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Vaselesa [24]
3 years ago
6

How do you write 3 2/4 as a fraction greater than 1

Mathematics
2 answers:
Ymorist [56]3 years ago
6 0
You can write it as 14/4 

Alenkinab [10]3 years ago
6 0
 the correct answer is 14/4
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Please help ASAP 40 pts
natali 33 [55]

Answer:

x=2

Step-by-step explanation:

We know AC = BC since this is an isosceles triangle

6 = 2x+2

Subtract 2 from each side

6-2 =2x+2-2

4 =2x

Divide each side by 2

4/2 =2x/2

2=x

4 0
3 years ago
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What is nine times for a total loss of 15 3/4 yards
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Try feet /cm or inches 
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Find the linear approximation of the function g(x) = 3 root 1 + x at a = 0. g(x). Use it to approximate the numbers 3 root 0.95
Virty [35]

Answer:

L(x)=1+\dfrac{1}{3}x

\sqrt[3]{0.95} \approx 0.9833

\sqrt[3]{1.1} \approx 1.0333

Step-by-step explanation:

Given the function: g(x)=\sqrt[3]{1+x}

We are to determine the linear approximation of the function g(x) at a = 0.

Linear Approximating Polynomial,L(x)=f(a)+f'(a)(x-a)

a=0

g(0)=\sqrt[3]{1+0}=1

g'(x)=\frac{1}{3}(1+x)^{-2/3} \\g'(0)=\frac{1}{3}(1+0)^{-2/3}=\frac{1}{3}

Therefore:

L(x)=1+\frac{1}{3}(x-0)\\\\$The linear approximating polynomial of g(x) is:$\\\\L(x)=1+\dfrac{1}{3}x

(b)\sqrt[3]{0.95}= \sqrt[3]{1-0.05}

When x = - 0.05

L(-0.05)=1+\dfrac{1}{3}(-0.05)=0.9833

\sqrt[3]{0.95} \approx 0.9833

(c)

(b)\sqrt[3]{1.1}= \sqrt[3]{1+0.1}

When x = 0.1

L(1.1)=1+\dfrac{1}{3}(0.1)=1.0333

\sqrt[3]{1.1} \approx 1.0333

7 0
3 years ago
5+<img src="https://tex.z-dn.net/?f=x%5E%7B2%7D" id="TexFormula1" title="x^{2}" alt="x^{2}" align="absmiddle" class="latex-formu
solong [7]
Answer is -44+x^2 . First you need to evaluate the equation and get 5+x^2-49 then u calculate it and your final answer is -44+x^2
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3 years ago
Which number line shows the estimate of square root of 47 in the correct interval of integers
mr_godi [17]

Answer:

B

Step-by-step explanation:

Because yes

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