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3 years ago

1. Find the slope of the line that goes through the given points. (-7, 6), (-7, -9)

Mathematics

2 answers:

Rama09 [41]3 years ago

7 0

This is a vertical line. Therefore, the slope of it is undefined

zvonat [6]3 years ago

3 0

The Slope would be the change in y over the change in x, or in simpler terms, rise over run. That would make it -15/0 and you cannot divide by 0, making it an undefined line.

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I need help

ankoles [38]

Your answer would be y + 9 = m(x + 6)

6 0

2 years ago

In a city the record monthly high temperature for july is 88. The record monthly low temperature is 30.

Zinaida [17]

Answer:

58 degrees or [30, 88]

Step-by-step explanation:

In such a context, the word "range" can have different meanings. On the one hand it is the difference between high and low:

88 -30 = 58 . . . . range of temperatures

On the other hand, it is the interval between (and including) the high and low:

[30, 88] . . . . temperature range

3 0

2 years ago

How many different rectangles can you draw, that has an area of 28 cm sq. ? pls help me

jolli1 [7]

Answer:

You can draw three rectangles:

1- 28 * 1.

2- 14*2

3- 7*4

Step-by-step explanation:

We have a total area of 28. If we use only integer numbers, we can find all the divisors of 28. The possible rectangles will be organized with them, taking into account that the product of them, which means the area should be 28.

28 = 1*2*2*7

We can organize them as follows:

R1: 28 = 1*28

R2: 28 = 2*14

R3: 28 = 4*7

Finally, we can conclude that there are only three possibilities

1- 28 by 1.

2- 14 by 2

3- 7 by4

The perimeters will be:

Perimeter 1 = 2x1 + 2x28 = 58

Perimeter 2 = 2x2 + 2x14 = 32

Perimeter 3 = 2x4+2x7 = 22

3 0

3 years ago

Write the equation of the quadratic function whose graph passes through <img src="https://tex.z-dn.net/?f=%28-3%2C2%29" id="TexF

blagie [28]

Answer:

$f(x)=x^2+3x+2$

Step-by-step explanation:

We want to write the equation of a quadratic whose graph passes through (-3, 2), (-1, 0), and (1, 6).

Remember that the standard quadratic function is given by:

$f(x)=ax^2+bx+c$

Since it passes through the point (-3, 2). This means that when $x=-3$ , $f(x)=f(-3)=2$ . Hence:

$f(-3)=2=a(-3)^2+b(-3)+c$

Simplify:

$2=9a-3b+c$

Perform the same computations for the coordinates (-1, 0) and (1, 6). Therefore:

$0=a(-1)^2+b(-1)+c \\ \\0=a-b+c$

And for (1, 6):

$6=a(1)^2+b(1)+c\\\\ 6=a+b+c$

So, we have a triple system of equations:

$\left\{ \begin{array}{ll} 2=9a-3b+c &\\ 0=a-b+c \\6=a+b+c \end{array} \right.$

We can solve this using elimination.

Notice that the b term in Equation 2 and 3 are opposites. Hence, let's add them together. This yields:

$(0+6)=(a+a)+(-b+b)+(c+c)$

Compute:

$6=2a+2c$

Let's divide both sides by 2:

$3=a+c$

Now, let's eliminate b again but we will use Equation 1 and 2.

Notice that if we multiply Equation 2 by -3, then the b terms will be opposites. So:

$-3(0)=-3(a-b+c)$

Multiply:

$0=-3a+3b-3c$

Add this to Equation 1:

$(0+2)=(9a-3a)+(-3b+3b)+(c-3c)$

Compute:

$2=6a-2c$

Again, we can divide both sides by 2:

$1=3a-c$

So, we know have two equations with only two variables:

$3=a+c\text{ and } 1=3a-c$

We can solve for a using elimination since the c term are opposites of each other. Add the two equations together:

$(3+1)=(a+3a)+(c-c)$

Compute:

$4=4a$

Solve for a:

$a=1$

So, the value of a is 1.

Using either of the two equations, we can now find c. Let's use the first one. Hence:

$3=a+c$

Substitute 1 for a and solve for c:

$\begin{aligned} c+(1)&=3 \\c&=2 \end{aligned}$

So, the value of c is 2.

Finally, using any of the three original equations, solve for b:

We can use Equation 3. Hence:

$6=a+b+c$

Substitute in known values and solve for b:

$6=(1)+b+(2)\\\\6=3+b\\\\b=3$

Therefore, a=1, b=3, and c=2.

Hence, our quadratic function is:

$f(x)=x^2+3x+2$

5 0

2 years ago

PIC BELOW <br> Need answer ASAP

MArishka [77]

Option D. 17/10 or 1.7 is correct or 1 7/10 is also correct :)

8 0

2 years ago

Read 2 more answers