (2,1,-1)$ ∉ C

this inequality is not hold

The set C is not a part of the subspace $R^3$

for point D:

$\to D={(-4,y,z)|\ y,\ z \ arbitrary \ numbers)$

The scalar multiplication s is not to hold

$\to for (-4, 1,2)\varepsilon D\\\\\to -1(-4,1,2) = (4,-1,-2)$ ∉ D

this is an inequality, which is not hold

The set D is not part of the subspace $R^3$

For point E:

$\to E= {(x,0,0)}|x \ is \ arbitrary) \\\\\to for (x_1,0 ,0) ,(x_{2},0 ,0) \varepsilon E \\\\\to a(x_1,0,0) +b(x_{2},0,0)= (ax_1+bx_2,0,0)\\$

The $x_1, x_2$ is the arbitrary, in which $ax_1+bx_2$ is arbitrary

$\to a(x_1,0,0)+b(x_2,0,0) \varepsilon E$

The set E is the part of the subspace $R^3$

For point F:

$\to F= {(-2x,-3x,-8x)}|x \ is \ arbitrary) \\\\\to for (-2x_1,-3x_1,-8x_1),(-2x_2,-3x_2,-8x_2)\varepsilon F \\\\\to a(-2x_1,-3x_1,-8x_1) +b(-2x_1,-3x_1,-8x_1)= (-2(ax_1+bx_2),-3(ax_1+bx_2),-8(ax_1+bx_2))$

The $x_1, x_2$ arbitrary so, they have $ax_1+bx_2$ as the arbitrary $\to a(-2x_1,-3x_1,-8x_1)+b(-2x_2,-3x_2,-8x_2) \varepsilon F$

The set F is the subspace of $R^3$

5 0

2 years ago

Which option best describes the polynomial?

liq [111]

Degree of ploynominal is the largest exponent you find, which is here the 4x to the 6th. there are only two terms here.

4 0

2 years ago

Disks of polycarbonate plastic from a supplier are analyzed for scratch and shock resistance. The results from 100 disks are sum

IrinaK [193]

Answer:

0.84

Step-by-step explanation:

Given that Disks of polycarbonate plastic from a supplier are analyzed for scratch and shock resistance. The results from 100 disks are summarized as follows:

P(A) = 0.86, P(B) = 0.79, P(A') = 0.14, P(AUB) = 0.95

We are to find out P(A'UB)

We have

$P(AUB) =P(A)+P(B)-P(A\bigcap B)\\0.95=0.86+0.79-P(A\bigcap B)\\P(A\bigcap B)=0.70$

$P(A'UB) = P(A')+P(B)-P(A' \bigcap B)\\= 1-P(A) +P(B)-[P(B)-P(A \bigcap B)]\\= 1-0.86+0.79-P(B)+[tex]P(A'UB)=0.14+0.79-0.79+0.70\\=0.84$ P(A \bigcap B)[/tex]

7 0

3 years ago