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ICE Princess25 [194]
3 years ago
7

each marble bag sold by Lisas marbles company contains three red marbles for every four orange marbles if a bag has 24 red marbl

es how many orange marbles does it contain
Mathematics
1 answer:
garik1379 [7]3 years ago
7 0

Answer:

32 orange marbles.

Step-by-step explanation:

First set up the proportion given in the problem, since it states that there are three red marbles for every four orange marbles. The ratio between them is 3:4.

This means that you can multiply the given amount of red marbles by 4/3 to find the amount of orange marbles and the given amount of orange marbles by 3/4 to find the amount of red marbles. Because there are 3 red marbles for every 4 orange marbles.

You can divide 32 by 3 to find the factor which is 8 and multiply by 4 to get 32. Because 8 × 3 = 24,

and 8 × 4 = 32.

It can also be thought of as fractions because the proportion is a factor which means that you can multiply and divide to get to a certain value.

So:

3/4 = 24/O

Red marbles / orange marbles = red marbles / orange marbles.

(cross multiply ; numerator times denominator, and denominator times numerator; numerator is the top number, and the denominator is bottom number of a fraction).

3O = 96

÷3 ÷3

O [orange marbles] = 32

You might be interested in
Dad is twice as old as Jon. Grand-Pa is seven years older than three times Jon's age. The sum of their age is 145 years. How old
Alenkasestr [34]

Step-by-step explanation:

let jon age be x

dad age is 2 of x i.e 2x

grandpa age is 7 + 3(x)

The sum of jon dad and grandpa age is 145 i.e

x + 2x + (7+3x) = 145

3x +. 7 + 3x = 145

6x = 145- 7

6x = 138

divide both sides by 6

x = 23

jon age is 23

while

grandpa age is

7 + 3(x)

7 + 3(23)

7 + 69

76.

3 0
3 years ago
Last one for finding lengths side, please answer correctly, thank you
Georgia [21]

∑ Hey,  shanellmccullough ⊃

Answer:

(a) Width of the garden : 64 m

(b) Length of the window: 95 cm

Step-by-step explanation:

<u><em>Given:</em></u>

(a) The perimeter of a rectangular garden is 306 m. If the length of the garden is 89 m, what is its width?

(b) The area of a rectangular window is 5130 cm². If the width of the window is 54 cm, what is its length?

<u><em>Solve:</em></u>

(a) The perimeter of a rectangular garden is 306 m. If the length of the garden is 89 m, what is its width?

Formula for perimeter: P = (L + W) × 2

Which;

P = Perimeter

L = Length

W = Width

Which also could be written as:

L + L + W + W = P

Hence, since it given that;

306 = P and L = 89

Then

We know that ;

89L + 89L + W + W = 306

So we can add 89 + 89 or 89 × 2

which is 178

Now we know that;

178 + 2w = 306

Solving for 2w:

178 + 2w = 306

178-178 + 2w = 306 - 178

2w = 128

W = 64

Hence, width is 64.

Check Answer:

64 × 2 + 89 × 2 =  306

-------------------------------------------------------------------------------------------------------------

(b) The area of a rectangular window is 5130 cm². If the width of the window is 54 cm, what is its length?

Formula for area: A = Lw

Which;

A = Area

L = Length

W = Width

So, since we know the width we can just do 5130/54 to find Length

5130/54 = 95

Hence, the length of the window is 95 cm.

Check Answer :

A = Lw

A = 95 × 54

A = 5130 cm²

-------------------------------------------------------------------------------------------------------------

<u><em>xcookiex12</em></u>

<u><em></em></u>

<em>8/18/2022</em>

7 0
1 year ago
PLEASE HELP, GOOD ANSWERS GET BRAINLIEST. +40 POINTS WRONG ANSWERS GET REPORTED
MA_775_DIABLO [31]
1. Ans:(A) 123

Given function: f(x) = 8x^2 + 11x
The derivative would be:
\frac{d}{dx} f(x) = \frac{d}{dx}(8x^2 + 11x)
=> \frac{d}{dx} f(x) = \frac{d}{dx}(8x^2) + \frac{d}{dx}(11x)
=> \frac{d}{dx} f(x) = 2*8(x^{2-1}) + 11
=> \frac{d}{dx} f(x) = 16x + 11

Now at x = 7:
\frac{d}{dx} f(7) = 16(7) + 11

=> \frac{d}{dx} f(7) = 123

2. Ans:(B) 3

Given function: f(x) =3x + 8
The derivative would be:
\frac{d}{dx} f(x) = \frac{d}{dx}(3x + 8)
=> \frac{d}{dx} f(x) = \frac{d}{dx}(3x) + \frac{d}{dx}(8)
=> \frac{d}{dx} f(x) = 3*1 + 0
=> \frac{d}{dx} f(x) = 3

Now at x = 4:
\frac{d}{dx} f(4) = 3 (as constant)

=>Ans:  \frac{d}{dx} f(4) = 3

3. Ans:(D) -5

Given function: f(x) = \frac{5}{x}
The derivative would be:
\frac{d}{dx} f(x) = \frac{d}{dx}(\frac{5}{x})
or 
\frac{d}{dx} f(x) = \frac{d}{dx}(5x^{-1})
=> \frac{d}{dx} f(x) = 5*(-1)*(x^{-1-1})
=> \frac{d}{dx} f(x) = -5x^{-2}

