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alexgriva [62]
3 years ago
9

Help me please please now asap

Mathematics
1 answer:
Oduvanchick [21]3 years ago
6 0
120 different 3digit numbers
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What is the start time 30 minutes before 10:08am
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9:36 am (in the morning)
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(2t^(3)+3t-4)-(4t^(2)-6t)
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Step-by-step explanation:

3 0
3 years ago
01.05 what is the slope intercept form equation of the line that's passing through (1,3) and (3,7)
Butoxors [25]

Hello!

To find the slope-intercept form equation of the line that passes through the two points is found by first, calculating the slope using the slope formula and substitution, and secondly, finding the y-intercept of the equation through substitution.

Remember, slope-intercept form is written as: y = mx + b, where m is the slope and b is the y-intercept.

1. Find the slope

The slope formula is: \frac{y_{2}-y_{1}}{x_{2}-x_{1}}. To use this formula, you need two points and also, you need to assign these points to x_{1}, x_{2}, y_{1}, and y_{2}.

In this case, x_{1} and x_{1} is assigned to the ordered pair (1,3), while x_{2} and y_{2} is assigned to (3, 7). After assigning these points, you can substitute the pairs into the formula and simplify.

\frac{7-3}{3-1} =\frac{4}{2} = 2

The slope of these two points is 2.

2. Find the y-intercept

To find the y-intercept, you substitute an ordered pair into the slope-intercept equation with the slope that was just calculated. We can use either ordered pair because it will result in the same y-intercept.

y = 2x + b (substitute)

3 = 2(1) + b (simplify)

3 = 2 + b (subtract 2 from both sides)

1 = b

The y-intercept of the two points is (0, 1).

With the necessary values to complete the equation, we can write the final equation.

The slope-intercept form equation of the line that passes through the points (1, 3) and (3, 7) is y = 2x + 1.

7 0
3 years ago
What is the x and y intercept of 2x+5y=-6
kykrilka [37]

the answer is.... (-3,0)

4 0
3 years ago
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Find the product of z1 and z2, where z1 = 7(cos 40° + i sin 40°) and z2 = 6(cos 145° + i sin 145°).
Oduvanchick [21]
For two complex numbers z_1=re^{i\theta}=r(\cos\theta+i\sin\theta) and z_2=se^{i\varphi}=s(\cos\varphi+i\sin\varphi), the product is

z_1z_2=rse^{i(\theta+\varphi)}=rs(\cos(\theta+\varphi)+i\sin(\theta+\varphi))

That is, you multiply the moduli and add the arguments. You have z_1=7e^{i40^\circ} and z_2=6e^{i145^\circ}, so the product is

z_1z_2=7\times6(\cos(40^\circ+145^\circ)+i\sin(40^\circ+145^\circ)=42(\cos185^\circ+i\sin185^\circ)=42e^{i185^\circ}
3 0
3 years ago
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