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3 years ago

If the coordinates of point A are (-3, 2) and the coordinates of point B are (7, 6), the x-intercept of is . Point lies on .

Mathematics

1 answer:

solmaris [256]3 years ago

4 0

You would move five down and eight to the right, it depends how how the grid is

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Can someone please help me with this question?? It’s really hard and once you understand it, can you please give me an explanati

Thepotemich [5.8K]

We want to subtract 8x + 3 from -2x+5. We can create an expression to represent this.

-2x + 5 - (8x + 3).

After this, lets distribute the - sign (think of this like expanding something with -1).

-2x + 5 - 8x - 3

Lastly, we just need to combine like terms.

-2x + 5 - 8x - 3

Combine the 5 and -3 to get 2.

-2x + 2 - 8x

Combine the -2x and -8x to get -10x.

-10x + 2

The final answer to the question is therefore A.

3 0

3 years ago

What is the answer to all of these?

4vir4ik [10]

$810000\\
0.000621\\
3040000000\\
0.0000043\\
0.00172\\
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0.0965\\
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3 0

3 years ago

My neighbors daughter was playing with a book and tore out the pages7,8,100,101,222 and 223. How many sheets did she tear out?

monitta

Answer:

Step-by-step explanation:

3 0

4 years ago

What are the zeros of this function?

Marina86 [1]

Where is the function?

7 0

3 years ago

30 points!!<br> What is the sum of the first six terms of the series?<br> 48 - 12 + 3 - 0.75 +...

Lunna [17]

Answer:

The sum of the first six terms is 38.39

Step-by-step explanation:

This is a geometric sequence since the common difference between each term is $-\frac{1}{4}$

Thus, $r=-\frac{1}{4}$

To find the sum of first six terms, we need to find the fifth and sixth term of the sequence.

To find the fifth term:

The general form of geometric sequence is $a_{n}=a_{1} \cdot r^{n-1}$

To find the fifth term, substitute $n=5$ in $a_{n}=a_{1} \cdot r^{n-1}$

$\begin{aligned}a_{5} &=(48) \cdot\left(-\frac{1}{4}\right)^{5-1} \\&=(48) \cdot\left(-\frac{1}{4}\right)^{4} \\&=(48)\left(\frac{1}{256}\right) \\a_{5} &=0.1875\end{aligned}$

To find the sixth term, substitute $n=6$ in $a_{n}=a_{1} \cdot r^{n-1}$

$\begin{aligned}a_{6} &=(48) \cdot\left(-\frac{1}{4}\right)^{6-1} \\&=(48) \cdot\left(-\frac{1}{4}\right)^{5} \\&=(48)\left(-\frac{1}{1024}\right) \\a_{5} &=-0.046875\end{aligned}$

To find the sum of the first six terms:

The general formula to find Sn for $|r|$ is $S_{n}=\frac{a\left(1-r^{n}\right)}{1-r}$

$\begin{aligned}S_{6} &=\frac{48\left(1-\left(-\frac{1}{4}\right)^{6}\right)}{1-\left(-\frac{1}{4}\right)} \\&=\frac{48\left(1-\frac{1}{4096}\right)}{1+\frac{1}{4096}} \\&=\frac{48(0.95)}{5} \\&=\frac{48(0.9998)}{5} \\&=\frac{48(0.9998)}{5} \\&=\frac{47.9904}{5} \\&=38.39\end{aligned}$

Thus, the sum of first six terms is 38.39

5 0

3 years ago