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jasenka [17]
3 years ago
15

Treats are being made by the Baker's Club for a local carnival. A recipe calls for 5 parts chocolate chips and 2 parts peanut bu

tter chips.If Rhonda has 14 cups of peanut butter chips, how many cups of chocolate chips will she need?
O A 5.6
B. 17
C. 24
D. 35
Mathematics
1 answer:
olga55 [171]3 years ago
7 0

Answer:

D 35

Step-by-step explanation:

2/5 = 14/x

solve for x by cross multiplying

5x=70

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https://learnzillion.com/lesson_plans/7928-use-models-for-division-of-fractions-by-fractions/


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Use the digits 2, 3, and 5 to create a fraction and a whole number with a product greater than 2
kupik [55]
The fractions which cannbe formed are: 2/3, 3/2, 2/5, 5/2, 3/5 and 5/3.
Now, we find the products with the third number:
2/3*5= 10/3 = 3.33 > 2
3/2*5 = 15 >2= 7.5 > 2
2/5*3= 6/5 (rejected)
5/2*3 = 15/2 > 2
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2x + 8y = 28<br> 2x + 8y = 29<br> e?
egoroff_w [7]

Answer:

The question is unclear. what is e?

6 0
3 years ago
When adding 35.64 to a certain number, the sum is 38.24, as seen below. What number should go in the box to complete the additio
Lunna [17]

Answer:

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6 0
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Butyric acid ( c3h7cooh) is a weak acid . what is the ph of a 0.0250 m solution of this acid. ka = 1.5 x 10-5 (2 decimal places)
Wewaii [24]

The pH of the weak acid is 3.21

Butyric acid is known as a weak acid, we need the concentration of [H+] formula of weak acid which is given by this equation :

[H^{+}]=\sqrt{Ka . Ma}

where [H+] is the concentration of ion H+, Ka is the weak acid ionization constant, and Ma is the acid concentration.

Since we know the concentration of H+, the pH can be calculated by using

pH = -log[H+]

From question above, we know that :

Ma = 0.0250M

Ka = 1.5 x 10¯⁵

By using the equation, we can determine the concentration of [H+]

[H+] = √(Ka . Ma)

[H+] = √(1.5 x 10¯⁵ . 0.0250)

[H+] = 6.12 x 10¯⁴ M

Substituting the value of [H+] to get the pH

pH = -log[H+]

pH = -log(6.12 x 10¯⁴)

pH = 3.21

Hence, the pH of the weak acid c3h7cooh is 3.21

Find more on pH at: brainly.com/question/14466719

#SPJ4

8 0
1 year ago
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