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3 years ago

10

Jason worked 3 hours more than Keith. Jason worked 12 hours. Which equation represents this situation, if k is the numbers of ho

urs Keith worked?

1 answer:

ycow [4]3 years ago

8 0

Answer:

j=k+3 (j=jason k+kieth)

Step-by-step explanation:

he did thrhee more so it will be adding 3 to k

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If Angela $98,760 home appreciates 3% a year will she have enough appreciation to try to sell a home for a $15,000 profit in fiv

olasank [31]

Answer:

Yes

Step-by-step explanation:

So, first, in 5 years, the home will have appreciated by 15%. (5 years times 3%). Once you find 15% of 98760, which is 658400, you have to add it on to the original price of the house. At this point, the house costs 757160 dollars. You then subtract the original price of the house from the price of the house 5 years from now. (757160-98760) and you get 658400. As you can tell, 658400>15000. Therefore, the answer is yes.

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2 years ago

Multiple Choice What is the sum of five-sixths + start fraction 2 over 3 end fraction? A. one and one-third B. One and one half

polet [3.4K]

Answer:

B

Step-by-step explanation:

$\frac{5}{6} +\frac{2}{3} \\\\=\frac{5}{6}+\frac{2*2}{2*3} \\\\=\frac{5}{6}+\frac{4}{6} \\\\=\frac{5+4}{6} \\\\=\frac{9}{6} \\\\=\frac{3*3}{3*2} \\\\=\frac{3}{2}$

$=\frac{2+1}{2} \\\\=\frac{2}{2}+\frac{1}{2} \\\\=1+\frac{1}{2}$

B.One and one half

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3 years ago

Which statement is true? A. In any right triangle, the sine of one acute angle is equal to the cosine of the other acute angle.

mr Goodwill [35]

That would be choice 2. One acute angle is the complement of the other

eg in a 90-60-30 triangle sine 30 = cos 60

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2 years ago

Read 2 more answers

A box contains three pen, two markers and a highlighter. Tara selects one item at random and does not return it to the box. She

kakasveta [241]

Two out of six probability

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3 years ago

Write an equation of the parabola that opens up whose vertex (−1, 2) is 3 units from the focus. (Vertex form or Parabola Form is

TEA [102]

Answer:

y2 = 4ax (opens right, a > 0)

y2 = -4ax (opens right, a > 0)

x2 = 4ay (opens up, a > 0)

x2 = -4ay (opens down, a > 0)

Vertex at (h, k) :

(y - k)2 = 4a(x - h) (opens right, a > 0)

(y - k)2 = -4a(x - h) (opens right, a > 0)

(x - h)2 = 4a(y - k) (opens up, a > 0)

(x - h)2 = -4a(y - k) (opens down, a > 0)

Equation of a Parabola in Vertex form

Vertex at Origin :

y = ax2 (opens up, a > 0)

y = -ax2 (opens down, a > 0)

x = ay2 (opens right, a > 0)

x = -ay2 (opens left, a > 0)

Vertex at (h, k) :

y = a(x - h)2 + k (opens up, a > 0)

y = -a(x - h)2 + k (opens down, a > 0)

x = a(y - k)2 + h (opens right, a > 0)

y = -a(y - k)2 + h (opens left, a > 0)

Step-by-step explanation:

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2 years ago