hello can anyone help with this?

Answer:

To find the surface area of a cuboid we can also label the length, width and the height of the prism and use the formula SA=2LW+2LH+2HW to find the area of a cuboid

Answer:

202 cm²

Step-by-step explanation:

The opposite faces of a cuboid are congruent , then

SA = top/bottom + front/ back + sides , that is

SA = 2(9 × 4) + 2(9 × 5) + 2(4 × 5)

     = 2(36) + 2(45) + 2(20)

     = 72 + 90 + 40

    = 202 cm²

Answer:

.0421

about 4.21%

Step-by-step explanation:

[tex]\frac{{8\choose3}}{{21\choose3}}=\frac{56}{1330}=.042105263[/tex]

Answer:

a)  b^5/12

Step-by-step explanation:

b^2/3 ÷ b^1/4    [if bases are the same then subtract the exponents]

2/3 - 1/4 = 5/12

Answer:

<A = 47.7°

<B = 99.4°

<C = 32.9

Step-by-step explanation:

When given the measurements of all three sides, you can calculate the angles using the Cosine Law.

c² = a² + b² - 2ab cos C

(based on Pythagorean Theorem)

If we say: a = 15

b = 20

c = 11

11² = 15² + 20² - 2(15)(20) cos C

121 = 625 - 2(15)(20) cos C

121 = 625 - 600 cos C

⁻504 = ⁻600 cos C

cos⁻¹ (504 ÷ 600) = C

< C = 32.9°

a² = b² + c² - 2bc cos A

15² = 20² + 11² - 2(20)(11) cos A

225 = 521 - 2(20)(11) cos A

225 = 521 - 440 cos A

⁻296 = ⁻440 cos A

cos⁻¹ (296 ÷ 440) = A

<A = 47.7°

Then, since we know the sum of all three angles of a triangle equals 180°:

180° - 32.9° - 47.7° = 99.4°

<B = 99.4°

Answer:

[tex]b_0 = 16.471[/tex]

Step-by-step explanation:

Given

[tex]\begin{array}{ccccc}x & {15} & {12} & {10} & {7} \ \\ y & {5} & {7} & {9} & {11} \ \end{array}[/tex]

Required

The least square estimate [tex]b_0[/tex]

Calculate the mean of x

[tex]\bar x = \frac{\sum x}{n}[/tex]

[tex]\bar x = \frac{15+12+10+7}{4} =\frac{44}{4} = 11[/tex]

Calculate the mean of y

[tex]\bar y = \frac{\sum y}{n}[/tex]

[tex]\bar y = \frac{5+7+9+11}{4} =\frac{32}{4} = 8[/tex]

Calculate [tex]\sum(x - \bar x) * (y - \bar y)[/tex]

[tex]\sum(x - \bar x) = (15 - 11) * (5 - 8)+ (12 - 11) * (7 - 8) + (10 - 11) * (9 - 8)+ (7 - 11) * (11 - 8)[/tex]

[tex]\sum(x - \bar x) = -26[/tex]

Calculate [tex]\sum(x - \bar x)^2[/tex]

[tex]\sum(x - \bar x)^2 = (15 - 11)^2 + (12 - 11)^2 + (10 - 11)^2 + (7 - 11)^2[/tex]

[tex]\sum(x - \bar x)^2 = 34[/tex]

So:

[tex]b = \frac{\sum(x - \bar x) * (y - \bar y)}{\sum(x - \bar x)^2}[/tex]

[tex]b = \frac{-26}{34}[/tex]

[tex]b_0 = y - bx[/tex]

[tex]b_0 = 5 - \frac{-26}{34}*15[/tex]

[tex]b_0 = 5 + \frac{26*15}{34}[/tex]

[tex]b_0 = 5 + \frac{390}{34}[/tex]

Take LCM

[tex]b_0 = \frac{34*5+ 390}{34}[/tex]

[tex]b_0 = \frac{560}{34}[/tex]

[tex]b_0 = 16.471[/tex]

Answer:

6 : 24

Step-by-step explanation:

If we are in the ratio of 1 to 4, the total is 1+4 = 5

Divide 30 by 5

30/5 = 6

Multiply each term in the ratio by 6

1  :4

1*6 : 4*6

6 : 24

Answer:

total ratio:

[tex] = 1 + 4 \\ = 5[/tex]

For the portion of 1:

[tex] = 30 \div \frac{1}{5} \\ = 30 \times 5 \\ = 150[/tex]

For the portion of 4:

[tex] = 30 \div \frac{4}{5} \\ = 30 \times \frac{5}{4} \\ = 37.5[/tex]

= 30 : 7.5

Answer:

9:2

Step-by-step explanation:

Answer: 54 SQUARE UNITS ITHINK!!

