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oksano4ka [1.4K]

4 years ago

8

In how many different ways could Alice, Ben, and Carlos be positioned second, third, and fourth?

1 answer:

saul85 [17]4 years ago

4 0

Six different ways, for small word problems like this, you can just write it out like (A,B,C), (A,C,B), (B,C,A), (B,A,C), (C,A,B), and (C,B,A) where A=Alice, B=Ben, and C=Carlos

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The base of a prism is a regular hexagon with the perimeter of 78mm the height of the prism is 16mm what is the area of one of t

kifflom [539]

The perimeter is the sides of the hexagon. One of the hexagon edges is shared with a rectangular face. Solve for that edge. 78/6=13. The height is the length and edge is the width. Area is length(16) times width(13). 16•13=208. The answer is 208mm^2

3 0

4 years ago

If the container on the left holds 2 liters ,about how many liters does the other container hold

Anarel [89]

The other holds 4 liters

8 0

3 years ago

Diagram 5 shows a triangle ABD such that AD = 17 cm, AB = 10 cm and BD = 21 cm. The straight line AC is perpendicular to the str

OlgaM077 [116]

Answer:

8

Step-by-step explanation:

100 = x^2 + AC^2

17^2 = AC^2 + (21 - x)^2

289 = AC^2 + 21^2 + x^2 - 2*21*x

289 =<u> AC^2</u> + 441 +<u> x^2</u> - 42x

from 1st equation AC^2 + x^2 = 100

289 = 441 + 100 - 42x

289 = 541 - 42x

42x = 541 - 289 = 252

x = 252/42 = 6

so AC^2 = 100 - 6^2 = 100 - 36 = 64

AC = 8

7 0

3 years ago

The line $y = b-x$ with $0 < b < 4$ intersects the $y$-axis at $P$ and the line $x=4$ at $S$. If the ratio of the area of

olga55 [171]

Answer:

b=2.5

Step-by-step explanation:

The given line has equation: $y=b-x$

The x-intercept is (b,0)

The y-intercept is (0,b)

The area of triangle QOP $=\frac{1}{2} |OQ|*|OP|=$

$=\frac{1}{2}b*b$

$=\frac{1}{2}b^2$

The area of triangle QRS $=\frac{1}{2} |RQ|*|RS|=$

$=\frac{1}{2}(4-b)*(b-4)$

The area of triangle QRS and QOP are in the ratio 9:25

We must negate the area under the x-axis.

$\implies \frac{- \frac{1}{2}(4-b)(b-4)}{\frac{1}{2}b^2} =\frac{9}{25}$

$\implies \frac{-(4-b)(b-4)}{b^2} =\frac{9}{25}$

$\implies {-(4-b)(b-4) =\frac{9}{25}b^2$

$\implies b^2-8b+16=\frac{9}{25}b^2$

$\implies 25b^2-200b+400=9b^2$

$\implies 25b^2-9b^2-200b+400=0$

$\implies 16b^2-200b+400=0$

$\implies 2b^2-25b+50=0$

$(2b-5)(b-10)=0$

$b=2.5\:or\:b=10$

But 0<b<4

Therefore b=2.5

VERIFY

The area of triangle QOP

$=\frac{1}{2}*2.5*2.5=3.125$

The area of triangle QRS

$=|\frac{1}{2}(4-2.5)*(2.5-4)|=|-1.125|=1.125$

We must take absolute value because this area is below the x-axis.

$\frac{1.125}{3.125} =\frac{9}{25}$

6 0

3 years ago

Read 2 more answers

Change 5 into a mixed number

IRISSAK [1]

Five over one 5/1 5
_
1

6 0

3 years ago