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Rudiy27
3 years ago
10

Subtract the integer 14-17=​

Mathematics
2 answers:
lora16 [44]3 years ago
3 0

Answer:

14-17= -3

Step-by-step explanation:

shepuryov [24]3 years ago
3 0

Answer: 14-17=3

Step-by-step explanation: Use your brain ;-;

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7<4q-9 ALGEBRA:) Need help Please! Gracias!!
Vanyuwa [196]

Answer:

q > 4

Step-by-step explanation:

Step 1: Write inequality

7 < 4q - 9

Step 2: Solve for <em>q</em>

<u>Add 9 to both sides:</u> 16 < 4q

<u>Divide both sides by 4:</u> 4 < q

Here, we can see that any number greater than 4 will work. So even 23573489079357 would work, as long as it is greater than 4.

6 0
3 years ago
Express 36 mins as a fraction of an hour​
Oxana [17]

Answer:

36/60

Step-by-step explanation:

There are 60 minutes in an hour

7 0
3 years ago
If i triple my number and add 5,i get 17 .what is my number
Sergeu [11.5K]
4.
4x3=12
12+5=17
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3 years ago
1. 648,400 seconds into days
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Step-by-step explanation:

6 0
3 years ago
Read 2 more answers
1. Type an equation in the equation editor that uses 2 fractions with parentheses around one of them. Example: <img src="https:/
avanturin [10]

Answer:

i) \frac{3}{5} + (- \frac{1}{2}) = \frac{6}{10} - \frac{5}{10} = \frac{1}{10}    \Rightarrow \frac{3}{5} + (- \frac{1}{2}) = \frac{6}{10} - \frac{5}{10} = \frac{1}{10}

ii)4^{3} + 8^{2} + \sqrt{9}   \Rightarrow  4^{3} + 8^{2} + \sqrt{9}

iii) (\frac{4}{5})^{2}. \sqrt[3]{8} \leqx^{3} - 3x + 6 \leq \sqrt{\frac{1}{3}} \Rightarrow \hspace{0.2cm}    (\frac{4}{5})^{2}. \sqrt[3]{8} \leqx^{3} - 3x + 6 \leq \sqrt{\frac{1}{3}}

Step-by-step explanation:

i) \frac{3}{5} + (- \frac{1}{2}) = \frac{6}{10} - \frac{5}{10} = \frac{1}{10}    \Rightarrow \frac{3}{5} + (- \frac{1}{2}) = \frac{6}{10} - \frac{5}{10} = \frac{1}{10}

ii)4^{3} + 8^{2} + \sqrt{9}   \Rightarrow  4^{3} + 8^{2} + \sqrt{9}

iii) (\frac{4}{5})^{2}. \sqrt[3]{8} \leqx^{3} - 3x + 6 \leq \sqrt{\frac{1}{3}} \Rightarrow \hspace{0.2cm}    (\frac{4}{5})^{2}. \sqrt[3]{8} \leqx^{3} - 3x + 6 \leq \sqrt{\frac{1}{3}}

7 0
4 years ago
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