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2 years ago

6

OMG HELP PLS IM PANICKING OMG OMG I GOT A F IN MATH AND I ONLY HAVE 1 DAY TO CHANGE MY GRADE BECAUSE TOMORROW IS THE FINAL REPOR

T CARD RESULTS AND I DONT WANNA FAIL PLS HELP-

1 answer:

patriot [66]2 years ago

6 0

Answer:

3 tens

2 hundreds

5 thousands

7 thousands

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The no. Of adel beads to that of claire's beads was the ratio 7 :4 .When adel gave 68 beads to claire the ratio become 25 :41 .

Novosadov [1.4K]

Answer:

37 beads

Step-by-step explanation:

Given that:

Adel : Claire = 7 : 4

Let Adel be = A; &

Let Claire be = C

A: C = 7 : 4

$\dfrac{A}{C} =\dfrac{7}{4}$

7C = 4A

$C = \dfrac{4A}{7}$ --- (1)

When Adel gave 68 beads to claire, we have:

C + 68 = 25:41

$\dfrac{A}{C+68}= \dfrac{25}{41}$

25(C + 68) = 41A

C + 68 = $\dfrac{41}{25}A$

$C = \dfrac{41}{25}A - 68$ --- (2)

Equating (1) and (2) together;

$\dfrac{4 * A}{7} = \dfrac{41 *A}{25} - 68$

$\dfrac{4 * A}{7}-\dfrac{41 *A}{25} = - 68$

$\dfrac{100A - 287 A}{175} =- 68$

$\dfrac{-187A}{175} = -68$

- 187 A = 175 × (-68)

-187 A = - 11900

A = - 11900/ -187

A ≅ 64

From; $C = \dfrac{4A}{7}$

$C = \dfrac{4*64}{7}$

C = 36.57

C ≅ 37 beads

4 0

3 years ago

Leezas aquarium holds 55 gallons of water. She is filling the tank and has already put in 22 gallons. Write and solve an inequal

svp [43]

Leeza can put 33 or less gallons of water in aquarium.

Step-by-step explanation:

Given,

Capacity of aquarium = 55 gallons of water

Water already in aquarium = 22 gallons

Let,

x be the number of gallons that can be added in aquarium.

As aquarium can hold 55 gallons of water, therefore, we cannot add more than 55 gallons of water.

$x+22\leq 55$

Subtracting 22 from both sides

$x+22-22\leq 55-22\\x\leq 33$

Leeza can put 33 or less gallons of water in aquarium.

Keywords: inequality, subtraction

Learn more about inequalities at:

#LearnwithBrainly

6 0

3 years ago

What is the quadrilateral

Dahasolnce [82]

Answer:

a quadrilateral is a four sided shape that has four straight sides

8 0

2 years ago

Find the two smallest possible solutions to part 1a

bixtya [17]

<h3>Answer: A. 5/12, 25/12</h3>

============================

Work Shown:

12*sin(2pi/5*x)+10 = 16

12*sin(2pi/5*x) = 16-10

12*sin(2pi/5*x) = 6

sin(2pi/5*x) = 6/12

sin(2pi/5*x) = 0.5

2pi/5*x = arcsin(0.5)

2pi/5*x = pi/6+2pi*n or 2pi/5*x = 5pi/6+2pi*n

2/5*x = 1/6+2*n or 2/5*x = 5/6+2*n

x = (5/2)*(1/6+2*n) or x = (5/2)*(5/6+2*n)

x = 5/12+5n or x = 25/12+5n

these equations form the set of all solutions. The n is any integer.

--------

The two smallest positive solutions occur when n = 0, so,

x = 5/12+5n or x = 25/12+5n

x = 5/12+5*0 or x = 25/12+5*0

x = 5/12 or x = 25/12

--------

Plugging either x value into the expression 12*sin(2pi/5*x)+10 should yield 16, which would confirm the two answers.

7 0

2 years ago

Read 2 more answers

Find the unknown value of 5/25 = 13/

Snowcat [4.5K]

65, since 25/5=5.
13*5=65

7 0

2 years ago

Read 2 more answers