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attashe74 [19]
3 years ago
14

I need it ungently it’s due in 20 mins

Mathematics
1 answer:
Ipatiy [6.2K]3 years ago
5 0

The answer is c you must look at key words i learned my key words

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Please Help me Please Please the question and options are in the picture
Vladimir [108]

The correct answer is B)x = \frac{y - p + mq}{m}.

In order to find this, we need for follow the order of operations.

y - p = m(x - q) ----> Divide both sides by m

\frac{y - p}{m} = x - q ----> add q to both sides

\frac{y - p}{m} + q = x ---> now we need to give it a common denominator. Multiply the q term by m/m

\frac{y - p + mq}{m} = x

This is our final answer.

5 0
3 years ago
What is the rate of change and initial value for the linear relation that includes the points shown in the table?
uysha [10]
X   1 2 3
y 10 8 6
So find the slope m:  (6-10)/(3-1) = -4/2 = -2
Going backwards when x = 0, y will be 12 (you can see this)
So
<span>Initial value: 12, rate of change: −2

</span>
4 0
3 years ago
4x^2 + 7x - 2 = 0 all the steps using quadratic formula ​
Natali5045456 [20]

the answer is 55

hope this helped

7 0
3 years ago
Solve the inequality for x. Show each step of the solution.10x &gt; 2(8x – 3) – 18
Luden [163]

Answer: x<4

Given the inequality:

10x>2(8x-3)-18

We want to solve the inequality for x.

First, distribute the bracket on the right side of the inequality.

\begin{gathered} 10x>2(8x)-2(3)-18 \\ 10x>16x-6-18 \\ 10x>16x-24 \end{gathered}

Next, subtract 10x from both sides of the inequality.

\begin{gathered} 10x-10x>16x-10x-24 \\ 0>6x-24 \end{gathered}

Add 24 to both sides of the inequality.

\begin{gathered} 0+24>6x-24+24 \\ 24>6x \end{gathered}

Divide both sides of the inequality by 6.

\begin{gathered} \frac{24}{6}>\frac{6x}{6} \\ 4>x \\ \implies x

The solution to the inequality is x<4.

6 0
1 year ago
A coin is tossed 600times with the frequencies as:
brilliants [131]

Answer:

  • see below

Step-by-step explanation:

Total number of trials =600.

Number of heads = 342.

Number of tails = 258

On tossing a coin,

let  \: E_1  \: and  \: E_2  \: be  \: the  \: events \:  of \:  getting  \: a  \: head

and of getting a tail respectively.

then,

1)  \: P \:  (getting \:  a  \: head) = \: P \:  (E_1)

\longrightarrow \:  \frac{number \: of \: heads \: coming \: up}{total \: number \: of \: trials}

\longrightarrow \:  \frac{342}{600}  =  \frac{57}{100}

\longrightarrow \: 0.57

\red{ \rule{150pt}{3pt}} \:

2) P (getting \: a\: tail) =P (E_2)

\longrightarrow \:  \frac{number \: of \: tails \: coming \: up}{total \: number \: of \: trials}

\longrightarrow \:  \frac{258}{600}  =  \frac{43}{100}

\longrightarrow \: 0.43

6 0
3 years ago
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