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3 years ago

Please help 60 points!!

Mathematics

2 answers:

Katen [24]3 years ago

4 0

Answer:

The answer to your question is FJ = 18

Step-by-step explanation:

Data

FG = 30

FH = 12

HE = 8

JH ║GE

These triangles are similar, so we can use proportions to find the length of FJ.

$\frac{FJ}{FH} = \frac{FG}{FH + HE}$

Substitution

$\frac{FJ}{12} = \frac{30}{12 + 8}$

Simplification

$\frac{FJ}{12} = \frac{30}{20}$

Solve for FJ

$FJ = 12 \frac{30}{20}$

Result

FJ = 18

cricket20 [7]3 years ago

4 0

Answer:

baby girl idk but good luckn on eerrrthing in life

Step-by-step explanation:

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Help pretty pretty please :(

Evgen [1.6K]

Answer:

27/20

Step-by-step explanation:

The absolute value of any expression is always positive or zero

3/5 + 3/4

Add the fractions

27/20

Solution

27/20

Alternate form

1 7/20, 1.35

8 0

3 years ago

Read 2 more answers

2. Dilute hydrochloric acid reacts with zinc to form zinc chloride and hydrogen.

DIA [1.3K]

Step-by-step explanation:

Zn + 2HCl => ZnCl2 + H2

Moles of Zn = 13g / (65.38g/mol) = 0.198mol

Volume of HCl = 0.396mol / (0.25mol/dm³) = 1.58dm³.

Volume of H2 = 0.198mol * (22.4dm³/mol) = 4.43dm³.

6 0

3 years ago

A box contain 12 balls in which 4 are white 3 are blue and 5 are red.3 balls are drawn at random from the box.find the chance th

Anastasy [175]

Answer:

3/11

Step-by-step explanation:

From the above question, we have the following information

Total number of balls = 12

Number of white balls = 4

Number of blue balls = 3

Number of red balls = 5

We solve this question using combination formula

C(n, r) = nCr = n!/r!(n - r)!

We are told that 3 balls are drawn out at random.

The chance/probability of drawing out 3 balls = 12C3 = 12!/3! × (12 - 3)! = 12!/3! × 9!

= 12 × 11 × 10 × 9 × 8 × 7 × 6 × 5 × 4 × 3 × 2 × 1/(3 × 2 × 1) × (9 × 8 × 7 × 6 × 5 × 4 × 3 × 2 × 1)

= 220 ways

The chance of selecting 3 balls at random = 220

To find the chance that all the three balls are selected,

= [Chance of selecting (white ball) × Chance of selecting(blue ball) × Chance of selecting(red balls)]/ The chance/probability of drawing out 3 balls

Chance of selecting (white ball)= 4C1

Chance of selecting(blue ball) = 3C1 Chance of selecting(red balls) = 5C1

Hence,

= [4C1 × 3C1 × 5C1]/ 220

= 60/220

= 6/22

= 3/11

The chance that all three are selected is = 3/11

4 0

3 years ago

If m 2 BOC = 27 and m ZAOC = 61, what is the measure of ZAOB?

ivanzaharov [21]

The correct answer is D.

3 0

3 years ago

A baker made 20 pies. A Boy Scout troop buys one–fourth of his pies, a preschool teacher buys one–third of his pies, and a cater

n200080 [17]

Answer:

The number of pies does the baker have left is 15.

Step-by-step explanation:

Given : A baker made 20 pies. A Boy Scout troop buys one–fourth of his pies, a preschool teacher buys one–third of his pies, and a caterer buys one–sixth of his pies.

To find : How many pies does the baker have left?

Solution :

Let x be the number of pies does the baker have left.

According to question,

Number of pies buys,

One-fourth = $\frac{1}{4}$

One-third = $\frac{1}{3}$

One-sixth = $\frac{1}{6}$

i.e. $x=20-(20\times \frac{1}{4}+20\times \frac{1}{3}+20\times \frac{1}{6})$

$x=20-(5+20\times \frac{1}{3}+10\times \frac{1}{3})$

$x=20-(5+\frac{20+10}{3})$

$x=20-(5+ \frac{30}{3})$

$x=20-(5+10)$

$x=20-(15)$

$x=5$

The number of pies does the baker have left is 15.

7 0

3 years ago