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ki77a [65]
3 years ago
12

Can someone please help me with my math problem?

Mathematics
1 answer:
elena-s [515]3 years ago
8 0
The answer to Part A is 2 15/32
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Help pls I really need it
Talja [164]

Answer:

Infinite number of solutions

Step-by-step explanation:

when you solve for V, you notice that the term in "v" goes away, and you end up with a true statement:

2 = 2

This is a true statement no matter what values the variable V has, so it is true for all possible (infinite) values of "v".

8 0
3 years ago
Withdrawing $10 every week from an outstanding balance<br> of $400
SSSSS [86.1K]
400-10x. X is how many weeks
4 0
3 years ago
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What is the equation of a line that has a y-intercept of 4 and an x-intercept of -2?​
Furkat [3]

Answer:

y = 2x + 4

Step-by-step explanation:

Two points on this line are (-2, 0) and (0, 4).  Going from the first to the second, x increases by 2 (this is the 'run') and y increases by 4 ('rise').

Thus, the slope of this line is m = rise / run =  4/2 = 2

Using the slope-intercept formula, we get y = mx + b = 2x + b

Let x = -2 and y = 0 to find b:  0 = 2(-2) + b, so b = 4, and the desired equation is then:

y = 2x + 4

6 0
3 years ago
Hector gave of his money and an additional $1 to Calista. He
trapecia [35]
I think it’s $34 but i’m not sure
6 0
3 years ago
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Find the area of this regular polygon.<br> Round to the nearest tenth.<br> 8.65 mm<br> [? ]mm2
nadezda [96]

Answer:

Actually it's not polygon. it's a nonagon. With r=8.65mm″, the law of cosines gives us side a:

a=√{b²+c²−2bc×cos40°}

a=√{149.645−149.645cos40°}

Area Nonagon = (9/4)a²cos40°

=9/4[149.645−149.645cos40°]cot20°

=336.70125[1−cos(40°)]cot(20°)

Applying an identity for the cos(40°) does not get us very far…

= 336.70125[1−(cos2(20°)−1)]cot(20°)

= 336.70125[2−cos2(20°)]cot(20°)

= 336.70125[2−(1−sin2(20°))]cot(20°)

= 336.70125[1+sin2(20°)]cos(20°)sin(20°)

= 336.70125[cot(20°)+sin(20°)cos(20°)]mm²

3 0
3 years ago
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