12/3m +2 = 21/6*<br> what is the answer

Three more than a certain number is 12 less than four times a number

k0ka [10]

Answer1.7 Step-by-ste explipcación:

4 0

3 years ago

Fill in the blanks below.

Crank

The slope for the first one would be:
y2-y1/x2-x1, so replace those with the coordinates and you'll get:
-3-1/-7-(-7) => -4/0 so I guess the slope is zero

the slope for the second one would be:
-3-(-3)/5-(-4)=> 0/9 I think this one would be undefined.
Check to make sure, though!

4 0

3 years ago

Read 2 more answers

What is the length of the shadow cast by a 12-foot lamp post when the angle of elevation of the sun is 60º? (to the nearest tent

elena-s [515]

Answer:

6.9 foot.

Step-by-step explanation:

Given: Length of lamp post= 12-foot.

The angle of elevation of the sun is 60°.

∴ The length of Lamp post, which is opposite is 12-foot.

Now, finding the length of shadow cast by foot lamp.

Length of shadow is adjacent.

∴ We know the formula for $Tan\theta$ = $\frac{Opposite}{adjacent}$

Next, putting the value in the formula.

⇒ $Tan 60° = \frac{12}{adjacent}$

Cross multiplying both side and using the value of Tan 60°

∴ Adjacent= $\frac{12}{\sqrt{3} } = 6.92$ ≅ 6.9 (nearest tenth value)

6.9 is the length of shadow cast by a 12-foot lamp post.

3 0

3 years ago

Which expression is equivalent to *picture attached*

DiKsa [7]

Answer:

The correct option is;

$4 \left (\dfrac{50 (50+1) (2\times 50+1)}{6} \right ) +3 \left (\dfrac{50(51) }{2} \right )$

Step-by-step explanation:

The given expression is presented as follows;

$\sum\limits _{n = 1}^{50}n\times \left (4\cdot n + 3 \right )$

Which can be expanded into the following form;

$\sum\limits _{n = 1}^{50} \left (4\cdot n^2 + 3 \cdot n\right ) = 4 \times \sum\limits _{n = 1}^{50} \left n^2 + 3 \times\sum\limits _{n = 1}^{50} n$

From which we have;

$\sum\limits _{k = 1}^{n} \left k^2 = \dfrac{n \times (n+1) \times(2n+1)}{6}$

$\sum\limits _{k = 1}^{n} \left k = \dfrac{n \times (n+1) }{2}$

Therefore, substituting the value of n = 50 we have;

$\sum\limits _{n = 1}^{50} \left k^2 = \dfrac{50 \times (50+1) \times(2\cdot 50+1)}{6}$

$\sum\limits _{k = 1}^{50} \left k = \dfrac{50 \times (50+1) }{2}$

Which gives;

$4 \times \sum\limits _{n = 1}^{50} \left n^2 = 4 \times \dfrac{n \times (n+1) \times(2n+1)}{6} = 4 \times \dfrac{50 \times (50+1) \times(2 \times 50+1)}{6}$

$3 \times\sum\limits _{n = 1}^{50} n = 3 \times \dfrac{n \times (n+1) }{2} = 3 \times \dfrac{50 \times (51) }{2}$

$\sum\limits _{n = 1}^{50}n\times \left (4\cdot n + 3 \right ) = 4 \times \dfrac{50 \times (50+1) \times(2\times 50+1)}{6} +3 \times \dfrac{50 \times (51) }{2}$

Therefore, we have;

$4 \left (\dfrac{50 (50+1) (2\times 50+1)}{6} \right ) +3 \left (\dfrac{50(51) }{2} \right )$ .

4 0

3 years ago

Write the equation of a line that passes through the points in the table.

lord [1]

Answer: do it have a answer choice if it do then its B,C and e

Step-by-step explanation:

6 0

3 years ago

12/3m +2 = 21/6* what is the answer