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uysha [10]
3 years ago
9

A 3-ounce serving of tuna provides 21 grams of protein. Use equivalent ratios to find how many grams of protein are in 9 ounces

of tuna
Mathematics
1 answer:
laila [671]3 years ago
6 0

Answer:

63 grams of protein.

Step-by-step explanation:

3 ounces = 21 grams of protein.

Question is how many grams are in 9 ounces, you then have to find the ratio that you need to use to find out how many grams of protein are in 9 ounces. Therefor you divide the amount that we don't know the equal amount of grams by the amount we do know the equal amount of grams, giving us the ratio.

9 divided by 3 = ratio (3)

Then use the ratio for the grams of protein.

21 x 3 = 63.

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How many solutions can be found for the equation 5x + 3(x − 1) = 10x − 2x − 3?
ahrayia [7]

Answer:

4. Infinitely many

Step-by-step explanation:

5x + 3(x − 1) = 10x − 2x − 3

5x+3x-3=8x-3

5x+3x-8x=3-3

0=0

x∈R

6 0
3 years ago
Write 24:14:6 in its simplest form.​
Iteru [2.4K]

Answer:

12 : 7 : 3

Step-by-step explanation:

24 : 14 : 6

12 : 7 : 3

Always start with smaller number to simplify it. Since 2 is the smaller number and it simplifies 24, 14, and 6 I start with it.

24 ÷ 2 = 12

14 ÷ 2 = 7

6 ÷ 2 = 3

We would stop here since it would not further simplify, therefore our answer is 12 : 7 : 3.

Hope this helps, thank you :) !!

3 0
2 years ago
Identify the self assessment test that each
ladessa [460]

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5 0
3 years ago
Find the nth taylor polynomial for the function, centered at c. f(x) = ln(x), n = 4, c = 5
masha68 [24]

The nth taylor polynomial for the given function is

P₄(x) = ln5 + 1/5 (x-5) - 1/25*2! (x-5)² + 2/125*3! (x-5)³ - 6/625*4! (x - 5)⁴

Given:

f(x) = ln(x)

n = 4

c = 3

nth Taylor polynomial for the function, centered at c

The Taylor series for f(x) = ln x centered at 5 is:

P_{n}(x)=f(c)+\frac{f^{'} (c)}{1!}(x-c)+  \frac{f^{''} (c)}{2!}(x-c)^{2} +\frac{f^{'''} (c)}{3!}(x-c)^{3}+.....+\frac{f^{n} (c)}{n!}(x-c)^{n}

Since, c = 5 so,

P_{4}(x)=f(5)+\frac{f^{'} (5)}{1!}(x-5)+  \frac{f^{''} (5)}{2!}(x-5)^{2} +\frac{f^{'''} (5)}{3!}(x-5)^{3}+.....+\frac{f^{n} (5)}{n!}(x-5)^{n}

Now

f(5) = ln 5

f'(x) = 1/x ⇒ f'(5) = 1/5

f''(x) = -1/x² ⇒ f''(5) = -1/5² = -1/25

f'''(x) = 2/x³  ⇒ f'''(5) = 2/5³ = 2/125

f''''(x) = -6/x⁴ ⇒ f (5) = -6/5⁴ = -6/625

So Taylor polynomial for n = 4 is:

P₄(x) = ln5 + 1/5 (x-5) - 1/25*2! (x-5)² + 2/125*3! (x-5)³ - 6/625*4! (x - 5)⁴

Hence,

The nth taylor polynomial for the given function is

P₄(x) = ln5 + 1/5 (x-5) - 1/25*2! (x-5)² + 2/125*3! (x-5)³ - 6/625*4! (x - 5)⁴

Find out more information about nth taylor polynomial here

brainly.com/question/28196765

#SPJ4

3 0
2 years ago
Which value of x makes x-3/4 + 2/3=17/12 true?
Bingel [31]

Answer:

x = 3/2

Step-by-step explanation:

x- 3/4 + 2/3 = 17/12

x = 17/12 + 3/4 -2/3

x = 3/2

7 0
2 years ago
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