Now at x = -1:
\frac{d}{dx} f(-1) = -5(-1)^{-2}

=> \frac{d}{dx} f(-1) = -5 *\frac{1}{(-1)^{2}}
=> Ans: \frac{d}{dx} f(-1) = -5

4. Ans:(C) 7 divided by 9

Given function: f(x) = \frac{-7}{x}
The derivative would be:
\frac{d}{dx} f(x) = \frac{d}{dx}(\frac{-7}{x})
or 
\frac{d}{dx} f(x) = \frac{d}{dx}(-7x^{-1})
=> \frac{d}{dx} f(x) = -7*(-1)*(x^{-1-1})
=> \frac{d}{dx} f(x) = 7x^{-2}

Now at x = -3:
\frac{d}{dx} f(-3) = 7(-3)^{-2}

=> \frac{d}{dx} f(-3) = 7 *\frac{1}{(-3)^{2}}
=> Ans: \frac{d}{dx} f(-3) = \frac{7}{9}

5. Ans:(C) -8

Given function: 
f(x) = x^2 - 8

Now if we apply limit:
\lim_{x \to 0} f(x) = \lim_{x \to 0} (x^2 - 8)

=> \lim_{x \to 0} f(x) = (0)^2 - 8
=> Ans: \lim_{x \to 0} f(x) = - 8

6. Ans:(C) 9

Given function: 
f(x) = x^2 + 3x - 1

Now if we apply limit:
\lim_{x \to 2} f(x) = \lim_{x \to 2} (x^2 + 3x - 1)

=> \lim_{x \to 2} f(x) = (2)^2 + 3(2) - 1
=> Ans: \lim_{x \to 2} f(x) = 4 + 6 - 1 = 9

7. Ans:(D) doesn't exist.

Given function: f(x) = -6 + \frac{x}{x^4}
In this case, even if we try to simplify it algebraically, there would ALWAYS be x power something (positive) in the denominator. And when we apply the limit approaches to 0, it would always be either + infinity or -infinity. Hence, Limit doesn't exist.

Check:
f(x) = -6 + \frac{x}{x^4} \\ f(x) = -6 + \frac{1}{x^3} \\ f(x) = \frac{-6x^3 + 1}{x^3} \\ Rationalize: \\ f(x) = \frac{-6x^3 + 1}{x^3} * \frac{x^{-3}}{x^{-3}} \\ f(x) = \frac{-6x^{3-3} + x^{-3}}{x^0} \\ f(x) = -6 + \frac{1}{x^3} \\ Same

If you apply the limit, answer would be infinity.

8. Ans:(A) Doesn't Exist.

Given function: f(x) = 9 + \frac{x}{x^3}
Same as Question 7
If we try to simplify it algebraically, there would ALWAYS be x power something (positive) in the denominator. And when we apply the limit approaches to 0, it would always be either + infinity or -infinity. Hence, Limit doesn't exist.

9, 10.
Please attach the graphs. I shall amend the answer. :)

11. Ans:(A) Doesn't exist.

First We need to find out: \lim_{x \to 9} f(x) where,
f(x) = \left \{ {{x+9, ~~~~~x \textless 9} \atop {9- x,~~~~~x \geq 9}} \right.

If both sides are equal on applying limit then limit does exist.

Let check:
If x \textless 9: answer would be 9+9 = 18
If x \geq 9: answer would be 9-9 = 0

Since both are not equal, as 18 \neq 0, hence limit doesn't exist.


12. Ans:(B) Limit doesn't exist.

Find out: \lim_{x \to 1} f(x) where,

f(x) = \left \{ {{1-x, ~~~~~x \textless 1} \atop {x+7,~~~~~x \textgreater 1} } \right. \\ and \\ f(x) = 8, ~~~~~ x=1

If all of above three are equal upon applying limit, then limit exists.

When x < 1 -> 1-1 = 0
When x = 1 -> 8
When x > 1 -> 7 + 1 = 8

ALL of the THREE must be equal. As they are not equal. 0 \neq 8; hence, limit doesn't exist.

13. Ans:(D) -∞; x = 9

f(x) = 1/(x-9).

Table:

x                      f(x)=1/(x-9)       

----------------------------------------

8.9                       -10

8.99                     -100

8.999                   -1000

8.9999                 -10000

9.0                        -∞


Below the graph is attached! As you can see in the graph that at x=9, the curve approaches but NEVER exactly touches the x=9 line. Also the curve is in downward direction when you approach from the left. Hence, -∞,  x =9 (correct)

 14. Ans: -6

s(t) = -2 - 6t

Inst. velocity = \frac{ds(t)}{dt}

Therefore,

\frac{ds(t)}{dt} = \frac{ds(t)}{dt}(-2-6t) \\ \frac{ds(t)}{dt} = 0 - 6 = -6

At t=2,

Inst. velocity = -6


15. Ans: +∞,  x =7 

f(x) = 1/(x-7)^2.

Table:

x              f(x)= 1/(x-7)^2     

--------------------------

6.9             +100

6.99           +10000

6.999         +1000000

6.9999       +100000000

7.0              +∞

Below the graph is attached! As you can see in the graph that at x=7, the curve approaches but NEVER exactly touches the x=7 line. The curve is in upward direction if approached from left or right. Hence, +∞,  x =7 (correct)

-i

7 0
3 years ago
Read 2 more answers
3. A town had a population of 25,000 in the year 2000. Each year, the population is estimated to increase at a rate of 5%.
leonid [27]
A. p=25,000*(1+0.05)^t
b. p=25,000(1+0.05)^20=66,332
7 0
3 years ago
I need help with #65 to #70 ASAP please and thank you …. Could someone please help me with it… I need to get it done ASAP… it is
Alinara [238K]

Answer:

i cant see the questions,if you bring it a little closer please

8 0
3 years ago
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