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4 years ago

I need help with finding the least number that 12 14 and 16 go into equally

Mathematics

2 answers:

Korvikt [17]4 years ago

8 0

Answer:

2 .they all can be divided by 2

Step-by-step explanation:

dimulka [17.4K]4 years ago

6 0

Answer:

1 or 2

Step-by-step explanation:

Write down the numbers that go into each one

12: 1, 2, 3, 4, 6, 12

14: 1, 2, 7, 14

16: 1, 2, 3, 6, 8, 16

The numbers that all three have in common are 1 and 2

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Write an equation for the line that passes through (-2,4) and (0,7)

xz_007 [3.2K]

Answer:

y = 3/2x + 7

Step-by-step explanation:

The slops is rise over run, so in this case the slope is 3/2 because it goes up 3 units while going 2 units to the right.

The points (0,7) is on the y axis. This means that 7 is the y intercept

With this information, we can follow the format of the slope intercept equation: y = mx + b

m stands for the slope and b stands for the y intercept. Plugging the information in, our equation is:

y = 3/2x + 7

8 0

3 years ago

Please help me to solve the problem.

Semenov [28]

Answer:

(i) anti clockwise, at point (1，0)

(ii) (5,2)

7 0

3 years ago

The center of an ellipse is (-9,3). one focus is (-7,3). The major axis is 24 units long. What is the equation of the ellipse in

CaHeK987 [17]

Answer:

$\frac{(x+9)^2}{144}+\frac{(y-3)^2}{140}=1$

Step-by-step explanation:

The standard equation of the ellipse is

$\frac{(x-\alpha)^2}{a^2}+\frac{(y-\beta)^2}{b^2}=1\;\cdots(i)$

where, $(\alpha, \beta)$ is the center and a,b are semi axes of the ellipse along x-axis and y-axis respectively.

Given that, $(\alpha, \beta)=(-9,3)$ and major axis, $2a=24$ .

So, semi major axis, $a=24/2=12$ .

Now, focus of the ellipse is (-7,3).

Let $e$ be the eccentricity of the ellipse,

So, $e=\frac c a$

where $c$ is the distance between the center and focus of the ellipse.

By using the distance formula,

$c=\sqrt{(-9-(-7))^2+(3-3)^2}=2$

$\Rightarrow e=\frac {2}{12}=\frac{1}{6}\;\cdots(ii)$

Again, the relationship among $a,b$ and $e$ is

$e=\sqrt{1-\frac{b^2}{a^2}$

$\Rightarrow \frac{1}{6}=\sqrt{1-\frac{b^2}{(12)^2}$ [from equation (ii)]

$\Rightarrow \frac{1}{36}=\sqrt{1-\frac{b^2}{144}$ [squaring on both the sides]

$\Rightarrow \frac{b^2}{144}=1-\frac{1}{36}$

$\Rightarrow b^2=\frac{35}{36}\times144=140$

So, the value of square of semi-minir axis, $b^2=140$ .

Hence, from equation (i), the equation of required ellipse is standard form is

$\frac{(x-(-9))^2}{144}+\frac{(y-3)^2}{140}=1$

$\Rightarrow \frac{(x+9)^2}{144}+\frac{(y-3)^2}{140}=1$

8 0

3 years ago

Can anyone help me here?

Murrr4er [49]

Answer:

3.14 x 4

Step-by-step explanation:

5 0

3 years ago

Write these numbers in standard form.

Wewaii [24]

Answer:

A ) -5

B) -4

C) -9

D) -12

3 0

3 years ago