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3 years ago

Write the equation of a line that is perpendicular to y=−1 and that passes through the point (8,-4)

Mathematics

1 answer:

Vera_Pavlovna [14]3 years ago

5 0

Answer:

perp. slope: 1

y + 4 = x - 8

y = x - 12

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Write a mixed number

djyliett [7]

Answer:

2¾ , 3¾

Step-by-step explanation:

Anything with a combination of fraction and whole number is a mixed number. so possibilities are pretty much endless

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3 years ago

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When you purchase an item with a credit card, you might not ever have to pay for it. True False

Sergio039 [100]

False. You get billed by the end of the month the money is “borrowed” money from the bank until you have to pay it all back, if it’s not paid on time you have to pay extra (interest). With debit card your money is taken out immediately.

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2 years ago

Let G be a connnectde graph with n vertices and m edges. supposed also that m = n. prove that G contains exactly one cycle

ElenaW [278]

Answer:

Contradiction

Step-by-step explanation:

Suppose that G has more than one cycle and let C be one of the cycles of G, if we remove one of the edges of C from G, then by our supposition the new graph G' would have a cycle. However, the number of edges of G' is equal to m-1=n-1 and G' has the same vertices of G, which means that n is the number of vertices of G. Therefore, the number of edges of G' is equal to the number of vertices of G' minus 1, which tells us that G' is a tree (it has no cycles), and so we get a contradiction.

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3 years ago

I really need help again

Lelechka [254]

Answer:

$\overline{a} = \frac{v - v_0}{t} = \frac{\Delta v}{\Delta t}$

Step-by-step explanation:

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3 years ago

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Whoever checks my work and tell what I did wrong and the right answer will

gulaghasi [49]

Answer:

See below

Step-by-step explanation:

I believe that you only had to do letters F, H, and J. In that case, let's go over each one!

F: For isolating $d_{1}$ , we need to get rid of the 1/2 first. Let's multiply each side by 2:

$2*m = 2*\frac{1}{2}(d_{1}+ d_{2}) \\2m = d_{1}+ d_{2}$

After this, we just subtract $d_{2}$ from each side to get $d_{1}=2m- d_{2}$ . Dark Blue is correct! Let's now plug in those numbers below:

$d_{1}=2(10)- 13 = 20-13=7$

G: Let's isolate the vw^2 on one side by subtracting y from each side:

$vw^{2} +y-y=x-y\\vw^{2}=x-y$

Let's now divide each side by v, then put each side under a square root to get our final answer:

$\frac{vw^2}{v} = \frac{x-y}{v}\\ w^2 = \frac{x-y}{v}\\\sqrt{w^2} = \sqrt{\frac{x-y}{v}} \\w=\sqrt{\frac{x-y}{v}}$

Orange is correct! Again, let's solve the problem underneath:

w= $w=\sqrt{\frac{38-(-7)}{5}} = \sqrt{\frac{38+7}{5}} = \sqrt{\frac{45}{5}}=\sqrt{9}=3$

H: This one has some stuff that we haven't worked with quite yet (like terms), but our approach is the same: isolate c on one side of the equation.

$2a-2a+3c=17a-2a+21\\3c = 15a+21\\\frac{3c}{3} = \frac{15a+21}{3}\\c = \frac{15a}{3}+ \frac{21}{3} \\c = 5a+7$

Purple is correct! Let's solve the problem:

$c = 5(\frac{16}{5})+7 = 16+7 = 23$

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3 years ago