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Andrew [12]
3 years ago
6

Help on Number 8? I know Y=5 and the slope is undefined and x=2 but I don’t know what to do now.

Mathematics
1 answer:
Stels [109]3 years ago
3 0

To find the slope, use the equation of a line: y = mx + b

Plug in the values for x and y:

5 = 2x + b

We know that 2 can fit into 5 twice, which means 2*2 is 4.

We now have 5 = 4 + b.

If 5 is b more than 4, then this means that 4 is b less than 5. Subtract 4 from both sides, and you will have b = 1.

This means your equation is y = 2x + 1.

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Which quadratic equation is equal to (x+2)^2+5(x+2)-6=0
ch4aika [34]

Answer:

(x+2)^{2} +5(x+2)-6=0

Let

u=(x+2)

Substitute in the expression

(u)^{2} +5(u)-6=0

therefore

the answer is

(u)^{2} +5(u)-6=0 is equivalent to (x+2)^{2} +5(x+2)-6=0

brainly.com/question/6501456

7 0
3 years ago
Read 2 more answers
Mr. Claussen needs a total of 15 pounds of ground beef to make hamburgers for a company picnic. He finds that 2 pounds of ground
Marat540 [252]

Answer: $33.00

Step-by-step explanation:

Given

Mr. Claussen needs 15 Pounds of ground beef

It is stated that 2 pounds of beef cost $4.40

Using unitary method

1 Pound cost  \frac{4.4}{2}=\$2.2

For 15 pounds it is

\Rightarrow 15\times 2.2=\$33

Thus, he has to spend $33 for 15 pound meat

6 0
3 years ago
Slope review help needed ASAP will give brainliest
Nutka1998 [239]

Answer:

positive angle

Step-by-step explanation:

6 0
2 years ago
5 (3y-4) and 15y -20 equivalent or not equivalent
kiruha [24]

Answer:

The  answer and step by step explanation is in the picture below. Thank you! :)

Step-by-step explanation:

5 0
3 years ago
A long distance runner starts at the beginning of a trail and runs at a rate of 4 miles per hour. Two hours later, a cyclist sta
kenny6666 [7]
Recall your d = rt, distance = rate * time.

the runner takes off and goes at 4mph.

the cyclist takes off 2 hours later, and goes 14mph.

now, when they both meet, namely the cyclist comes from behind and meets ahead the runner, the distances both of them travelled, is say "d" miles for both, since both of them are "d" miles from the starting point.

if by them the cyclist has been going for say "t" hours, we know she took 2 hours later, so by the the runner has been running for "t + 2" hours then.

\bf \begin{array}{lcccl}
&\stackrel{miles}{distance}&\stackrel{mph}{rate}&\stackrel{hours}{time}\\
&------&------&------\\
Runner&d&4&t+2\\
Cyclist&d&14&t
\end{array}
\\\\\\
\begin{cases}
d=4(t+2)\\
\boxed{d}=14t\\
-------\\
\boxed{14t}=4(t+2)
\end{cases}
\\\\\\
14t=4t+8\implies 10t=8\implies t=\cfrac{8}{10}
\\\\\\
t=\cfrac{4}{5}~hr\impliedby \textit{or 48 minutes}
4 0
3 years ago
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