Step-by-step explanation:

Answer:

"D" is the correct answer

Step-by-step explanation:

Divide total miles by speed:

360 / 50 = 7.2 hours

Answer:

the answer is 8

Step-by-step explanation:

Total amount Ted wishes to spend this month is $5000.

Mahogany costs $450 per block, and he has already bought 7 blocks of spruce at $200 each.

So, money spent on spruce =  dollars

Money left after buying spruce =  dollars

Now as $450 will be spent on buying 1 block of mahogany.

So, $3600 will be spent

and then your answer will be 8

Let f(x) be the sum of the geometric series,

[tex]f(x)=\displaystyle\frac1{1-x} = \sum_{n=0}^\infty x^n[/tex]

for |x| < 1. Then taking the derivative gives the desired sum,

[tex]f'(x)=\displaystyle\boxed{\dfrac1{(1-x)^2}} = \sum_{n=0}^\infty nx^{n-1} = \sum_{n=1}^\infty nx^{n-1}[/tex]

Answer:

[tex]\frac{20}{40} \ and \ \frac{4}{8} \ is \ equivalent[/tex]

Step-by-step explanation:

1.

[tex]\frac{2}{3} \ and \ \frac{12}{9} \\\\\frac{2}{3} \ and \ \frac{4}{3}\\\\Not \ equivalent[/tex]

2.

[tex]\frac{20}{40} \ and \ \frac{45}{55}\\\\\frac{1}{2} \ and \ \frac{9}{11}\\\\Not\ equivalent[/tex]

3.

[tex]\frac{20}{40} \ and \ \frac{4}{8}\\\\\frac{1}{2} \ and \ \frac{1}{2} \\\\Equivalent[/tex]

4.

[tex]\frac{5}{5} \ and \ \frac{25}{50} \\\\\frac{1}{1} \ and \ \frac{1}{2} \\\\not \ equivalent[/tex]

Answer:

I don't think it's possible.

Answer:

b=2

Step-by-step explanation:

Answer:

(D.) 3x^2 + 6x

Step-by-step explanation:

the other options aren't complete and some don't even create a parabola.

Answer:

B.

Step-by-step explanation:

[tex] \frac{4 + 4i}{5 + 4i} [/tex]

[tex] \frac{4 + 4i}{5 + 4i} \times \frac{5 - 4i}{5-4i} [/tex]

[tex] \frac{(4 + 4i)(5 - 4i)}{(5 + 4i)(5 - 4i)} [/tex]

[tex] \frac{20-16i+20i-16i^2}{(5) {}^{2} - (4i) {}^{2} } [/tex]

(combining like terms)

[tex] \frac{20+(-16i+20i)-(-16)}{25-(-16)} [/tex]

[tex] \frac{(20+16)+4i}{25+16} [/tex]

[tex] \frac{36+4i}{41} [/tex]

distributing the denominator

[tex] \frac{36}{41} + \frac{4}{41}i [/tex]

That is, option B.

Answer:

Step-by-step explanation:

Total sales = x

Correct choice is C

Answer:

£2120.27

Step-by-step explanation:

A = P (1 + r100)

A = 2000 (1+ 0.03/365)^365(2)

A = 2000 ( 1.00008)^730

A = 2000 (1.060)

A = £2120.27

Answer:

0.15866

Step-by-step explanation:

6-7/1

=-1

p(x>-1)=1-p(x<1)

=0.15866

Answer:

$2,981.67

Step-by-step explanation:

Answer:

Step-by-step explanation:

This is a beginning counting problem, or is it?

If order matters then the answer is

17P3 = 17! / (14!) = 17 * 16 * 15 = 4080

However if it does not matter then

17C3 = 17!/(3! * 14!)

680

Answer:

Between 50 and 60

Step-by-step explanation:

4,346/82 is 53 which is between 50 and 60.

Hope this helps!

9514 1404 393

Answer:

  2 -i

Step-by-step explanation:

The two diagonals have the same center point. Here, it is ...

  (1 +2) +(-1 -2i) = 0 +0i

The fourth point is found by reflecting the 2nd point across the origin:

  p4 = -p2 = -(-2 +i) = 2 -i . . . . . fourth corner of the square