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Zepler [3.9K]
2 years ago
7

Can anyone help me with my homework

Mathematics
1 answer:
enyata [817]2 years ago
8 0
Answer-
1. 3(2) + 5=6+5 = 11
2. 4(3) - 7 =12-7= 5
3. 2(3) - 4 =6-4 = 2
4. 6(4) - 2(3) =24-6 = 18
5. 2+3-4+5 = 5-4+5= 1+5 = 6
6. 7(4) - 2(2) = 28-4= 24
7. 2(3)(4) + 1 = 24+1 = 25
8. 2(3)(4) - 4 = 24-4 = 20
9. 8(2) /4 = 16/4 = 4
10. 6(3)/2 =18/2 = 9
11. 4(2) + 7 = 8+7 = 15
12. 2(4) - 7 = 8-7 = 1
13. 2(3) + 3(2) = 6+6 = 12
14. 6(2) - 1 = 12-1 = 11
15. 3(4) - 2 = 12-2 = 10
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Solve for y:<br> y+5x=6;x= -1,0,1
AlekseyPX
When x = -1

y + 5x = 6

y = 6 - 5x

y = 6 - (5)(-1)

y = 6 - (-5)

y = 11

when x = 0

y = 6 - 5x

y = 6 - (5)(0)

y = 6 - 0

y = 6

when x = 1

y = 6 - 5x

y = 6 - (5)(1)

y = 6 - 5

y = 1
7 0
3 years ago
Read 2 more answers
What's the value of<br><br><img src="https://tex.z-dn.net/?f=%20%5Csqrt%5B111%5D1%7B%7D%20" id="TexFormula1" title=" \sqrt[111]1
timurjin [86]
The answer is 111

Calculate square root of 1 to get 1

Multiply 111 x 1

=111
6 0
2 years ago
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Danny made a mistake in the following problem.
Olenka [21]
The mistake is made on line 3, after calculating 35/7, he should have started with the multiplication 6(2)
7 0
3 years ago
The value of (2i3)3 is<br> IS
snow_lady [41]

Answer:

Assuming you mean (2^3)3, the answer is 24

Step-by-step explanation:

using pemdas you do 2 to the third power first

2x2=4 4x2=8

then you mutiply by the number not in parenthesis; 3

8x3 = 24

4 0
2 years ago
Is square root 1 minus sine squared theta = cos Θ true? If so, in which quadrants does angle Θ terminate?
Cloud [144]

Answer:

True; quadrants I & IV

Step-by-step explanation:

We know the relation between sine and cosine function which is given by

\sin^2 \theta +\cos^2 \theta = 1

Let us solve this equation for cosine function.

\cos^2 \theta = 1-\sin^2 \theta

Take square root both sides. When ever we take square root we need to write the solution in plus minus form

\sqrt{\cos^2 \theta}=\pm\sqrt{1-\sin^2 \theta}

\cos \theta=\pm\sqrt{1-\sin^2 \theta}

\cos \theta=-\sqrt{1-\sin^2 \theta}, \sqrt{1-\sin^2 \theta}

If Θ is in quadrants I and IV then the value will be positive and if Θ is in II and III quadrant then the value is negative.

Hence, if Θ is in quadrants I & IV, then we have

\cos \theta=\sqrt{1-\sin^2 \theta}

Thus, the correct option is: True; quadrants I & IV


6 0
3 years ago
Read 2 more answers
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