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4 years ago

Is the relation a function?

Mathematics

1 answer:

Margaret [11]4 years ago

8 0

No; a domain value has two range values.

x = -2 then y = 1 and 2

they would form a vertical line, which tells us that it's not a function

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a bag of a certain type of sand weighs 42 pounds and has a volume of 0.5 cubic feet. a sandbox has a volume of 16 cubic feet. ap

Aleonysh [2.5K]

Answer:

It will take 1344 pounds of sand to fill the sand box.

Step-by-step explanation:

I'm sorry If I got this wrong, have a great day!

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3 years ago

3. Find the equation of the line that passes through the points (-80, 43) and (-62, 52) in point slope form!

leonid [27]

Answer:

Step-by-step explanation:

$slope=\frac{52-43}{-62+80}=\frac{9}{18} =\frac{1}{2}\\reqd. eq. is~ y-43=\frac{1}{2}(x+80)\\\\$

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3 years ago

Given the graph below, what is the y-intercept(s)? Give your answer(s) as an ordered pair.

Irina18 [472]

Answer:

(0, 4)

General Formulas and Concepts:

Algebra I

The y-intercept is the y value when x = 0. Another way to reword that is when the graph crosses the y-axis.

Step-by-step explanation:

According to the graph, the line passes the y-axis at y = 4. Therefore, our y-intercept is (0, 4).

5 0

3 years ago

Naomi deposited money into a savings account that is compounded quarterly at an interest rate of 6%. She thinks this quarterly r

N76 [4]

Answer:

She is not right.

Step-by-step explanation:

There are 3 months in a quarter so we divide the 6% by 3 to get the monthly value.

A quarter = 3 months so 6% quarterly = 6/3 = 2 % monthly.

4 0

3 years ago

Read 2 more answers

Use the graph of △ABC with midsegments DE, EF and DF. Show that EF is parallel to AC and that EF=1/2 AC

Vinvika [58]

According to the midsegment theorem, the midsegments are parallel to

and half the length of the opposite side.

The completed statement are as follows;

Because the slopes of $\overline{EF}$ and $\overline{AC}$ are both -4, $\overline{EF}$ ║ $\overline{AC}$

EF = $\underline{\sqrt{17}}$ and AC = $\underline{2 \cdot \sqrt{17} }$

Because $\underline{\sqrt{17} }$ = $\underline{\frac{1}{2} \cdot 2 \cdot \sqrt{17} }$ , $EF = \frac{1}{2} \cdot AC$

Reasons:

From the given graph of ΔABC, we have;

Coordinates of the points A, B, and C are; A(-5, 2), B(1, -2), and C(-3, -6)

The coordinates of the point D and E on $\mathbf{\overline{DE}}$ are; D(-4, -2), and E(-2, 0)

The coordinates of the point F is; F(-1, -4)

$Slope, \, m =\dfrac{y_{2}-y_{1}}{x_{2}-x_{1}}$

$\displaystyle Slope \ of \ line \ \overline{AC} = \mathbf{\frac{(-6) - 2}{-3 - (-5)} = \frac{-8}{2}} = -4$

$\displaystyle Slope \ of \ line \ \overline{EF} = \frac{0 - (-4)}{-2 - (-1)} = \frac{4}{-1} = -4$

$Length \ of \ segment,\ l = \sqrt{\left (x_{2}-x_{1} \right )^{2}+\left (y_{2}-y_{1} \right )^{2}}$

Length of EF = √((-1 - (-2))² + (-4 - 0)²) = √(17)

Length of AC = √((-3 - (-5))² + (-6 - 2)²) = √(4 × 17) = 2·√(17)

Therefore, we have;

Because the slope of $\mathbf{\overline {EF}}$ and $\mathbf{\overline {AC}}$ are both , -4, $\overline {EF}$ ║ $\overline {AC}$ . EF = $\underline{\sqrt{17}}$ , and AC

= $\underline{\frac{1}{2} \cdot 2 \cdot \sqrt{17}}$ ,. Because $\underline{\sqrt{17}}$ = $\underline{\frac{1}{2} \cdot 2 \cdot \sqrt{17}}$ , $EF =\mathbf{ \frac{1}{2} AC}$

Learn more about midsegment theorem of a triangle here:

brainly.com/question/7423948

8 0

3 years ago