Fruit:sugar in 2:3 . If 8kg of sugar is used how much fruit ?

Answers

Answer 1

Answer: F:S = 2:3

F:8 = 2:3

F/8 = 2/3

F = 2/3 * 8

F = 16/3

Therefore F, Fruit = 16/3 kg.

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Need help pleeeeeeeeeeeeeeease

Answers

Answer:

I think it's a.)

Step-by-step explanation:

It says that they already started off with $400 and raise $80 a day, and d stands for the number of days.

It's basically $80 a day being added on to the $400 that was already collected.

So it's 80d+400=7,600

Gavin read 6 books in 3 months. If he reads at a constant rate, how many books did he read each month? Give your answer as a whole number or a FRACTION in simplest form.

Answers

Answer:

2 books in one month

Step-by-step explanation:

Answer:

2 books per month .

Step-by-step explanation :

6 divided by 3 = 2

So 1 book per month = 2 , 2 books per month = 4 , 3 per month = 6 ( counting by 2 )

Hope that helped :)

Evaluate (2 – 5i)(p + q)i when p = 2 and q = 5ia. 29i
b. 29i-20
c.-21i
d.29

Answers

Answer:

The value of the (2 – 5i)(p + q)i is 29i .

Option (a) is correct .

Step-by-step explanation:

As the expression given in the question.

= (2 - 5i) (p + q)i

Simplify the above

= (2 - 5i) (pi + qi)

= 2pi + 2qi - 5pi²- 5qi²

As i² = -1

Put in the above

= 2pi + 2qi + 5p +5q

= 2i × 2 + 2i × 5i + 5 × 2 + 5 × 5i

= 4i + 10i² + 10 + 25i

As i² = -1

= 4i - 10 + 10 + 25i

= 29i

Thus the value of the (2 – 5i)(p + q)i is 29i .

Option (a) is correct .

29i (A.) hope this helps

A candy jar contains 120 pieces of candy, including 45 red candies. If the ratio of red candies to the total number of candies in the jar remains the same, how many red candies would there be in a jar containing 200 pieces of candy?

Answers

Answer:

75 red candies

Step-by-step explanation:

200 candies divided by 120 = 1.666666666666667

Simply times 45 by 1.666666666666667

bisects EOG. EOF = y + 30 and FOG = 3y – 50. Solve for y. 20
70
40
50

Answers

Answer: the correct option is (C) 40.

Step-by-step explanation: As shown in the attached figure below, the line OF bisects the angle EOG, where

$m\angle EOF=y+30,\n\nm\angle FOG=3y-50.$

We are to find the value of y.

Since OF is the bisector of the angle EOG, so it divides the angle EOG into two congruent angles.

Therefore, we get

$m\angle EOF=m\angle FOG\n\n\Rightarrow y+30=3y-50\n\n\Rightarrow 3y-y=30+50\n\n\Rightarrow 2y=80\n\n\Rightarrow y=(80)/(2)\n\n\Rightarrow y=40.$

Thus, the value of y is 40.

Option (C) is CORRECT.

since given it bisects EOF must be equal to FOG

y+30 = 3y-50

2y = 80

y = 40

there are two numbers that are 3 times as far from 19 as they are from 11. the sum of the two numbers is?

Answers

To find one number, solve (x - 19) = 3(x - 11) .

To find the other number, solve (19 - x) = 3(x - 11) .

It turns out that the numbers are 7 and 13 .

Their sum is 20 .

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